seymour
schuster)
B) A)
\037)
------) - --- N
-........--
-L)
u)
A')
B)
c')))
-)
'\\..
ELEMENTARY
GEOMETRY)
VECTOR
SEYMOUR
SCHUSTER)
DOVER PUBLICATIONS,INC. Mineola,
New
York)))
Bibliographical Note Dover
This
the work
first published in 2008, by John Wiley published
edition,
originally
of Congress
Library
is an
unabridged
& Sons,
vector geometry /
Elementary
p. em. Originally Includes
ISBN-13:
published:
of 1962.)
Cataloging-in-Publication Data
Seymour.
Schuster,
republication
Inc., New York, in
Seymour
New York:
John
Schuster.
Wiley
- Dovered.
& Sons,
Inc., 1962.
index.
978-0-486-46672-9
ISBN-lO: 0-486-46672-8
1.Vector
QA433.S38
analysis.
2. Geometry.
I. Title.
2008
515' .63-dc22)
2007050541) Manufactured
Dover
Publications,
in the United States of America Inc., 31 East 2nd Street, Mineola, N.Y.
11501)))
to
my
parents)))
preface)
This short presented at a
is
work
held at lectures
outgrowth
of
pattern
teachers.
geon1etry
mathematics
in
events
of
the material covered is
material
essential
rather,
but,
that were
of lectures
Institute
Foundation
Science
Carletol1 College ill the summer were to serve the purposeof
backgrounds
that
the
National
The
1959.
of
the
\"enriching\"
However,
no longer
the
indicates
education
enrichment
for every
knowledge
teacher.
It
ago that linear algebra'was a course for beginning graduate students and vector analyclass course by taken as an upper sis was typically The students. and engineering mathematics, physics, was
just
a few
years
last decade has brought mathematics vectors quite
is
\037
revolution
in
undergraduate
today the knowledgeof earlier stage. Indeed,it is to study vectors and freshmen
and education, a much at acquired for
usual
matrices,
particularly
college as
more, the studies and
applied
to geometry.
recommendatio11s made
v)))
Furtherby
the)
PREFACE)
VI)
Commission Group,
Study
Programin
on Mathematics, the School M\037thematics on the Undergraduate and the Committee all in the direction Mathematics of point
getting some of these
into the c011cepts
school
high
I have curriculum. 1011g felt that vector techniques find ,viII their should way into the high school n otas of the mathecurriculum-perhaps integral part matical training of all studentsbut, at the as least, work to excite and challenge superior students. and
an
very
On
a
level,
elementary
very
marily development mathematical tool in greater insight into the
theorems
proofs proofs,
a
knowledge
prerequisite.
oped slowly-more so on vectors. Simple
vector
natural
vector any
are develstandard works
algebra
of the
as
explanations,
well
as
two
(in
geometry
developedasa
gain a
are used. Beyond this, some and three dimensions) is
illustrations,
analytic
is to
attempting
by
of in
than
geometric
numerous
aim
proofs in contrast to the synthetic of which the reader brings as a
elements
The
vector The
geometry.
and analytic
deals prialgebra as a
this textbook of
the
with
of the
outgrowth
vector treatment. to assist in other
In addition, the vectorapproach is used areas of elementary mathematics: algebra,trigonometry In short, and and higher geometry. (plane spherical), it was felt that whenever vectorshave been employed in facilitating they would aid in gaining and/or insight to develop and proofs. I have tried very computations this small little machinery but to go a long way with himself amount. the reader will. find Accordingly, with such topics as linear inequalities,convexity, dealing
linear because
involutes,
programming,
As for I
and
projective
theorems.
prerequisites, they are not listed do not claim to have given a logical
development
of
of
Loosely
geometry.
assumed that the and concepts
formally
(axiomatic)
reader
Ellclidean
is
familiar geometry
speaki11g, I have the definitions and with the bare)))
\\vith
VII)
PREFACE)
area
and
parallelism,
angle,
For example, the notionsof are assumed. It is further
of trigonometry.
essentials
assumedthat the knows sine, cosine, and tangent
functions
as ratios
sides
the
of
resliits
from
of a
(in
the
right triangle).
and
geometry
of the
definitions
the
reader
naive
I haye
trigonometry,
sense,
In regard to indeed
taken very little for of. geometric granted. Samples information that are calledupon are: formulas for the area of a parallelogram and volume of a parallelopiped\037 the fact that two points determinea unique and the line, result that opposite sides of a parallelogram are equal. 3 I use the)aw of cosinesfor motivatio11, In Chapter but who has not seen it before will be consoledby the reader a
shortly
given
proof
thereafter.
It is entirely possibleto give a vector of development Euclidean from \"scratch.\" In fact, some geometry believe that a first course in geometry should people Others believe that the coordinate beginwith vectors. method should be given at the and still others olltset, of have faith in a combinatio11
coordinate
development,
beenadopted several in rewriting the by
high
approaches.
The
in various forms, has recently of the current groups interested school
curriculum.
mathematics
the reader interested in seeing how a strict vector would do the job, I strongly recomme11d the \"Geometric excellent Vector and the Analysis paper Conceptof Vector Space\" by Professor Walter Prenowitz. For
approach
This fine expositionconstitutesoneofthe chapters of the Yearboolc Third of the of National Council Twenty-
Teachersof Sincere who
of
came comfortably
l\\lathematics.
thanks to
and
Carleton learning
are appreciation i11 the SlImmer mathematics
due to the teachers of 1959 in the hope in cool
Minnesota
the strains but who, instead, laboredal1dperspiredunder of vector geometry and the 96% humidity. For reading the and for their valuable suggestiol1SI am))) n1a11uscript
PREFACE)
VIII)
grateful especially School in Lincoln of
the
University
the
of
Chicago
to Mr. Saul Birnbaum of the New New York City, ProfessorRoy Dubisch of Washington, Professor J. M. Sachs Technical College, and my Carleton
colleague,Professor
B.
William
special
thanks
courageously Science
National
1961.
This
tribute
substantially
Jr.
Houston,
Also,
who go to ProfessorDick Hall, used the text in mimeograph form at a Wick
Institute
Foundation
enabled
experience by
pointing
out
Sllmmer of Hall to conerrors ill judg-
in the
Professor ID.y
ment and typography.) SEYMOUR
N orthfield,
January,
Minnesota 1962)))
SCHUSTER)
contents)
1
Chapter
Fundamentalproperties4.
\302\267
of
5. Auxiliary
vectors. of
Uniqueness
\037
Chapter
IN
combinations
point technique.6.
COORDINATE
SYSTEMS
40)
systems and orientation.8.
7. Rectangular and
vectors
Linear
representations.)
VECTORS
Basis
of vector. 3.
2. Definition
Introduction.
1.
1)
OPERATIONS)
ELEMENTARY
applications
\302\267 9.
The
complex
plane.)
Chapter 10.
3
INNER
Definition.
12.
Components.
14.
Work.)
60)
PRODUCTS)
11. Properties of inner product \302\267 13. Inner product formulas.
IX)))
CONTENTS)
x)
ANALYTIC GEOMETRY
4.
Chapter
15. Our pointof view A,nalytic
Distance
\302\267
16.
geometry a
from
The
to
point
line
straight
21. method of proof. 20. Circles. 24.
line in
straight
two lines
plane
\302\267 28.
Chapter 5 29.
a point to a plane three dimensions \302\267 26. Angle
\302\267 25.
\302\267 27.
a line
\302\267
The
be-
a
line with
and a plane.) 135)
PRODUCTS)
a
from
to
point
a plane
\302\267 32.
Dis-
cross
lines. 33. Triple
two
between
scalar product.
30. Triple
products.
31. Distance tance
of a
Intersection
between
Angle
CROSS
Cross
by points on it
from
Distance
tween
a plane
Det\037rmining
17.
22.
Spheres.
\302\267 23.
Planes
\302\267
line continued. 18. a line. 19. Analytic
the
of
76)
products.)
Chapter
151)
TRIGONOMETRY)
6
34. Plane
trigonometry .,35. Spherical
trigonometry.)
Cl\037apter
7
36. Loci defined booby traps. 38. Linear
more parametric
by
inequalitjes.
37.
A
few
Segments and convexity.39. 40.
programming.
general
160)
GEOMETRY)
MORE
Theorems
arising
in
geometries. 41. Applicationsof
equations
to
locus
problems.
42.
Rigid motions.) APPENDIX
204
ANSWERS
fJ06
INDEX)
\03711)))
elem.entary
operations)
1.
INTRODUCTION
The
history
of the
of mathematical ideas
development
indicatesthat abstractconcepts arise
from
generally
in some
problemsof
counting,
roots
Arithmetic stemmedfrom
\"practical\" problem.
arose
geometry
from
problems
of
surveying land in Egypt, and calculusdevelopedprincito solve the problems of motion. from the efforts pally goes quite beyond the point of however, Mathematics, the that initiate the particular solving merely problems a for is with mathematics concerned building study, deductive science that is general and abstract,that may science have a wide range of application. By a deductive that a we mean, logical development beginswith briefly, of a set of assumptions a basic framework consisting and a set of terms used in or (calledaxioms postulates)
stating the assumptions. the of the assumptions are then the All
theorems
science,
which is
logical of
consequences deductive
the
concerned with abstractions or idealiza1)))
ELEMENTARY
2)
GEOMETRY)
VECTOR
tions of concepts from the original rather than problem the original problem itself. For example, the study is based on a set of assumptions of geometry that deal with lines and POil1tS than rather with fences essentially Line is an abstract concept; and fenceposts. an idealization of the fence, and it admits to all sortsof other of light, the edge of a board, ray.
with
it is
a interpretations:
the path of a
some
under
molecule
circumstances,
still others. Thus geometry,with finds application in a variety surveying,
a host of
and
in
origins
its of
problems.
The
however, goes on-and far
mathematician,
pure
oncehe a mathematical beyond. Because free to exercisehis imagination making
study,
begins
he is
logical
by
deductiollS
and developing theories from the realities of the motivating is a reality mathematician there theorems)
(proving
that qllite apart the problem. within his deductivescience. drawing metry for illustration, we can point to the are
li\"'or
from
Again
dimensional
fOllr
of
geo-
developments
n-dimensional
geometry-even
world is spite of the fact that our physical non-Euclidean or to the of invention dimel1sional,
geometry-in three
geometries,that contradictEuclid'sParallelPostulate for
(which,
2000
over
mathematicaltruth). were
consequences the
beyond
consideration
was accepted
years, Such
as absolute
by mathematicians
creations
of strong imagination and quite of any elementary problem in
the physical
world.
Vector
in
physical
is
analysis problems.
also It
a subject that was developed
has its roots primarily
handle problemsin physics, problems chanics but, later, problems in various other
in
initially
physical
twentieth of
a vector
science.
Developments
have resulted in a consequently,
centuries and,
of
th\037
branches
nineteenth
to meof
and
in the abstractconcept wide range of interpre-
tationsofthisabstractconcept.
The
result
is that
vec-)))
OPERATIONS)
ELEMENTA.RY
tors
now
to name
3)
a prominent role in a variety a few: just physicalchemistry,fluid-flow play
studies, theory,
and
psychology,
economics,
theory,
electro-magnetic
of
electrocardiography.
with illustrations filled in Geometry books aI,vays fact that the circle of are abstract point,line, spite that do 110texistin physicalreality in spite concepts of the fact that beginning students are apprisedof the are
and
and
the at of abstract subject their course. The reasonis
Abstract
simple.
quite
is difficult;
reasoning
beginning of
the very
nature
students therefore need-or
leastassisted by-the
of
help
some
real
model
are at
(or inter-
a pretation) of the abstract concepts.Consequently, with a sharp pencil is a convenient dot marked of point, and a sharppencildrawn for the concept model a ruler leaves a n1arkthat isusedasa of the along edge of line. for the Such pencil marks are a model concept until they get to feel at for convenience beginl1ers great and in the home subject begin to feel that there is a in itself. Later in their mathematical reality geometry
small
studies students
other
encounter
abstract
concepts,
but
models to by this time they can, and do, use geometric assist in still more abstract reasoning. This them is precisely of what occurs in the pattern development the concept of vector can of vectors. Although study a geometric be made abstract, model (directed line segthat assists the beginner in development) is the crutch ing steadylegsinthe field that is new to him. that It is the author'sview steady legs in abstract and that vector algebra are developedslowly reasoning in the model (now geometry)for some extended period should be done preliminary to engagingin the abstract
study.
this
Therefore
with a geometricstudy
entire of
vectors
textbook (i.e.,
co\037cerns itself the application
vectors to geometry),in contrastto the general study
of vectors.
Let us
begin.)))
abstract
of
2. DEFINITION Earlier
we
OF VECTOR out
pointed
from
originally
GEOMETRY)
VECTOR
ELEMENTARY
4)
physics.
the idea of vector came let us considerTherefore,
that
from the physicist's point of view-the statement of the television announcer who, beforegiving his final \"Good and the wind night,\" states, \"The temperature is 110W 37\302\260 is' 12 miles per hour in a northeasterly In direction.\" this simple weather announcementwe observe examples different types of quantities in the sense of two distinctly that the first (temperature) requires only a single num-
units, of course-for its description, (wind velocity) requires two facts, quantity
ber-with
whereas
second
the
magnitude
These
dire-ction.
and
the quantities encounteredin Hence, quantities
called tude
examples
are typical
elementary
physics.
of
classification is made: that are singled out and possess only magnitude whereas that both quantities possess scalars, magniare called vectors. and direction the
simple
following
In additionto of scalar examples quantities are mass, length, area, in addition volume; and, to velocity, examples of vectors are force,acceleration, temperature,
and
a11delectricalintensity.
Just as the
for his means
trained which he can A convenient the
geometry-to by
of
reasoning.
is a directedline
a
desires
mathematician
general concepts, so
segment
model
geometric
doesthe physicist.
For
has a reality
physicist-also
\"visualize\" and be aidedin his geometric model for a vector (/)
because
this
possesses
both magnitude (length) and direction,simultaneously. which suits the needs of physicists, is also This model, our for for it is our aim to quite satisfactory purposes, mearlS of vectors). study (by geometry Hence, for our
mathematical development,
we
make
the
for-
following
mal definition.
Definition. use
boldface
A
vector
is
a directed
type to indicate a
line segment.
vector. The
W
e shall
symbol1AI)))
OPERATIONS)
ELEMENTARY
5)
Q
Terminus
or
endpoint)
c) p
Origi n)
.0)
(b))
(a))
FIGURE
that the
event
we
PQ,
P
being
of
PQ
to designate
used
be
will
(see
the length of vector A.
In
vector we speak of isthe directedsegment its
emphasize origin
and
Figure
1a).
the
1)
the \037
nature by writi11gPQ, Q being the terminus or endpoint Another useful conventionwhen vector
\037
--7
\037
a vectors OB, OC, and OD with several and common 0 is to call these vectors D, origin with concerned a is if discussion That respectively. is, from a single point, we may several vectors emanating
to
referring
B, C,
them
designate
merely
by
their
individual
endpoints
(see Figure1b).
The one is called a unit vector. of zero length (with direction), any conalthough peculiar, is actually a great apparently vector zero venience. We refer to such a vector as the The to time. and shall point to its usefulnessfromtime of the direction notation for the zero vector is o. The 4. zero vector is discussed further in Secti011 A
of
vector
notion
Scalars,
being
bered scalars they
length
vector
of a
merely
will
magnitudes,
be real-num-
mathematics (In more may be elements of the complexnumbers;indeed, be from any number field. Our needs, may
quantities.
advanced
how-)))
GEOMETRY)
VECTOR
ELEMENTARY
6)
not require such generalityand will therefore be scalars to the real by restricting numbers.) are designated by lower-case Latin letters: a, b, c,
ever, do satisfied They
or
by
3.
FUNDAMENTAL
numerals.)
Our desire end we must tors equal.
Definition. A
=
(ill)
vectors
Two
and only if I
B) if
(i) (ii)
PROPERTmS
is to build an algebraof first present a criterion
A
is
A
and
the same sense of the length of A equals \\BI, i.e., B
=
possess
It cannot be emphasized toostrongly
even
be equal
space.
a
A
vector
if
may
do
they
hold:
not
direction; and the that
vectors
the same
possess
of B.)
length
may
position in
our definition indicates that relocated provided that we move it to its original position and parallel of fact,
matter
be
a position
to
rigidly
that (see
a
As
conditions
three
following
vec-
two
to B;2
parallel
IAt
calling
called equal (written
B are
A and the
for
to this
and
vectors,
its length or sense of direction It may therefore be relocated in a posi2a). Figure with its in space that we choose. origin anywhere are vectors this freedom, they are termed given not
do
we
change
tion When
free. 1 The
phrase
is actually
tion
is,
(a) If
(b)
If 2
We
\"parallel in
high
where Section to
itself.
in
this
on the
A
=
conditions use or
the on
\"if and only if\" points a double implication, conditions B, then
(i), (ii), and (iii) word parallel in the same line.\"
up the
(i),
(ii)
,
more
Although
and
A =
then
hold, the
fact that the
defini-
or logicalequivalence. general this
(iii) hold;
That
and
B. sense to mean is not given
usage
it is quite common in analytic school geometry, geometry, two lines possessingequal slopesare called parallel (see a vector is equal and Thus a line is parallel to itself, 16). The latter would not be tru\037 if we didn't use \"parallel\" vectors sense. Figure 2b exhibits two equal generalized
same
line.)))
OPERATIONS)
ELEMENTARY
7)
\"
Y
\"
,
\"
y
, \"-
\"
Y)
,
\ '\\
,
(b))
(a))
FIGURE
,)
2)
study of geometry this libertyto make displaceis highly advantageous. In the applicato other this freedom is not sciences tionsof vectors for it is to restrict vectors to necessary always granted, in For the of rigid mechanics some example, degree. that a vector be confined to a bodies it isoftenrequired line;that is,it may be moved rigidly but only in the line This line is referred to as its line of it lay originally. In Figure 3a we show three vectors, F, G, and action. three forces of the same magnitude, which H, represent and haviIlg the same sense of) lines on parallel acting In the
of vectors
ments
/ //
/
/
z/)
(b))
(a))
FIGURE
3)))
(a))
(b))
4)
FIGURE
direction. our
effect
F represents
definition.
ofthe bar of
and
and
G
represents
F and
bar.
the
therefore
would
They
are
GEOMETRY)
VECTOR
ELEMENTARY
8)
be equal accordingto
a pulling force a pushing force
G would have considered
therefore
at the center at the center
the same
mechanical
mechanically equal.
a effect However,H appliedat theendofthebar of turning motion, which is quite different from the F = G. Thus H is not equal to the otherforces. it would be natural to insist that forces such studies different lines of action be unequal. Thisjustifies having would
effect
In
two
inthe the stipulation of of permitting a vectorto be theory
of rigid
mechanics
displaced
only
along
bodies, its
line
of action.
If the fieldof applicationwere the theory of elasticity, still would be necessaryto restrict(force) vectors both of the more. Figure 3b shows two forces J and K, of line same magnitude and directed along the same the K has on a soft material. acting plasticlike action, of effect of stretching the mass, whereas J has the effect This illustration indicates why, in the it. compressing then it
theory
of
elasticity,
two
vectors
applied
at
different)))
OPERATIONS)
ELEMENTARY
considered
not
are
points
9)
restrictedto its
position;
original
displace it. Such vectorsare
bound.
called
emphasis, once again: in accordance with our
with state, are free,
We
book
this
In this field a vectoris there is no freedom to
equal.
all
in
vectors
of
definition
equality. any two vectors (Figure 4a). for vector B so that its origin of A. Now we construct a third the terminus at placed A + B, whose origin coincides with the called vector, A with the terof and whose terminus coincides origin
Let
Addition.
B be
and
A
is
a location
select
can
We
minus of B.
The construction of
nal sense
(ii)
B =
+
A
Addition
(A
Part (ii)
the
indi-
both are the same diagothe same parallelogram, and they possessthe same Hence we have the followingresult. of direction. =
A
Theorem 1. (i) that is
A clearly
B +
of
4b)
(Figure
cates that B +
of
the of
definition
A +
B, for
Addition
of
is
vectors
commutative;
B + A. of
vectors
+
B)
+ C
theorem addition.
that is
is associative;
=
A
+
is easily
(B +
C).
established 5 illustrates
Figure
Co> \037
Co \037
\037 \037
\037 \037
\037)
B+C
\037)
A)
A)
FIGURE
6)))
by using the proof.)
GEOMETRY)
VECTOR
ELEMENTARY
10)
the However, the reader is advisedto phrase proof from deduction elementarygeometryindependent logical of
as a
a figure.)
EXERCISE
1. Give an elementarygeometry vectors areaddedto equal vectors, As indicated in Section1, the toricallYr notably
from
an It
force.
but nonethelessexcellent,Dutch experimented
(1548-1620),
replace the covered by the
two
by
sums
the
of a
notion
to characterize
attempt is interesting
a single
theorem: If equal are equal vectors.
of the
proof
his-
physical quantities,
to note that the little-known, in
forces
one, called the
the resultant was actually the diagonal of a parallelogram,of which
sented the two original forces (Figure6). of formulation of the principles of addition complete statics.
an effort He
resultant.
that
a
Stevin
Simon
scientist, two
with
used extensivelyin developing rium-the beginningof modern
vector arose,
force
the This
forces,
theory (It is for
to dis-
represented
sides repreto his led
which he of equilib-
this reason that the parallelogramsof Figure 6 are sometimes referred to as parallelogramsof forces.) the many other accomAmong of are his: work on hydrostatics,which Stevin (1) plishments of) laid for the reclamation of the below-sea-Ievel land plans
Force Fl)
Force
Fl)
Force
FIGURE
6)))
Fl)
and (2) developmentof
Holland, with
the
first
to
11)
OPERATIONS)
ELEMENTARY
a give entitled
chapter
systematic \"Stevin
for numbers
notation
decimal
He was the
for computation.
methods
consequent
treatment of decimals. (Seethe on Decimal Fractions\" in A Source
by D. E. Smith, or A History oj MatheJ. F. Scott, 1960.) of vector addition is consistentwith our Thus definition the desires of the physicist who is interestedin applying techthe of vector analysis to his problems. (The of student niques a critical science should constantly maintain attitude applied toward the mathematical definitions, care to see whether taking or not they accurateiy reflect situations.) given physical Before continuing, it should be mentioned that Galileo (15641642), quite independently, cameto the sameconclusionas did Simon two scientists discovered how vectors Thus Stevin. two centuries prior to the inven\"should\" add, approximately tion of vector algebra and vector analysis in the nineteenth in Mathematics,
Book
matics, by
century.
Our
can
addition
vectors:
n
of
Of course,
(ii) of
of
definition
sum
the
Al +
this can be doneby
Theorem 1) and applying the
However, as
simply
grouping the
pairs
definition
(note
part
repeatedly.
process might be described that its origin is at the so that its origin is at the Aa
geometric
follows: of
terminus
now be extended to find \302\267 \302\267 \302\267 + An. 2 + Ag +
A
AI;
move move
A 2 so
of A 2 ; continue this process until An is placed its origin at the terminus of An-I. The sum Al + \302\267 \302\267 \302\267 A is then the vector whose + + An g + origin
terminus
with A
2
coincides cides the with
with
the
origin
terminus
of Al
and whose
terminus coin-
of An.
What would be the sum of the vectors that form a closed polygon with arrowstakingusallthe way around? to find the answer before reading on.) Consider, (Try for A + B + C + D + E + F 'of Figure 7. example, This is the vector whose the coincides with origin origin of A and whose terminus with the terminus of coincides F as))) after are placed \"origin to terminus\" the vectors
FIGURE
sum
We then
length.
A
This answer
+
our
D+ E+ F=
C +
B +
query
for
a polygon
o.)
of n sides, so the
zero vector.
is: The
Multiplicationof a by convenient to introduce
a scalar.
vector
In arithmetic as
multiplication
of addition. 4 + 4 + 4.
the
zero
write
holds
argument to
the origil1 and terminus of same point, al1dthe vector is of
are the
vector
7)
Hence
above.
described
GEOMETRY)
VECTOR
ELEMENTARY
12)
For example, 3 X 4 may be
\037\037 rt,\037)
FIGURE
8)))
\037
of
thought
Similarly, we can-at least to begin
x
it is
extension
an
with-)
OPERATIONS)
ELEMENTARY
13)
think of multiplying a vector by a of vector addition. An illustration A +. A. that 2A should represent of
we
addition,
parallel to
the
however,
A +
vector
know
and
A
the
having
of A +
length
is
A
twice
is
8)
(Figure
From our definition
same sense A
extension might be
as an
scalar
a vector
actually
of direction as A; the
length
of A.
A by A, the result of multiplying 2 is a vectorparallelto A having the same sense of direction as A but with twice the length of A.
Therefore,
=
2A
if
A +
the scalar
Before proceedingto the general case of multiplying vector by a scalar, let us considerthe questionof would
be appropriate
f\037r
ble demand might
for the
In
parallel of
if
general,
A
X =
+
to A, (b)
direction
this
stipulate
moment,
opposite
+
A
and
see where (a)
(c) X must
=
IAI
(-A)
= 0;
0, we know that
\\xl, and to that of
A reasona-
of -A.
a definition
be that
A
(Figure
a what
9).
so let us,
it takes us. X
must
be
have a sense Thus-A
precisely the properties a, b, and c menin the previous sentence. (Alternatively, if tioned A + X = 0, then A followed by X can be thought of as a closed in which the origin of X is at the terpolygon minus of A.) Consequently,our definition should (and that a will) stipulate multiplying by negative scalarhas the effect of changing the sense of direction of a vector. We are now ready to present a definition for the multia of a vector scalar.) by plication should
have
FIGURE
9)))
VECTOR
ELEMENTARY
14)
GEOMETRY)
;1A)
FIGURE 10)
3
vector parallel to A with magnitude In \\ = the times of A. In symbols, InAI magnitude Inl iAI. to have the same sense of Further, if n > 0, nA is defined direction as A; and if n < 0, nA is defined to have a sense direction to that of A; finally, if n = 0, nA is of opposite to be the zero vector (which follows from the first defined Definition.
sentence
nA
of our n =
Figure definition). 3, n = -3, and n
for
init.ion
Theorem
2.
is a
m(nA)
(ii) (iii)
illustrate, a
=
nA 3
2A,
The
sy-mbol
(m + n)A
definition
=
=
(mn)A. mA
nA.
+
B) = mA + consider m = 5 and meA +
vector twicethe
Inl refers
as follows:If n > 0, then rfhe
i-.
If m and n are scalars,then (i)
To
10 illtlstratesthe def-
=
asserts that the absolute
n
=
n
of
length
to the absolute value = n; and if Inl
mB.
- 2.
A btlt
directed
of n,
which is
< 0,
then
value
of
always non-negative; e.g., 131 = 3, 1-31 = 3, and = shall need the fact that of \\mnl Imllnl, the truth he clear from definition.))) the
a 101
Then
Inl
defined =
number =
which
O.
-n.
is We
should
15)
OPERATIONS)
ELEMENTARY
(i) states:
oppositely to A.
5(-2A) = (5)(-2)A
=
(-2))A =
(5 +
states:
(ii)
3A =
5A
+
5A
+
-lOA.
or
(-2)A
(- 2)A.
(iii) states: 5(A + B) = 5A + 5B. Proof. (i) By examining the length of ber of (i), our definitionof multiplication a scalar yields) 1 m(nA)
I
=
=
ImllnAI
that the directions that
of
same
the
have
the definition (ii)
=
they are parallel follows multiplies of A. The reader
That length. that both are
a vector
by
= ImnllAl
ImllnllAI
that the vectors of
1 proves
Equation
mem-
left
the
of
I
are
(i)
(1)
(mn)AI.
in
equal
fact
from
the
is left
to check
sense.
Use
(Hint.
nA..)
+ n = 0, both sidesof (ii) point in the same direction \037s A. If m
vectors
represent
m +
If
n <
0,
that point in the direcboth sidesof represent vectors tion oppositeto that of their The A. comparison lengths is left to the reader. (Hint. Usethe (ii)
of
definition
of
nA.)
Let us
(iii)
We consider the
nonzero. A
+
B
suppose that
(see
--7
then P R
=
and
Bare
B.
and
nOllparallel
trianglePQR,-?
which
11), by having A
Figure A +
A
=
PQ,
defines
-?
B =
QR,
P' Q'
R'
triangle , -? \037 = = = mA mA where + mB. P'Q', mB Q'R', then P'R' is similar to triPQR SincemAilAandmBilB.triangle
-?
We
construct
-?
angle
is meA + B), and we have = mA + mB. + B) meA The reader should consider two questions cOllcerning of (iii). The first is: What of the direction proof
result
the
P'Q'R'.
Thus
P'R'
the
that
of)))
GEOMETRY)
VECTOR
ELEMENTARY
16)
p')
p) A)
mA)
11)
FIGURE
B) as comparedto that of mA + mB? The and with the proof explicitly concerned itself, lengths of the two sense parallelism vectors, but it didnotdiscuss of The direction. second is: Does the proof question
m.(A +
break
if
down
A\\.IB?
Since
NOTE.
part
(i)
change of parenthesis is legitimate,we be no confusion if we eliminate For entirely and write mnA. example,) = (3
3(2A)
=
\302\267
2)A
As il1 elementary
Subtraction.
that traction is operation we define subtractionof
is
an
2A
the
=
parenthesis
6A.)
arithmetic, where subthe inverse of addition, of' vector the inverse real number, we write the expresses equation and in terms of addition as
vectors
More precisely,if
addition.
\302\267
3
there
know
now
would
a
is a
a - a = a + (-a) O. This fact that subtractioll is of addition. that subtraction is the -a is the that the realilumber maticians say =
defined
Mathe-
inverse
a relative
is the
to the
operation inverse
additive
of
of
a
inverseof -a). carry defining subtraction.
Definition. to
mean
-lB.)))
A
-
B =
A
+
or
additioll
(similarly, these
We
a
that
states
2
Theorem
of
(-B)
ideas
simply
a is
that
of
- a
the additive
over to
where
inverse
-B
vectors in
is
written
ELEMENTARY OPERATIONS)
17)
of subtraction can take in 12. Note that the any preeented Figure A of the is equal to the sum diagonal parallelogram B (A - B) and also to the sum (A - B) + B. Such who is for the algebraic checking advised beginner the
Geometrically
operation
forms
the
of
+
is
having difficulty in finding the correctorientationfor the difference A-B. With these few tools of addivector and subtraction we can begin applying vectorsto tion geometry.
elementary
We shall use our vector operationsto work an states exercise, one equivalent to the theorem which elementary that the diagonals of a _parallelogram Let each other. bisect not on one line. Call M the mid0, B, and C be threepoints We shall prove that BC (see Figure 13a). point of segment 1.
EXAMPLE
--+--+
--+
OM
cussed
=
(OB \037
+ OC).
on page
In accordance with
5, we shallwrite
\037
=
B
OB,
convention
the --7
C = OCand M
Then) \037
M
= B
+ BM)
M
= C
- MC = C \037
and
\037 BM
\037
(since
Bill
\037
= MC).)
B)
-B)
B)
FIGURE 12)))
dis\037
= OM.
GEOMETRY)
VECTOR
ELEMENTARY
18)
B)
B)
D)
o)
o)
c)
c)
(a))
(b))
13)
FIGURE
we get)
Adding,
2M=B+C
M =
or)
If we considerour figure
(Figure13b),
then
o to the midpoint
to
the
result
of
diagonal
\037
for
=
IODI
we
conclude
may
eachother.
of a
t\\VO sides
any
one half of
that the line joining the triangle is parallel to the third
Prove
2.
EXAMPLE
be part of a parallelogram OBDC states that the line joining vertex BC is one half the diagonal OD,
In equivalent (and more usual) language CI. that the diagonals of a parallelogram bisect
+
\\B
it.
In triangle PQR (seeFigure of
PQ
A
=
and
\037
Then
=
.C
A
these
that
(Note
\037
B = PkI
= NR,
vector
Adding
and D
to
of
that
C
equality
=
A - B, we of vectors)
the segment
and
N be
C+
midpoints
\037
\037 C
J.lfQ, A
-
follow from
=
MN,
D
to
equal
and
D =
\037
QR.
- B = o. about
vectors
summing
quadrilateral JfNRQ, respectively.) both sides of the last equation, we get C +
Since
=
- Band equations
MNP
triangle
111 and
let
14)
midpointsof
Call)
respectively.
PR,
\037
PN
+ C).)
(B \037
A
- B = D.
(by the proves simultaneously thatNM\\lRQ have
2C
= D, which
NM equals one half
of
the
base
definition
RQ.)))
and
OPERATIONS)
ELEMENTARY
19) p)
R:)
14)
FIGURE
EXERCISES
the easily rememberedSHORTCUTLEMMA: \037 AB + BC = AC. (This lemma has also been appropriately the Bypass Lemma by Professor D. W. Hall.) named 2. Reproduce Figure 15 on another sheet. Then construct and labelthe vectorsC - A, B - C, B + C, and -B - C. 1. Establish \037
\037
\037
3. four
Show
that
arbitrarily
PQ + chosen
\037
TlS =
\037
where
2MN,
points
P,
and where
FIGURE
15)))
Q, R, \037f
and
and S N
are
are the)
midpoints of PR and QS, respectively. around the polygon NMRS and 4.
Draw
5.
\037
\037
point,
+ XR.
current. The 5 mph N. and give a the resultant. is
of the forces. (vectors) acting is the zero vector, solve the following
in equilibrium
a body
mid-
\037
force resulting the direction of
the
.of
geometricconstructionthat shows 6. Using the fact that the sum on
\037
X Q
+
XP
upon by the wind and mph E and the current velocity
the magnitude
Compute
of A.B, Q the
CA. If X is any
acted
is
is 8
velocity
\037
XB + XC =
NMPQ.)
midpoint
midpoint of
\037
sailboat
A
wind
P the
with
R the
and
XA +
that
show
ABC,
triangle BC,
the
Sum
(Hint.
vectors
point of
GEOMETRY)
VECTOR
ELEMENTARY.
20)
problem.
of 100 lb hangs by a wire and is pushed by a horiA weight zontal force until the wire makes an angle of 1r/4 (or 45\302\260) the vertical. with Find the magnitude of the horizontal force and the tension in the wire.
7.
8. If
the
that
Show
A,
D are
any four points \037 \037 AB + AD + CB + \037
that
plane),
prove
and Q
are the
of
a
quadri-
of a parallelogram.
and
C,
B,
sides
of consecutive
midpoints
are vertices
lateral
of
midpoints
and
A.C
in a
necessarily
(not \037
\037
=
CD
4PQ
where
(How does
BD.
relateto Exercise
P
this
2?)
Using associative (A
-
properties -
B) and
(A
10. Establish the
nology: the 0'
vector
0' =
4.
of subtraction, the
definition
the
9.
zero
of addition, B) +
B, do actually
that has the
reduce
to
equal
A.
the termithus justifying Consider the possibility of a properties of O. Then prove that
uniquenessof
vector.
and
commutative
show that the sums B +
0,
(Hint.
0.))
LINEAR
COMBINATIONS
OF
VECTORS
we have learned to add and subtract vectors, we can combine multiply vectors by scalars, to enrich and vectors these operations to generate new our algebra of vectors. For example,if we are given A)))
Now that and also to
21)
OPERATIONS)
ELEMENTARY
can perform our various operationsto get B, B, 2A - 3B, 5A + 6B, etc. Such combinationsof A and B are called linear combinations of A and B. The set of aillillear combinationsof A and B could be written {xA + yB Ix and y real}. 4 The defil1ition of linear combinations is now extended in the folB, we
and
A
+
A
If
way:
lowing
Xl, X2,
.
X3,
.
xiA
AI, ,X n
.
+
I
A 2, A 3 , . . . are n scalars,
X2A 2
+
X3
and
n vectors
the vector \302\267 \302\267 \302\267
+
A g
are
, An
+
xnAn
of AI, A 2, A 3 , . . to be a linearcombination of section is devoted to the study The present of certain sets of vectors, and the combinatiolls contained herein are perhaps the most difficult Thus we shall proceedslowly. el1tire book. The is cautioned to study the definitionsand to take
is said
.
,An. linear
ideas the
in
reader them
literally!
Iloted
We
that
X
tiplying
=
earlier that -A, that is,
Since
-1.
A by
could, in a sense,say
conversely,that
,ve began with
A
the equationA X is the vector X is
that
is
A
dependent
Y
=
A =
and)
X =
obtained
0 implies by
from
derived
mul-
A, we
depe11dent upon A; UpOll X. Similarly,
or, if
we could write
3Y = 0,
+
be
can
X
+
--lA
-
3Y.)
Y is shown to depel1d 011A and A to depel1d on A a11d to state that Y. It might be preferable simply that to observe Yare dependent. It is almost trivial if the scalar such dependency would be impossibleto sho,v Hellce we exclude coefficients of A and Y were both zero. Thus
this case
in
consideratiol1
from
making
the
following
definItion. 4
The
defining
z
satisfying
synlbolism sets in the the
{I
},
following
condition
borrowed \"ray:
or sentence
from {zIS(z)}
S (z).)))
set theory, is useful in the set of all represents
VECTOR
ELEMENTARY
22)
Two vectors
Definition.
dependent if and
not both
zero,
if
only
aA +
so that
A
there
are called two scalars
B
and
GEOMETRY)
exists
bB = o.
Remark:The studentshould
recognize
the
linearly
a and b, fact
that
of words are statements of logical definitions equivalence and its (see footnote 1, page 6); i.e., boththe statement hold. converse Using our present definitionto illustrate this explicitly, we would say that the definition states: dependent implies that and both of them zero, so that b, not aA aA + bB = 0 holds, + bB = 0; and (2) if a relation a and to not both b zero with equal (i.e., at leastonebeing A and B are linearly dependent. then nonzero), is equivalent to saying that Algebraicallyour definition B and being (1) there existscalarsa A
linearly
two vectors are linearly dependentif and only if (see \"if and only if\ footnote 1 for explanation of the phrase one of them is a scalar multiple of the other (show this!). A geometric would be the following:Two interpretation if and only if they are vectorsare linearly dependent
parallel. The
can
reader
verify
these
interpretations
of the form formally by constructing a general that preceded our definition. One point, however, must be mentioned; this concerns the presenceof 0 as argument fine
one
of
that 0 10 +
vectors
the and
any
under vector
consideratio11. A are linearly
First,
we note
dependent, for
OA 0 and for 0, which is the definingcondition be linearly depe11de11t nonzero (1 is the required would say that Our scalar). geometric i11terpretation o isparalleltoA, where A may be any vector. This may howto the beginning student of vectors; appear strange it is a of conmatter convenience to retain the ever, great vention that the zero vector is parallel to every vector! as having no direcInstead of regarding the zero vector tion (as would to some), we regard 0 as having appeal and direction, Vectors direction. any specify magnitude so we chooseto say that 0 has any or all directio11s simul-)))
A to
=
OPERATIONS)
ELEMENTARY
the zero vectoristhe only
Of course,
taneously.
23) vector
a property, for any other vector (line segment has a u11iquedirection. Later,when of nonzero length) work we with perpendicular vectors, we shall have occasion regard zero vector as perpendicular the to every such
with
to
vector.
zero
has
vector
In the we
ent, we
the idea
with
consistent
is
this
Again,
that the
direction.
any
event that two vectors arenot linearly dependcall them linearly independent. Summarizing,
that:)
say
of vectors,
A pair
(1)
is a
one
which
of
is a
zero vector,
linearly dependentsetof vectors. A
(2)
(3)
of
pair
vectors is a linearly
nonzero
parallel
dependentset. A
of
pair
a linearly
vectors is
nonparallel
nonzero,
independent
set.)
merely the generalizing would leave us
contradict fllrthermore, of linear
and, theory
extend Xn,
so
where not all
If,
that
A 2,
. . if
example,
of
the
are
x's
the
zero).
implies
one
definition.
following
. .
. , An a set
exist
X2 A 2
+
of
is called
n vectors
of scalars Xl,
Xl
. , An three
\302\267 \302\267 \302\267
+
(i.e. when
zero
X2A 2 = is
X2 = said
+
XnAn
+
on the other +
xlAl
AI,
A 2,
vectors
.
X2,
.
.
,
that)
xlAI
equal
the
with
Definition. A set AI, linearly dependentif there
spirit
of
mathematics
of
with a rather meager We therefore proceed to
dependence.
notions
Ollr
to pairs
attention
our
Confining
would
hand,
the
\302\267 \302\267 \302\267
+
\302\267 . = \302\267
to be \037
= 0,
at least one doesnot equation
xnAn
Xn =
= 0
0, . then
linearly independent.
vectors, AB, cases must following
\037
set
the
For
\037
CD, and EF,
occur: (a) It
are
is
given,
possible)))
three scalars a,
to find
\037
\037
not
is
It
to
possible
The
shall
we
of
a
geometric
3.
any
vector
third
bination of A
are
the they
It dealing
segments
be
can 16).
Figure mille
form
step ill the
A
C;
and
B
and
which
are
is parallel
B, can
be
then
independent,
linearly
to (or
expressed
as
in)
the
plane
com-
a linear
B.
be should be recalled that vectors may they possess a commonorigin, that is, with free vectors. Therefore, even when A, B, and C might be in space, representing to positions in the same plane (see moved A and B with a commonorigin deterFor
so that
arranged we
If
by A and
Proof.
o.
may be rather
theorem.
Theorem determined
to be linearly dependto be linearly independent. a = 0, b = 0, and c =
an intermediate
provide
In
scalars.
said
to n vectors
vectors
two
from
jump
steep,so
if
hold
will
that)
(*))
such
three
find
the three vectorsare In case (b) they are said
In both cases(*)
such
\037
case (a)
ent.
all zero,
not
c,
+ cEF = o.)
bC D
+
aAB
(b)
and
b,
GEOMETRY)
VECTOR
ELEMENTARY
24)
a
alld
pla11e;
C (being
parallel to this plane) may
then be displacedsothat it is
actually
the
in
plalle
A and B.)
'B) \037)
A)
B
\037
)) A)
FIGURE
16)))
of
OPERATIONS)
ELEMENTARY
I
C
I
yBII
25)
B xA
\037----
A)
0
/
and
0
oX <
A
> a
y
/x >o
is a
if C
nOllzerovector, there isa parallelogram
C alld
diagonal
y>O
17)
FIGURE
Now,
and
(b))
(a))
with
A \037
I
A)
'\\vith
B.
A and
along
edges
An
in explicit constructioll of this parallelogramis given 17 and call be as described follows:Call0 the Figure common of the vectors A, B, and C, and call A, origin and C the respective B, endpoints of the three vectors. Constructa line \302\243through C parallel to B, alld call D
theintersectionof
the
\302\243 with
line
of actiol1
of A
the
(when
\037
of A
origin
of
multiple
is 0). A; let
\037
=
DC
yB.
OD =
us say
xA.
In the
trivially
\037
DC,
let
B;
us
say
Then)
C = xA
+
(2))
yB,)
of C to be a linear combination the event that C is the zero vector, true, for)
shows
which
some
is
l\037\"urthermore,
is some multipleof
to B,
parallel
being
Then OD, being parallelto A, \037
o
=
A
B.
and
is
theorem
OA + OB,)
(3))
and our proof is complete.
Toassociate more
the
concept
Corollary.
strollgly
of lillear Any
linearly depertdent.)))
three
the
idea
3 '\\vith
we state the
dependence, vectors
of Theorem
in
the
same
plane. are
Proof. Equati
For,
C
when
\037
zero. When C = 0,
number, say
=
Xg
one of
and
exist
(2) is
non-
to any
real
y in
Xg equal X2 = o.
=
Xl
X
set
can
we
1, and
there
that
= O.
+ xgC
least
at
0,
X2B
+
xlA
GEOMETRY)
VECTOR
ELEMENTARY
26)
EXAMPLE 3. We shall use the concept of linear combination to achieve another view of Example 1. Suppose we wish to bisect each other. prove that the diagonalsof a parallelogram Let the parallelogrambe OA.BC (see Figure 18), with P the intersectionof the diagonals. Again, using the convention of \037
\037
writing
= to
meA
be
= OP, A
=
A
+ C)
meA
and grouping the A-terms
Because
in
m
Thus m
-
+
-
n(A
C),)
and C-termsgives
- n)C = o. C are linearly independent, their scalar coeffi.A and the last equation must both be zero. Hence + n
(m
cients
\037
= OA, and B = OB, we have P = mB \037\"'hd \037 - C), where m and n are scalars + C) and PA n(A \037 Since A = P + P A, we may write determined. P
=
n =
+ , \037
n = which
1
l)A
+
(m
and proves
m
that P
-
n =
o.
bisects both diagonals
simultaneously.)
B)
o)
FIGURE
18)))
OPERATIONS)
ELEMENTARY
tant
in
vectors
the
that
theorem
with
work
the development on linear combinawe next establish an imporplane, serves as a very strong instrumentfor
with
Continuing
tionsof
problems.
geometric
Let A, Band C
4.
Theorem
a common origin O. If
(i)
C has
80 that
located
be
on the line joiningthe end-
its endpoint
points
A
and
C =
lA
+
mB,
where
if
C has
a representation
+
mB,
where
C=
lA
they have
Then:)
of
Conversely,
(ii)
27)
B,
l + m
l + m
= 1. in the form
= 1,
C has its endpointon the line joiningthe end-
pointsof that
in
same
the
in .
B
and
and
B.
the readershould with three vectors (often called coplanar vectors), and plane nonzero and nonparallel, then every vector is a linear of A and B (C being one
with the proof, proceeding is concerned the theorem
Before observe
A
A
the
are
plane
if
such)
of
combination
A
and
theorem singlesout
B,
by
Theorem linear
particular
3.
However, the
combinations
by
coefficients. condition on the scalar of For the converse point view, we would state (and should the reader verify) that any linear combinationof A and B, is equal to a two independent linearly vectors, A B of in the and when the two are situated vector plane a common origin. Once again, the theorem to possess
means
of a
states
that
Proof: (i) Here
that
and
B,
A,
C has
that
points
of
A
and
li\037\037ar
combinations
have
implications.
geometric
interesting
and
of these
certain
we
have
as
our
hypothesis
the fact
C all emanate from the same point 0, its endpoint on the line joiningthe endWe follow the convention of calling))) B.
GEOMETRY)
'VECTOR
ELEMENTARY
28)
o)
FIGURE
B,
A,
(see Figure
Let e
endpoints of A,
e the
and
19)
and
B,
C, respectively
19).
divide
the
in
BA
segment
\037
C
=
= Thus C =
B + +
(1
lA
+
mB.
(ii) For the and C emanate
\037lgebraically
=
lBA
B +
leA
- B)
- l)B.
converse, our hypothesisstates that from the same point and C = lA
where l + m =
1.
point on the line C= Now
We
jOillillg
lA
or
+
that
himself
\037
Be = B
lA
l:m, where
ratio
l + m = 1. (The reader should convince a given ratio can always be transformed so that the two parts sum to Ullity.) TheIl)
+
show
must the
mB
that
endpoints
= lA +
C = B+
leA
-
C has of A
(1 -
A,
B,
+ mB,
its end-
and B.
l)B;
B).)
A geometric examination of this comlast equation for the be statesthat the equation pletes may proof, 0 reached from to and BA.))) then line by traveling B, along
e
OPERATIONS)
ELEMENTARY
29)
EXERCISES
1. What
happens
2.
What
are
land
3.
What
can
be
stated
4.
(a) l = (b)
5. (a)
land
(b)
Do the
(c)
Do
the
positive?
for cases
where
=
j;
=
-i.
= l:m,
is 2:3
division
of
m so
that 1 +
=
m
same for the ratio same for the ratio
1.
4: 3. 5: -3.
4 reduces Example. 1 (p. 17) to a and C be threepoints not on one line, M the JJI divides of BC (see Figure 20). Then midpoint in the ratio I: I( =i-:\037), and Theorem 4 allows us to write 4.
For,
triviality.
BC
C if
\037 ;
If the ratio
EXAMPLE
with
=
m
! and m = i and m
find
of
O?)
1 =
(c) l
zero vector? the location
regarding
are
constructions
Give
is the
= A?
m if C
(a) both land m (b) l is negative?
(c) 1 =
or B
A
if
Theorem let
0, B,
M
=
i-B +
i-C.)
o)
c)
FIGURE
20)))
GEOMETRY)
VECTOR
ELEMENTARY
30)
A)
c)
21)
FIGURE
POINT TECHNIQUE
5. AUXILIARY
The introduction of
an
to play
point
auxiliary
the role
of commonorigin for severalvectorsunder consideration often the use of Theorem 4. The following facilitates three are devoted to an exploration of this examples
technique. 2.
of Example
problem
new approach Let
shall
We
5.
EXAMPLE
be
may
new
to
approach
difficult
more
perhaps,
the
the
instructive.
ABC be
triangle
a
now provide Although
N midpoints
and
jJ{
with
given,
of
sides AB and AC, respectively. Let 0 be a point in general position(not coincidingwith any of the already named points or lines). Applying 4 to the three vectors A, C, and Theorem N that emanate from 0 (see Figure21),we get =
N and
similarly,) Since
M =
- M. N
which parallel
proves to
C
+
C;) \037
\037A
+
\037B
to compare MN
we desire
vector N
\037A
.
with
must
we
BC,
examine
Thus
- M
both
=
desired
- B (=
-
C \037
B
\037
(C \037
-
results: that N
\037 BC)
=
\037
and
that
MN
B),)
-
\037 M
\037
=
BC.))) \037
(=
]jfN)
IS
EXAMPLE 6. a
of
the
31)
OPERATIONS)
ELEMENTARY
again prove that the diagonals each other, but this time by using Theorem 4,. and still another technique,
bisect
parallelogram
point
auxiliary
once
shall
We
tool that has heretoforebeenunemployed. the
position
we
vention,
\037
A
=
OA,
write
B = our
Establishing
\037
OB, C =
\037
\037
=
D
OC,
in vector
hypothesis
=
BC or)
-
D
A
=
=
and P
OD,
\037
\037
AD
intersection
be ABCD, calling P the parallelogram let 0 be a point in general Furthermore, diagonals. Once again, according to our con(see 22). Figure
the
Let of
language, we
\037 OPe)
write
- B.)
C
(4))
let us pause to discussthe approach. further, proceeding We in finding the precise ratio in whichP divides are interested if we can AC and also the ratio in which P dividesBD. Thus, get a representation of P, say Before
P = then
nA
we would know,
(Note how the
+
where n +
mC,
by Theorem 4, \037
\037
IPCI:
IAPI in
coefficients
the
a linear combination
of
A
=
1,
that)
= n:m. statement
related to the vectors in Figure19.)
P as
m
and
of Theorem
Consequently,
C so
that the sum
c)
A)
o)
FIGURE
22)))
4 are seek
we of
the)
GEOMETRY)
VECTOR
ELEMENTARY
32)
coefficients be unity. _1\\ fu\037ther observation, which is the basis the new tool we promised, is that P has two possible repreas a linear combination of two vectors sentations emanating be n, m, r, and 8 so that) from 0; namely, there must for
= nA
determined,
we
where)
can
If we 8 are
and
= rB + sD, (5)) = = n + m 1 and r + s 1.) succeed in producing theserelationssothat n, m, r, P
our end.
achieved
have
will
(5) is to add
OUf first step toward of
+ mC
+ B
A
D =
B +
A
+
members
C.)
(6))
the sum of coefficientson each side of we divide both membersby 2, obtaining)
Since
Both members
of
left
=
D
+
\037B
4.
to both
getting)
(4),
\037
with
comply
(7)
+
\037A
(6)
equation
is 2,
(7))
tC.)
of Theorem
the conditions
a vector
member
from represents emanating whose endpoint must be on BD; and the right member from emanating 0, and whose endpoint representsa vector the
Thus
0,
and
is
on
the vector (on
Therefore
AC.
for P is the only point onboth BDand AC. 4 once again, we concludethat divides the ratio \037: \037, which is the desired result.
P
EXAMPLE7. Employing to prove the familiar result: a point two-thirds the way
Let
ABC
triangle
AC, respectively medians AM and Applying
Theorem
the
The
BD
and
AC in
we attempt techniques, of a triangle meetin medians
vertex to the oppositeside.
N midpoints the
be
the fact
OP, Theorem
....L\\.pplying both
(see Figure 23). CallP BN, and let 0 a point in 4 to
be
same
from a
M and
have
\037
each side) must
of
sides
BC
and of
intersection
general position.
that M is the midpoint
of
BC,
we have) \037
OM
Similarly,
N being
=
the midpoint \037
ON Attempting
and also
M. =
=
N =
tB + of
AC
tA +
C. \037
(8))
yields)
tC.)
P as a linear combination of A and to achieve as a linear combination of Band N, \\ve subtract
(9))
M, (9))))
eliminate C:
(8) to
from
M
adding
Then,
33)
OPERATIONS
ELEMENTARY
\037A
-
N to
+
N
of
M
=
both members by i.
multiply
l TA
+
\037B
+
on each
coefficients
the
2\"\"
N.
side of
2M --
-g-
(10) is
, \037
we
so
IB + \037. 2N
-g-
member of (11) representsa and whose endpoint is on AM,
o and
(10))
Thus
left
The
l
A 2\"\".
both members, we get
iA + The sum
IB -
--
(11))
vector whose the
right
origin
member
is
of
is 0 and whose endpoint origin (11) representsa vector whose is on BN. But the left and right members of (11)are different This forces us to conclude of the same vector. representations this
that
is P,
vector
that
is,)
+ iM = j-B + iN. (12) us that P divides AM in the ratio i:j-. in the equation allow how the coefficients
P = j-A tells
12
Equation
(Noticeonce
again
us
to
Query: in a
\037
\037
\037
intersect
point?
Remark. of
\037
that lAP!: IPMI = !BPI:!PN!= t:t. Why is it now clear that all three medians
deduce
common
In the examples presented, the point 0 of the vectors ,vas always selected in) origin A)
B)
c)
M)
FIGURE
23)))
This is, of course, not necessary. point of common origin in some
position.
general
the
placing
Moreover,
special,judiciouslychosellposition
leads
often
siderable
to
con-
Exercises 1, 2, and 3, which to illustrate this point.) provided
simplification. been
have
follow,
GEOMETRY)
VECTOR
ELEMENTARY
34)
EXERCISES
1.
out
Carry
5 when
Example
point
is the
A
point
of emanation
of the severalvectorsin the proof.
2. Do
same
3. Do the
same for
the
for
Example
6.
Example
7.
that the midpointsof lateral are verticesof a parallelogram 4. Prove 4
with
(compare
5.
Given
7 on
Exercise ABC
triangle
with
MN is parallelto BC,prove
consecutive
by
p. 20).
N on AB that
of a quadriof Theorem use making sides
AN:NB
and
on AC, such that ilf = AM:MC. What and NM/BC? If>!B
is the relationshipbetweenthe fractions AN 6. Prove that the line joining a vertex of a the midpoint of an oppositeside trisectsthe
to
parallelogram diagonal
crossed
by it.
7. Generalization of 6. Let ABCD be a parallelogram wit.h interon AD such that AP = (l/n)AD. Prove that BP A is sects diagonal AC in a point Q, whose from distance
P
[l/(n + l)]AC.
8.
the of Theorem 3, namely: If A, B, analogue three nonzero, noncoplanar (cannot be placedin one in space can be expressed as a plane) vectors, any vector linear combination of A, B, and C. (Hint. A parallelepiped is the space analogue of the parallelogramin Theorem3.) of part a, prove that any four vectors the result (b) Using in space form a linearly dependentset.) and
6.
(a)
Prove
C are
OF
UNIQUENESS
REPRESENTATIONS
The importance of Theorem 3 liesin the statement in a plane call be expressed as a linear any vector bi11ation
of
two
given
that com-
linearly independent vectors
in)))
35)
OPERATIONS)
ELEMENTARY
The question we now pose is: Is such a A B More let and be representation unique? precisely, the given linearly independent vectors and C somearbiin the plane of A and B. chosen vector trarily by Then, m and n so Theorem 3, we know that there exist scalars that
plane.
that)
=
C
But is it
possiblethat
+ nB.)
mA
is
there
(13))
another
representation,
different from (13), of C as a linear
perhaps of A and B?
there
Suppose, then, that
=
C
Then
(m -
and)
+
r)A
(n
that)
sB.
= rA +
nB
+
mA
rA +
s so
rand
scalars
are
combination
(14))
sB
- s)B =
O.)
B being linearly independent implies = rand m n = s. 0 and n - s = 0;hence, C Thus (14) is precisely the same of as is represelltation (13). As a result, we say that the representation of C as a linear combination of A and B is unique. that statement Similarly, we could prove the general the representation of of a vector as a linear combination vectors is unique. Suppose linearly independent A and
However,
that
-
m
C=
r =
a2 A 2
+
alAI
+
\302\267 \302\267 \302\267
+
anAn
= blAl +
where
{AI,
A
of vectors.
(al -
bl)A
2,
.
.
.
,An}
5
is
a linearly
I
Again, the the
bnAn,
set
independent
+
(a2
-
b
2 )A 2
+
\302\267 \302\267 \302\267
+
(an
the linear indepelldence of the A's = b and an = bn . al l , a2 = b 2 , . . ., 5
+
Then)
Again,
roster of
+ b2 1\\2 \302\267 \302\267 \302\267
brace symbolism
elements
represents
of the set.)))
a set,
-
bn)An
=
O.
\037mplies that reader) The
with a listing
or
36)
the
observe
of the
uniqueness
B.
A and
of
combination
linear
construction in Theorem3 to of C as a representation
the geometric
check
should
We now employthesefacts geometry.
EXAMPLE8. We return problemin orderto prove to
Referring
once the
some
attack
to
that
24, we
Figure
GEOMETRY)
VECTOR
ELEMENTARY
of
problems
to the parallelogram bisect each other.
more
diagonals
write)
\037
PT
= meA
+ B).)
(15)) \037
one side
as
it
\037
But
since
QT
\037
nQS
=
another
achieve
can
We
\037
(15)
and
(16) we
\037
PT
= mA
= A
+ neB -
PT
as
PQ
QT. QT =
+
\037
write
(16))
A).
conclude that) + mB =
(1 -
B are linearly independent, a linear combination of A and
Since A and \037
=
Thus)
PT From
considering \037 \037
by
\037
of triangle PQT, that is, PT \037 is part of diagonal QS, we may
- A).
nCB
of PT
representation
n)A
+
nB.)
the
representation
B Inust
be unique.
R)
FIGURE
24)))
of That)
37)
OPERATIONS)
ELEMENTARY
. IS,)
which
=
m = n
that
imply
m =
and)
m=1-n)
the desired
proving
, \037
n,)
result.
The reader should comparethis approach with that of 3 to see how closelyalliedthey are (as are the of linear independence and uniqueness of concepts Example
representation).
We
9.
EXAMPLE
new approach. \037
is a
AP
to \037
\037
PN
=
\037
\037
=
= mAM.
\037
and
Simi-
n.)
\037
\037
= m( \037AB
mAM
the
that
\037
median AM, so we write AP \037 nBN. We seek to determinern AP
+
(17))
AC).) \037
\037
we
a pair
AC constitute
and
AB If
Figure
part of
\037
larly,
median problem, using 23 again, we note
the
to
turn
Referring
in
succeed
of
representation
\037
linear
and AC, we
To this end we
\037
AP
\037
employ
our
NP) \037
\037
\037
\037
- n(tBA
tAC \037
=
then
may
write)
- nBN)
AC \037
=
as a
---4
\037
=, AN + =
of AP
\037
of AB
combination
new technique.
vectors. \037
another
finding
independent
linearly
n\037
-
AC \037
2
BA
+ -
tBC)) \037
n\037
2
-
(AC
AB).)
Finally,) \037
=
AP
Comparing(17)with m\037 \"2 which-because
AB + of
I-n\037
2
\037
AC +
(18))
gives)
(18)
m\037 AC
\037
=
nAB
2\"
the
nAB.
uniqueness
1-n\037
+
2
AC,)
of representation-allows)))
state
us to
that)
m
pair
to
effort
linear
=-.1
n
3)
there was a persistent \037 of AP as a linear com-
observe that
two representations
get
is)
equations
\037
\037
A usual specific vectors AB and A C. encountered by the beginningstudent revolves
of the two
bination
difficulty
aroundthe
of
problem
the
getting
He knows that he will often get ships by summingvectorsaround e.g.
2)
and)
3)
should
reader
of
2
m=The
1- n
=
2
to this
solution
m -
and)
-=n 2)
The
GEOMETRY)
VECTOR
ELEMENTARY
38)
two
\"right\" relationships. relationindependent two
different
polygons,
,) \037
\037
=
AP
\037
\037
BP)
AB +
But where to go from
he
If
here?
\037
=
\037
+ NP.)
AN
bears
the
in mind ---?
get different
to
aim
general
AP
and)
representations
as a linear combination of instance) linearly independent vectors, he may employ
AP
of
same
the
the
(for set
of
technique
coefficients.)
of equating
EXERCISES
1.
medians
the
that
Prove
the method
of
Example \037
of NP.
sentations
Exercises 6 and 7 on page 34,
2. Re-do
method of
3.
that
Show
Given
bisects
AC,
intersection
use
making
of the
Example9.
line join.ing
the
vertex of a triangle
4.
of a triangle meet in a point by now by getting different repre...
9, but
F
of AD
of
a
to a
median
the
intersection
and Be.
the condition that of AB and CD; call G
with
ABCD,
quadrilateral call
the midpoints
trisects the side oppositethe vertex. Prove that AC is parallel
to
BD the FG.)))
(This is Let
a
\037
difficult \037
DG
= xDA
\037
and DF =
call CG \037
x in terms of . in
of u.
terms
tion
of DA
u.
n)
and
you
m: (1
m to
in
the
be sure that
the
have
will
and
CB
a linear
CD
point
AC' of
intersection
would
the
the
with
CP
(How would
bisect
a linear
desired
Let
- m). Let
ratio m: n.
as
of triangle
parallelogram
its sides. CallD the AB
expressGF
Theorem:
following
that A'CB'P is a divides
as
com-
\037
Finally, if you
divide the sides CA
n: (1 -
= mBG and expressBG
+
and DC, in two different ways, to determine Follow a similar procedure to determiney
and DC,
5. Prove the
\037
=.uDA
\037
\037
\037
\037
DB
that
Note
yDC.
\037
of DA
bination
the following hint.
provide
\037
\037
\037
Then
so we
problem,
\037
uDC.
39)
OPERATIONS)
ELEMENTARY
angle
combina-
result.)
points B' and A' ABC in the ratios P be chosen so and CB' as two of and AB. Then D you
C?))))
choose
nand
in
vectors
coordinate
systems)
'1.
AND
SYSTEMS
RECTANGULAR
ORIENTATION
is undoubtedly familiar with rectangular in the plane, where two perpendicular lines are chosen as axes. One is called the x-axis, the the of intersection and their other the y-axis, point on each is chosen arbitrarily direction origin. A positive The
coordinate
reader
systems
axis, and the
customary
correspondence
is made
between
the where numbers, points of each axis and the real real numbers are on that side of the origin positive as the positive part of the axis (see arbitrarily designated The standard convention (by no means 25). Figure is to have the horizontal axiscalledthe x-axis, binding) with its positive side to the right of the origin o. The axis is then vertical the y-axis, with its positive side the above 25b). Now, consider any origin (see Figure P in the. plane. From P we drop perpendiculars point to the axes. Call P x a11dP y the feet of these per40)))
IN
VECTORS
COORDINATE
41)
SYSTEMS)
y)
y)
P (x, y) Py
----1 (-3,2)
. (3,2)
.
t I I I
y
x\037
x
X
0
(a))
P%)
(b))
26)
FIGURE
x- and y-axes,respectively. The real P x on the x-axis is calledthe with P and the real number associx-coordinate of abscissa) (or ated with Py on the y-axis is called the y-coordinate(or P is then designated by a11orderedpair of P. ordinate) of real the abscissa occupying the first numbers: y) (x, in the pair and the ordinate occupyingthesecond position position. If the point P is on the x-axis, we give it a if P is on the y-axis, we give of it y-coordinate 0; and an x-coordinate of o. Thus the origin 0 has coordinates . of associating pairs of real numbers This method (0, 0). with points has the advantage of associating one exactly with and each one of the pair exactly point plane point, with each ordered pair of real numbers. pendiculars number
on the
associated
Rectangularcoordi11ate
systems
in
space
follow
the
in the same general pattern as do such systems plane. Three mutually perpendicular (intersecting) lines are selected as axes: the x-axis, the y-axis, and the z-axis, with of the three axes being the of intersection point a called the Again, positive direction is chosen))) origin.
on each axis, and the usual the points of each axis
arbitrarily
between
made
bers. The zeropointoneach and the positivereal numbers
origin
is
axis
are
has
been
the
same pattern
which
is
correspondence
the real numtaken at the origin, on that side of the and
as the
designated
GEOMETRY)
VECTOR
ELEMENTARY'
42)
positive side
of
axis.
the
Following
plane, we consider
point
any
of
P
in the
as
development
in
From P we space. we call P x, P y and
to axes, and drop perpendiculars P z the feet of these perpendicularson the x-, and real number The (see Figure 26a). respectively z-axes, associated P x on the x-axis is called the x-coordi11ate the
y-,
with
number associated with Py on the y-axis y-coordinate of P; and the real number associated with P z on the z-axis becomes the z-coordinate of P. If P isonany of the axes, not all the perpendiculars may be drawn. Thus we state further that if P is are on the x-axis, its y-coordinateand its x-coordinate if P is xand both on the its zero. Similarly, y-axis, both zero. z-coordinates are Finally, the coordinates
of
of
real
the
P;
becomes
the
the
are
origin
all zero.
It is quite convenientto visualize
P
not on solid, with the
an axis) at the of a rectangular as shown origin 0 at the opposite This may suggest to the other might prefer in finding the coordinates corner
in Figure
corner,
reader
of
P
if
its
coordinates
are
given.
P is
(when
26b. he
approaches P,
or in
For example,
locating
suppose we
know the coordinatesof P. We can then locate P by to its that finding the point on the x-axis corresponds a disx-coordinate, then moving parallelto the y-axis tance that corresponds to the y-coordinateof P, and then to the parallel to the z-axis a distancethat corresponds z-coordina
te
of P.
in space, When we designate a pointby its coordinates the order of the triple of numbersfollows the alphabetic in dealing with that is: (x, y, z). We stated order, that,
plane coordinates, it
is
customary
to
take
the
x-axis
as)))
COORDINATE
IN
VECTORS
43)
SYSTEMS) z)
z)
/)
y)
y)
A) x)
x)
(a))
(b))
FIGURE
In space, the conventionis to have the upward and the xy-plane horito have one positive axis customary
etc.
horizontal,
z-axis
positive
26)
pointing
zontal. It isalso
pointing toward us
and the other pointingto
our
right.
to toward us and which our our influences vector right seriously development. before on the orientation we decide Consequently, finally of the axes, it would to explain be well the notions of and right-handed left-handed triples of vectors. Let {A, B, C} be an ordered set of three linearly indecan be considered as pende11tvectors (which always
However,
axis points
which
emanating from the
same
Moreover, no two
the
first
a11d
are
second
0).
point
are linearly independent,they
do
not
Since
vectors one plane.
the
all lie in
same or parallel lines. A and B, form an vectors,
on
the
Thus
a11gle
Remember! It's the order of first, r).l 1 is in measurement Here, and throughout the entire book,angle terms of radians. The use of degrees is due to an. unfortunate historical when accident and serves to confusestudents,especially in reach calculus. the of functions they study trigonometric in good Because radian measure serves the mathematician stead to propathroughout the whole field of mathematics, we prefer gandize by making exclusiveuse of it.))) (J
(0
<
8 <
VECTOR
ELEMENTARY
44)
GEOMETRY)
c)
o)
{A, B,
C} right-handed
triple
(a))
0\\ \\ \\ \\ \\)
C
{A, B,
C} left-handed
triple
(b))
FIGURE
second,
tllird
il1 the
triple that
27)
counts. Now, consider
an observerstatio11ed onthe side
of
the
plane
of A
and B
that allowshim to walk from a point on A through the right arm angle 8 to a pointon B, with his outstretched That 27). always pointing away from 0 (seeFigure is, If on his left. he walks about, keeping point 0 always side of))) of the plane the observer's head is on the same
45)
SYSTEMS)
COORDINATE
IN
VECTORS
A)
28)
FIGURE
A and
B as or
ha'nded
{B,C,
the vector C, we triple.
positive
say
B, C} is a rightbe clear, then, that
{A,
It should
are also right-handed triples, whereas{B,A, C}, {A, C, B} and {C, B, A} are leftIf you think of the vectors as handed or negative triples. Ai
seats at a
and
{C,
A, B}
circulartable,all
clockwise
{A,B, C} right-handed)
{A,B,
C} left-handed) (b)')
(a))
29
FIGURE
(a) ()rdinary
(starti11g)
readings
screw goesin.
(b)
Ordinary
screw
comes
out..)))
GEOMETRY)
VECTOR
ELEMENTARY
46)
and all letter) yield triples of the same orientation of the readings yield triples opposite This situation is described succinctly by orientations. a cyclic permutation of the vectorsof a triple that stating or rightdoes not change its orientation as a left-handed as the case be. handed triple, may mnemonic is found A useful by studying a screw. Think of the first two vectors of the triple as being
at any
counterclockwise
on the the
of
head
ordinary (right-handed) into the second through
vector
first
than 1r. If
thishasthe
effect
the general
direction
the
of
of
the angle less screw we say
the
driving
third
Turn
screw.
an
vector,
in
the
triple is right-handed.) 8.
AND APPLICATIONS
VECTORS
BASIS
Let i, j, and k positive x-, y-,
be
to be a positivetriple,
a right-handed triple, we call otherwise right-handed;
system
i, j, and k are linearly (by Exercise 8, p. 34) that
expressed as a P
We
may addition,
linear
\037
z)-coordinate
is
left-handed
(0,0,0)
space
and
P
can In
them.
of
know
we in
vector
any
combination
=
be par-
(x, y, z);
referred to as the position P (see Figure 30c). of point to extend our algebra of vectors so that we wish with the vectors i, j, and k. To explore work and multiplication by scalars in subtraction, yj
+
zk.
P is
this form, let) A and)
y,
(x,
we say it independent,
where 0 =
= OP
then P = xi + vector
form
z-directions
and
30.).
Figure
let
rightis,
That
system.
the
Since
ticular,
respective {i, j, k} the
establishes
this
and
take
We
z-directions.
handedness of tbe (x, y, z)-coordinate if three vectors in the positivex-,y-,
(see
in the
vectors
unit
three
and
=
a1i +
B = b1i+
a2i + a3k b
2j
+
bgk.)))
47)
SYSTEMS
COORDINATE
IN
VECTORS
z
z)
k) :y)
:y) j)
(x,y, z)-system
{i,j,k} right-handed) :JC)
right-handed x)
(a))
(b))
z)
y)
x)
(c))
FIGURE
Then by A
+
30)
Theorem 1, B =
(al +
and by Theorem2 mA
=
b1)i + (a2 +
b2)j
+
(a3 +
b 3)k,
(iii)
mali
+ ma.2j
+
ma3k,)
and)
mA
+
nB =
(mal + nb1)i+
(ma2
+
nb 2)j
+ (ma3 +
nb
3 )k.)))
VECTOR
ELEMENTARY
48) We
linear
general,
sums, scalar multiples, and, in of vectors expressed in terms
that
therefore,
see,
GEOMETRY)
com1?inations
also be expressedas linear combinations and k. That is, every vector in space of these can be expressed as a linearcombination three unit vectors. Mathematicians describe this situationby the the vectQrs saying i, j, and k span or generate space the set of under consideration. vectors Furthermore, the A basis space. {i, j, k} is said to serve as a basisfor the space. is a linearly independentset that generates The need for a basis to generate the is space quite clear, but
of
i,
of
k can
j, and
vectors
i, j,
what is the purposeof pendence? The \"answer
expressed a that is our in
(in
A =
the
manner
present
discussion),
we may conclude,as we
li11ear
existence
V
cli
V
=
Thus,
(VI +
if h
= vIi
+
v2j
C2j +
+
vgk +
+
Cg.)
V4
h .)
(19))
set {i,j, k, h} C4 (not
cgk +
h
C4
=
the
implies
all zero)
such
o.)
that)
(20))
alld (20)yields
cl)i + \037
0,
(V2
we
+
C2)j +
(V3
+
cg)k
for
+
(V4
+
c4)h.
the existen:ceof combination of are an infinite number of)))
demonstrated
have
distinct representations V there h. i, j, k, Actually and
=
a3
C2,
of the dependence of scalars CI, C2, C3, and
Addi11g (19)
that)
earlier,
{i, j,
v rfhe
cgk,
hand, that we used a linearly k, h} to generate the space. Then could be written in the form)
set
vector
any
C2j +
c1i +
other
the
on
dependellt
did
if
=
agk
CI, a2 =
al = Suppose,
desire to have a vector to a given basis, relative
in the
lies
unique
a1i + a2j +
inde-
linear
of
stipulation
as
a linear
49
SYSTEMS
COORDINATE
IN
VECTORS
such representations, which can be demonstrated by scalar an before arbitrary (20) by multiplying adding it
to (19).
has no to reference
reader
The
Remark.
avoided
have
any
dimensional
and
reasoningis
in
being
space
with
accordal1ce
doubt observed that we the plane as being two three dimensional. Our the thinking of mathe-
so-called \"vector maticians who deal with spaces.\" to assume of do not choose the C011any They knowledge related to cept of dimension. They see it as naturally In fact, they other involving vectors. define concepts do so in one of two comof a space-and the dimensiol1 pletely equivalent ways:)
(1) The
dimension
a
basis
vectors
in
is
for the
be
to
shown
(Of course, this number choice of the basis.)
space.
of the
independent
of a space is the maximum vectors in the space.) independent
The dimension
(2)
of linearly
Consequently,
see
we
sional. is
as the numberof
is taken
a space
of
lishesthe
the
line would
be one dimen-
its corollary imply that 8 on p. Exercise Finally,
dimensional.
two
that
3 and
Theorem
of
character
three-dimensional
number
the plane 34 estab-
what
we have
called \"space.\" The foregoingdiscussion,which notes that and that i, j, and k are linearly independent, adding would result in a any other vector to this set of three further t hat establishes linearly dependent set, \"space\" is
dimensional.)
three
EXAMPLE A
=
(1, 3,
10. Let 4), B
points be given
= (1,5, 2), C =
and E = (0,2, -2).
(1)
The
4k
and
as
D
6),
position
How
should
=
(-2,
(0, 0, 0), 5, -2),
of point A is given by A. = i vector of point E is E = 2j
- 2k.+
vector
position
the
0 =
follows:
(-2, 1,
3j +
\037
(2)
\037
31, we see
we write
that AB =
B-
A.
the vector AB? Hence
we
write)))
Noting
Figure
VECTOR
ELEMENTARY
50) \037
=. (i
AB
= (1 =
(3)
How
do
+ 2k) -
+ 5j
l)i +
- 2k
2j
-
(5
3j +
+
4k)
(2 -
3)j +
4)k
= E.
a vector
\\ve write
(i
GEOMETRY)
from
emanating
0,
pointing
\037
toward
C but
half the
length
of
C ?
We seek!
OC. This
is
merely)
C
=
i( -2i
C \037
=
-i
\037
(4)
How
do
segment BC?
= M
= iOB =
(i \037
=the
+ 3k. 0
from
midpoint
to the midpoint of segment BC,
of we
\037
+
OC \037
+ 5j \037i +
midpoint
(21)
+ 2k)
3j +
Observing that M is the position
coordinatesof
6k)
,,,,rite)
\037
\037
OM
the
it!
employ Theorem 4 to
+ ij
a vector
we write Calling
+ j +
+
( \037
- 2i
+ j + 6k)
4k. vector
of BC:
of M,
M =
we can
(-t,
state the
3, 4).
z)
y)
x)
FIGURE
31)))
This)
SYSTEMS)
COORDINATE
IN
VECTORS
51)
c)
D)
o)
FIGURE
ment. Call
PI = Then,
if ill
is the midpoint
= =
+
OP2 \037
[(xli \037
+
Ylj +
+
(5) What
the
are
medians
\037
\037
f(
OB \037
jOM +
'
YI +
write)
\037
Y2)j
Y2
2
coordinates
of triangle
coordinatesof the ratio 2: 1, and using
seek the =
l(YI +
X2
+
2
(
\037
Z2).
+ OP2)
(OPI \037
zlk) + (x2 i
x2)i +
Xl
=
M
OP
\037
=
OPI \037
(Xl \037
we can
2,
Y2,
(X2,
seg-
+
Y2j +
+
\037(Zl
z 2 k)] +
z2)k.
midpoint is
Thus the
of the
P IP
of
of any
midpoint
P2 =
\037
\037
=
the
find
and
Zl)
YI,
(Xl,
\037
0111
to
be generalized
can
procedure
32)
'
Zl
+
Z2
2
. )
of the
point
Referring
to
BCD?
of
intersection
Figure
32, we
DM in 4 once again, we write \037 \037 + -lODe Employing equation 21, we have OP = \037 \037
lOC)
P.
Recalling Theorem
that
P divides
+ iOD,
or) \037
OP
\037
=
(t)OB
\037
+ (t)OC
\037
+ (i)OD.)
(22))))
VECTOR
ELEMENTARY
52)
GEOMETRY)
Consequently,) \037
OP
=
+ 2k) + (j.)(-2i
+ 5j
(j.)(i
+j+
P
Then
=
(-1,
these methods of
the
points
It is hardly
11/3, 2).
are valid
are.
plane, the formula
That is, (22)
even if
- 2k).
to state that the relative positions C, and D are in one
0,
B,
true.
holds
still
+ 5j
necessary
what
matter
no
6k)
(j.)( -2i
+
EXAMPLE 11. Prove that the (radial) vectors drawn of a regular polygon to its verticessum to the center
from
the
zero
vector.
should-before reading the next paragraphto this problem when the polygon is a attempt He may choose to set the triangle, square, and pentagon. The reader a
polygon within
solution a
coordinate
resortto somehelp from
system
and perhaps, if
trigonometry
Before proceeding, the reader should does the original problem, which not
or elementary some give
specify
necessary, geometry. to
thought
any particular
case. regular polygon but concernsthe general the origin of a two-dimensional coordinatesystem Consider to be at the center of the regular polygon of n sides (see Figure 33) . Let S be the sum of the radial vectors. If S is not the
zerovector, it has a unique to
the
origin.
to the x-axis. Sincethe
of a
inclination
Rotate the x-axis. The sum vector S
radius
polygon 21r/n radians
is now
rotated
y)
respect about
the
21r/n radians precisely the same) a +
inclined
figure
with
has
y)
%)
\037x)
FIGURE
33)))
53)
SYSTEMS)
COORDINATE
IN
VECTORS
appearance on the coordinatesystem new sum 8' of the radial vectors
one, the
unrotated
the
as
again be inclined a
must
the x-axis. Thus we have two vectors Sand 8' in differently while both represent the single sum Since we know that the sum vector in unique, i.e., question. solution is the vector 8 = 5' = O.) S = S',the only compatible to
radians
inclined
regular
proof. on the lookout
of
student
The
such
for
to
that it was the symmetry of enabled us to constructthis natural science should always be intrinsic that may help properties as well as the geometric con-
be emphasized that polygon
should
It the
the
simplify
physical
siderations
his
of
encounter-and
Scientists
problems.
of-symmetry in such diversefields as geomebotany, zoology, electrical circuit theory, algebra,
make use try,
mechanics,
and
Hermann
Weyl, Princeton
metry,
k110W
in
symmetry
World
The
of
of
of Weyl's
A portion
Press.
University
book ,is reproducedin James R.
by
is referred to the fascinatinglectures which are contai11ed i11his book, Sym-
science
and
art
role played
prominent
to
who wishes
reader
The
optics.
the
about
more
Mathematics
by
Newman.)
EXERCISES
1.
(a)
on a
Locate
sheet
of
(0,
10,
paper
graph
(6,4,10), B = (-6,4, -10),
C
=
(4,
-6,
the
points
-10),
A =
and D =
4).
(b)
Write
(c)
Find
the position
Ii, j, k}-basis.
vectors A,
sum A + B + C + the computation in the
the
this against
B,
\037
C, and
D
graphically,
D in terms and
Ii, j, k}-system. \037
of the check
- B, D - C, BD, and AC. Find the AB. (e) midpoint of segnlent X that divides segment Find the of the coordinates (f) point ' . AB in the ratio 2: -1. B is the midY so that Find the coordinates of the point (g) . of AY. point segment Find the median point of the triangle ABD.))) (h) (d)
Compute
A
VECTOR
ELEMENTARY
54)
2. Let
=
A
2i
- 4j,
D = 2i - 3j + k. (a)
2A as
Determine
B = -i
-
a linear
combination of
=
C
2j,
i + j i, j,
GEOMETRY)
+ 3k,
and
and k..
-3B as a linear combination of i, j, and k. (c) Determine3B - 2A as a linear combination of i, j, and k. of i, j, and k. (d) Find A + B - C as a linear combination (e) Is {A, B, C} a linearly dependent set? (f) Is {A, B, C, D} a linearly dependent set? be a parallelogram, with 0 = (0, 0, 0), A = (g) Let OAXC = the and C Find fourth vertex. (Hint: (1, 1, 3). (2, -4, 0), (b) Determine
How is the sum
3. Do the A
(a) Is
=
B,
{A,
and C
A, B,
paralielogram?)
of Exercise1 form
a
set?
- j,
i
to a
related
vectors
two
vectors
position
dependent
linearly
4. Let
of
B=i+
j,
C} a linearly
(b) Expressthe vectorV tion of A, B, and C.
- k. set? dependent
=
=
C
and
2i +
4j
j
- k as a linear cOlnbina-
of the eight vectors from the center vertices is the zero vector. Do this by assignto the vertices, writing the vectors ing coordinates eight explicitly, and then summing. (b) Let \037 be a regular dodecahedron (12 faces) and sits circumscribedspllere. Prove that the sum of the vectors from the How center of S to the vertices of \037 is the zero vector. this radial such look vectors are there ? (You might many up in a solid geometry text.))
5. (a) Provethat the sum
of
a cube
to the
THE COMPLEX PLANE A two-dimensio11al space that naturally admits il1 terms of vectors is the complex plal1e,which analysis be familiar to the reader from his studies in algebra may and 111 order to see the complexplane trigonometry. 9.
from
the
vector
coordinate
system
point with
y-direction
consider a rectangular basis of two unit consisting il1 the positive x-directiol1arId
of view, a
1 is taken vectors:1 i. i il1the positive vector i is actually the and
to
(see imaginary
Figure
34). Ul1it
The i = VI
unit
-
1.)))
SYSTEMS
COORDINATE
IN
VECTORS
55)
Y)
i)
1) %)
o)
34)
FIGURE
That
role of basis vector i
is, the
played by
1, and the roleof j is now
J)
-
1
-
i)
\037
If
P
=
(x, y),
by
played
is now
8) i.)
New
Old i) .
Section
(of
the vector OP
=
+
xl
yi,
more
or,
\037
simply,
vector
the
be thought
OP may
of
as
the
complex
is actually number iy. Thus every complex a vector il1 the plane. If scalarsare taken to be real scalars cornumbers, the multiplication of vectors by numbers responds precisely to multiplicationof complex by real numbers (see Exercise 2, page 57). Is it also true that vector addition corresponds to addition of comTo answer this question we considertwo numbers? plex
number
complex
their
x +
numbers:
related
CI =
vectors
Xl
being
+
iYI Cl
=
and xII
C2 = X2 + y1i and
+
iY2, C2
=)))
GEOMETRY
VECTOR
ELEMENTARY
56)
y)
(Xl + X2,Yl
-
+
Y2)
-----------1
r
I I
C2
Yl + Y2
I
__.J
\"1------
i I I
I
Cl:) o)
Y
2 i.
numbers allows us
of complex
Addition
>1)
36)
FIGURE
X21 +
\037X)
X2
Xl + X2)
to
say)
CI
+
and addition of Cl + we see
Thus
Cl +
vectors
C2
permits
=
that
X2) +
+
(Xl
+
us to
x2)1 +
(Xl + Cl
i(Yl + Y2);
C2
(Yl
corresponds
write +
Y2)i.
to
the
vector
C2.
EXAMPLE
vector
this
C2 =
extremely
Let
12. Considering to the approach
simple and
Example
11 in the light
complex plane, we
elegantsolution.
can
reach
of
an
with one be centered at the origin, radial vectors lying on the x-axis. Figure36 illustrates the If the radial vectors are for the pentagon. approach chosen of unit from we know algebra that the n vectors, length, as complexnumbers, are the solutions of the equation))) simply
of its
the
regular
polygon
n
x = 1. for the
SYSTEMS)
COORDINATE
IN
VECTORS
of the roots of
the sum
But
Xn-l
the
of
coefficient
term is
the proof.)
57) this
is zero,
equation
zero. This completes
EXERCISES
1. Sketch
the vectors
+ 2i. traction, and
and 3
the
with
2. Exhibit
a
Give
representing the complexnumbers this
compare
2
- i
addition and subview of the operations
for their
construction
geometric
algebraic.
the following scalar
j(2 -
2(2 - i),
fact
What
geometric
scalar
multiples
Remark
i),
on
of a Length
-
\037 (2
multiplesof i),
2
- i:
-2(2 -
-(2 - i),
can one deduce about the set complex number? Absolute
and
our can
which
is
I real
\\,
or
complex)
y)
x)
36)))
(real)
Value. From vectors we
brief discussionof complex numbers as some the gain symbolism insight regarding used to denote the absolute value of a
FIGURE
of
i).
as well as
number
the
notions, absolute value and length, are actually more
or,
value is
a special caseof the
basisfor the The
stated,
accurately
related,
of
choice
same
37), then the
(see Figure
can be determined by lal +
are
we
R
=
if
the
of
the V
But
complex = al +
bi
theorem
fa +
with
dealing
may treat themas a subset
in of
length
2 + b
n umber
2 2 V a + b .
orean
Pythag
V a2
=
bit
In the event that
the
the
symbolism.
or modulus of a complex
absolute value
a + bi, written la + biL is defined as if we view the complex number asa vector plane
of absolute
This is
vector.
of a
length
inter-
notion
the
two
The
vector.
a
of
length
GEOMETRY)
VECTOR
ELEMENTARY
58)
bil.
real numbers, we numbers. complex
That is, =
R#
Thus
the
length
In both caseswe
of
see
{a + r
that
=
bit
al + the
b
=
O}.
Oi is
absolute
= Irl
v?
value of
iy)
a + bi)
b)
x)
FIGURE
37)))
=
lal.
a number)
< -2) , -6
,
-5
I
I
I
3)
>
1'1'1
I
-2 -1 0 1
-4 -3
59)
SYSTEMS)
COORDINATE
IN
VECTORS
2
4
3
I
I
5
6
::.-)
131=3=131
1-21=2=1-21) 38)
FIGURE
is merely
the length
with
associated
vector
the
of
that
One may also considerthe real numbers as a a one-dimensional as line constituting (see space, Le., with the same result: absolute value of the Figure 38), real number equals length of the vector. Fromthis geometric of absolute view value one can see number.
the plausibility of the
of
properties
following
absolute
value:)
(i)
I al
la +
(ii) (iii) (iv)
1- al >
=
lal \\lal
bl
\\b(
Ibll
0
<
lal +
<
\\a +
<
-
\\a
Ibl
bl
bl.)
EXERCISE
1.
Considering interpretations above.)))
a and b of the
to be complexnumbers, give geometric algebraic properties (i) through (iv)
\
.
Inner
products)
10.
DEFINITION
In
this
chapter
aspectsintoour
to introduce quantitative
we begin vector
Of
algebra.
particular
interest
notions of distance and angle. In orderto see how to introduce best such concepts, it might be advisabletohave a look at them in the framework of coordinate are the
geometry. Sincethe plane is
more
than
with
dealt
easily
three
dimensions, we let A = (aI, a2) and B = (bl , b 2 ) distance points in the (x, y)-plane (Figure39). The A and
B, (by the Pythagorean theorem)from
between
d or)
Note
\\vhich
2 =
(al -
d =
vi (al
w\037
b
-
l )2
b
denote
+ 1 )2
(a2 +
can
IABI, the
(a2
b
-
60)))
d,
be found
formula)
(23)
2 )2, b
(23a))
2 )2.)
that it makes no differencewhether to first, for d is alsoequal
is considered
be
A
or
B
PRODUCTS)
INNER
61)
V (b 1 -
al)2 +
2
(b
-
a2)2.
Expanding (23) results in
d
2
=
a1
2
+ a2
2
+
b1
2
b 22
+
- 2(alb 1
If we introduce--for convenience in notation
cussion-the (24)
=
*B
alb l +
as
=
IABI2
we have
Hence
in
completely
and
A
A *A
+ B*B
-
+
the
a 2b 2).
(24)
present
dis-
a2b2, we
can rewrite
*B.
2A
(25)
distance between two points described of the symbol * (al1d, of course, +
terms
-).
we examine Turning to anglemeasurement, Figure 39a, an for the seeking expression angle (J. By the law of cosines we write) - 21oAIIoBIcos(J;) = IOAI2 + IOBI2 IABI2
y)
x)
B
tal
-
bII
--------,F I
lI a 2
-
b21
I A)
(4))
(b))
FIGURE
39
N ate. In (b) above, it is possible that a - b is positive, negathe distance (non-negative) or We zero. therefore tive, designate between A and F (also B and F) by absolute value in order to assure the non-negative character of distance. Consequently - bll 2 + la2 - b21 2 .))) d = Vial (23a) can be written
VECTOR
ELEMENTARY
62)
our * notation,)
or, using = IABI2
Upon
GEOMETRY)
A *A
+ B*B
- 2V from
(26)
subtracting
*
A
A
v' B*B and
(25)
cos 8. (26) we
simplifying,
find)
A*B
8 =
cos
,
(27))
V A*A VB*B)
so we see that angles * notation.
this
from
motivation
Drawing
discussion-primarily
from (27)-we start afresh with vectors two dimensions) and definethe inner as) \302\267 B =
A
cos
IAIIBI
of the
in terms
be expressed
also
can
(not product
to
restricted of A
and B (28))
8,)
when the angle between the two vectors they from same to emanate the are arranged point. Accordof distaIlce (length) and angle are both the notions ingly,
8 is
where
incorporated in
for
cos Because
8 =
whether
cos ( -
of the
is alsocalledthe
of
definition
our
that it is immaterial
8)
=
cos
notation dot
8,
-
inner
product.
8, or 2r
- 8 is
another
name
Note chosen,
- 8). (SeeFigure 40.) employed, the inner product
product.
FIGURE
(2r
Still
40)))
is
the)
INNER
PRODUCTS)
scalar
product,
vectors these
63)
for this method a scalar
yields
two
\"multiplying\"
(examine equations 28). in
popularity
enjoy
terminologies
of
and physics textbooks, so we feel order that the reader may
shall
at
them
employ
home
All
mathematics
with
all in
anyone.
Two immediate corollariesto the defillitioll are) A
\302\267 B =
B
and
A
If
A
However,
is
\302\267 A \302\267 A =
IAI2
perpendicular \302\267 B =
if A
(1)
0, there =
A
(2) B (3)
of dot
(commutativity
0,
= 0, to B.
is perpendicular
A
product)
(since cos 0 = 1). to B, A. B = o. (Why?) are three possibilities:)
is perpendicular If we agree that the zero vector to every 22 of this for vector (see page justification convention), in a single statement: we can combine these conclusions B = (); 5. If A is perpendicular Theorem to B, then A \302\267 B = 0, then A is perpendicular to B. and conversely, if A \302\267 Word of Caution. Many beginning students have in space are perpel1in that vectors difficulty believing It must therefore be dicular if they do 110t intersect. emphasizedonceagain that the definition of equality of a free as long as it is vector vectors permits us to move this enables us to to its position; kept parallel original It is the sinof any two vectors as intersecting. think that the reader has long since of the author cere hope
understoodthis point
and
the word 11.
of
is thoroughly
hence
bored
with
caution.)
PROPERTIES
OF INNER
An examination
of
geometric interpretation
PRODUCT 41
Figure of
the
leads inner
to an product.
interesting From)))
64)
GEOMETRY)
VECTOR
ELEMENTARY
B)
o)
c) 41)
FIGURE
see that cos ()
OCB we
triangle
A)
=
or) :\037I
\037
\037
=
loci \037
B
on
= Band
A, we
write
OC is
projection
which smacks multiplying IAI of
by
of
(projection
IAI
which product
gives us concept.
of
the \"vector projectioll\"of A
BOll
inner
the
=
product ill the
results
IBI cos
8,1
(29)
A. B.
In fact,
right member
B on
A) =
lAllBI
cos
8 =
we must clarify
Note
that
(29)
\302\267
may
feel-and
(29) gives the may
projection
be positive,
the
contained
rather
perspecial
definition.)
vector
negative, or
inner
so-that
rightly
of one
(30)
B,
ideas
the
a from haps too much has been iIlferred We therefore the picture. provide following 1
A
a geolnetric association with
Before proceeding, in (30). The reader
as a scalar, which
of
That is,
A.B.
becoming
(J.)
\037
OB
Since
cos
lOBI
zero.)))
upon another
PRODUCTS)
INNER
65)
of B on
By the projection
Definition.
we mean the
written
A,
pr AB,
of orthogonal projection of obtained by dropping perpendiculars from the origin and endpoint of B to the line of act\037.on of A the The distance between 42). feet of these (see Figure the the (J is the of pr AB. If perp\037ndiculars magnitude angle B are arranged A and B (when A and to emanate between is acute, then prAB is positive; if (J from the same point) is obtuse then pr AB is negative. on
the line
actio\"n
pr AB is
The
A.
of
B
Now that the notion of projecting one vector has been made precise,we restate as (30) = Theorem 6. A. B = (prBA)IB\\ (prAB)\\A\\. as an exercise The completion of the proof is left
upon
another
for
the reader.
From this result we distributive with respect
Theorem 7. A.
+
(B
that
prove
C)
the
which
to addition,
B + A.
= A.
inner
product we state as
C.)
I I I I
-
\037
I
)
I I
A
rv \\ prA
<
/
y B>
A
II \\)
0)
-CL__ /)
y prA
B < 0)
\037)
_
I
I
I
I
I
I
_ -..t:L
_
ri
\\
I
A
I
>
__
_J:J__
/
I) y
prA B>
y
prA B>
0)
FIGURE
42)))
0)
is
GEOMETRY)
VECTOR
ELEMENTARY
66)
I I I I I
I I I I
A
I
'----v----/'--v---'
B
prAC)
\\.)
prA J)
v
prA(B +
\\)
(B
+
A B) 1)
\037
<
0)
43)
43.)
C)
= prA(B + C)IAI = (prAB + prA = (prAB)IAI + =
pr
pr A C
(See Figure
Proof. \302\267
(B -I: C)
C)) FIGURE
A
A)
___-1_____
11._____b...____
\037 prA
I I
A
C)
Theorem
IAI
(justify!)
6)
(prAC)IAI
+
A
+
D) =
\302\267 B
(by
\302\267 C
6 once
Theorem
using
(by
again).
Corollary .)
(A
+
B)
Proof. (A
and
+
leave
\302\267
(C
A
A
\302\267 C
+
+
B
+
(A +
\302\267 D
\302\267 C
+
B.D.)
By Theorem 7 we write) B)
\302\267
(C
+
D) =
the remainder
(A
+
B)
\302\267 C
B)
\302\267
D,)
of the proof for the reader.
by vector methods that the pertriangle meet in a point. we adopt what may seem Becausevectors be \"moved,\" may to be a strange approach. The reader should become familiar with when useful several it, for such an approach is extre\037ely lines meeting ill a point must be sho\\\\yn. Let the triangle be ABC, with the bisectors perpendicular of AB and BC meetingin point Let M, N, 0 (see Figure 44). and P be the midpoints of sides AB, BC, and AC, respectively. 1\"he approach. \\ve is to show that OP is actually adopt perto A C.))) pendicular
EXAMPLE
13.
pendicular
bisectors
vVe
prove
of a
PRODUC.TS
INNER
67)
by stating the hypothesis
We begin \037
OM.
\037
language.)
\037
= 0
AB
(by Theorem 5, for
Rewriting (31), using the we
vector
in
is the
M
that
fact
\037
(31)
..L AB).
OM
midpoint of
AB,
get.)
(}A
(B
to the
according
expanded
which,
- A)
\302\267
+ j-B)
=
0,)
corollary
Theorem
of
7,
becomes)
iB Hence
A.
=
A
B
-
\302\267 B
\302\267
iA
which
B,
\302\267 A =
expresses
length of A equals the length of B. that) to show in a similar manrier B
o.
\302\267 B =
C
We
the leave
fact that the the reader
it to
\302\267 c.)
-+
\037
Now which
what must
be shown is that OP is perpendicularto P
We
therefore
is)
vectorially
expressed
expand
hypothesis in the
AC,
\302\267
(C
p. (C
hopes of
- A)
-
A),
=
o.)
using
this
showing
relations
dot
our
from
product
to
be)
c)
A)
M)
B)
B)
(a))
(b))
FIGURE
44)))
zero.
P
Thus
which
from This
GEOMETRY
Thus)
P
\302\267
(C
\302\267
(C
- A)
we conclude
example
= (\037A +
!C)
:c: \037C
-
- A) :c: 0
of
\302\267 C
(for
- A)
\302\267
(C
tA
\302\267 A.
A \302\267 A =
C
\302\267
C),
the desiredresult.)
illustrates
judiciousselection
the
once point
vectors'are consideredto emanate solution to an otherwisedifficult 12.
VECTOR
ELEMENTARY
68
again from often
the fact that a the several
which leads
to a
simple
problem.)
COMPONENTS
In the study of mechanics it is frequently usefulas the and often necessary-to considera singlevector in sum or resultant of two other vectors. For example, F a lawn mower (see Figure 45a), the force pushing is exerted but the questions along the bar of the mower, is the force Fh that contributes) asked in physics are: What
Fh)
(b))
(a))
FIGURE
45)))
PRODUCTS)
INNER
force
motion
horizontal
the
to
69)
\"wasted\"
is
Fv
the
of
being
by
ward? Thesequestions
mower?
applied vertically down-
answered
are
what
And,
by
F
considering
is called the the vertical component of of F. 1'he the component problem determining in as is easily dot components is a problem products of a component is seen by noting that the magnitude the magnitude of the of F. For example, projection = 81, and this relation can easily be transIFllcos IFhl into a dot product formed by making use of a un\037t vector U along the horizontal. and Then IFni = IF \302\267 we ul,
sum
as the
of
Fv (Figure 45b). Fh and is called Fv F,
and
Fn
horizontal
may further
of
write)
Fn The unit vector U serves
=
(F. as
U) U)
In plotting
14.
(Figure 46) it is found
F = 3i +
that
\\tvind
the
in two
a \"gimmick\"
ties. First, to assistus in writing component in the language of dot it enables us to write Fh explicitly U itself, in (32),imparts a direction while it doesnot distortthe magnitude. EXAMPLE
(32))
capaci-
of a magnitude and second, product; as a vector because
forces
the
to the
right member
2
on
,vind force,
plane graph paper at present, is vector
basis vectors in the xy-plane). We of scalar products: (1) the magnitude of F; (2) the component of F in the direction of the x-axis; F the the and the of y-axis; (3) component along (4) component of F along vect or A = -i + 5j.
shall
(1)
IFI
than as
j are
in terms
=
yF.
(2) Fx = 2 Some
and
(i
4j
express,
F =- Y
(3i + 4j) \302\267 (3i + 4j)
(F. i)i = [(3i+ 4j)
\302\267
i]i)
authors prefer to define components a matter vectors. This is purely
as
of taste,.
scalars
rather
which is
ofteI;l
considerations. Readers of mathematicai do well to heed the advice to Alice by literature would given I use a word, it means just what I Humpty Dumpty: \"When choose it to mean-neither more nor less.\)
colored
by
pragmatic
ELEMENTARY
70 (3)
Fy
= (F.
(4)
The
unit
j)j =
[(3i +
A
GEOMETRY
\302\267
jJj
4j)
vector along
VECTOR
,
is
11
the
Thus
component
1
of F that we seek is) F.. =
F
13.
+ \302\2533i
and
= (F\" A)
)
IAI
4j)
\"
of
of
the
such
A
=
A (F\" A)
IAI2
+ 5j))
(-i
PRODUCT
INNER
The form
some
A
'AI
(
=
A
\"
A\" A \
(-i +
:( \037\037
\037i
+
5j)
FORMULAS in Example
answers
that it
a nature
14 is qllite cumber-
is difficult
gain any insight from the answers.
We
for
anyone
y)
x)
-3)
-4)
-5)
FIGURE
46)))
to turn)
therefore
PRODUCTS)
INNER
expressiollS that
to simplifying those ill of vectors product
attention inner
the
71)
directions
(in the
components
take advantage
We shall
terms
our
Thus
pla\037e.
First,
directly with of
the basis k = 1. Vk.
the length of
that
fact
the
note
we
in space vectors in
the possibility
admitting
in
axes).
than
will deal
computations
vectors three dimensions, i, j, and k components.
rectangular
vectors
handle
to
involve
rectangular
that
fact
are generally no moredifficult a
the
of
the
of
their
of
vectors is one implies V i. i =.V
\302\267 =
j
j
Hence)
i i
j
j
from the Now, let)
k. =
+ a2j
ali
Then, (33)
\302\267 =
follows
which
A
j
and A
\302\267 =
k
j
\302\267 k =
1.
(33))
\302\267 i =
0,)
(34))
that
we observe
Secondly,
and
\302\267 i =
+ a3k)
\302\267 k =
k
mutual perpendicularityof =
B
and)
b1i +
makillg use of Theorem 7 (and (34), we compute as follows: + a3k )
b.Gj
its
+
i,
j,
b 3k.)
corollary),
+ b 2j + b3 k) a1i \302\267 b3 k \302\267 \302\267 \302\267 b + a2j 2j + a2j b3 k b1i + a2j \302\267 \302\267 b 3k + a3 k b1i + a3 k b 2j + a3 k \302\267 \302\267 = a1i \302\267 b 2j + a3 k \302\267 b 3k. b1i + azj
\302\267 B =
+ a2j
(ali
=
a1i
\302\267
b1i
we have
Finally,
+
a1i
a -Simple
\302\267
(b1i
\302\267 b
2j +
for
formula
the
inner
product
of two vectors:) A
The length IAI
=
\302\267 B =
of
A
V A
a1b 1 + ca n
\302\267 A =
a 2b 2
ll0 \\V Va12
be
+
(35))
a3b3-)
found
+ a2
2
the
by 2
+ a3
.
formula (36))))
VECTOR
ELEMENTARY
72)
of A
definition
the
From
\302\267
B,
cosine of the anglebetween cos 8
=
A
2 Y a1 +
a2
2
a3
+
and
2
solve for
(36), we B, getting
and
(35)
a2 b 2
a1b 1 +
GEOMETRY)
a 3b 3
+
\302\267
Y b 12 + b2 2
2 + b3
(37)
are the formulas (Note that (35) and (36) precisely used in our heuristic reasoningon pages60-62, where we were a method for informally exploring introducing quan-
titative aspectsinto
vector
our
We return to Example 14 to make answers given there in unsimplified
15.
EXAMPLE
3
=
y
(2) Fz
=
[(3
(3) F y
=
[(3
=
(3(-1)
IF\\
Fa
(4)
explicit
of the
computations (1)
algebra.)
. #
4 .4
1) +
(4
+
(4
= 5
\302\267
3\"+
0)
\302\267
O)]i
=
. 1)]j =
+ 4,
form.)
3i.
4j.
-i + 5j (-1)(-1) + 5.5
5)
=
(
-i
\037\037
We now ask the
and A
of
cos
ex
is
the
angle
a between
F
14?
Example
Using (37),
question: What
+ 5j).)
we V\\Trite F.A
=
\037
yF
Y A
\302\267 F
\302\267 A)
+ 4j)
(3i
.3 +
y3 -3
5y
+ 20 26)
4
.4
Y
\302\267
(-i
+ 5j)
5 (-1)( -1) + 5 \302\267
17
5
y 26
')
or)
cos
ex
=
17 V26
.
130
If we were interested in a precisevalue for a in terms of degrees ex by or radians, it would be a simple matter to find now consulting a table for the values of the cosine function.)))
INNER PRODUCTS)
p
\\)
73)
J)
v
Q)
s)
prSF)
47)
FIGURE
14. WORK
applying a is defined fs. Thus work is done
The work done in a distance 8 through product occurs. only that
pute
to
work
the
an
object compute
If we
done. (see
object
Figure
be
the if motion to
only
be
that
emphasized
motion
have a
47) with
and
if
f
magnitude
physicists
by
Furthermore, it s\037ould force which produces the
of
force
used
is
to com-
force vector F applied
the effect
of
the
moving
along a straight line from P to Q,the force the work done is that of the component
used
to
of F
\037
That
PQ).
(along
is, calling =
W
S = PQ,)
(prsF)ISI.)
(38))
of F to allow (We write (38) in terms of the projection the possibility of work being negative. Although negative work may sound strange to the uninitiated, the is one of great practical vallIeto the physicist concept and who need it for an adequate mathematical engineer, laws of mechanics and formulation of the fundamental
electricity. )
But
formula
(38) can W
or,
=
be rewritten IFI cos
8\\sl;)
finally,)
W=F.S
,)))
that
which expresses the idea as work done by one vector
reader
or perhaps the
in
product
the
direction
dot
the
be viewed the of other; view work as a can
dot
to
prefer
might
physical interpretation of
GEOMETRY)
VECTOR
ELEMENTARY
74)
product.)
EXERCISES
1. Using the of
a
approachof meet
triangle
two medians
section; then sideAB.
2. Let one
a right
be the of
force
that
show
of
foot
triangle the
magnitude
with
inter-
D
and
BC,
from A.
acting in the
-.:
point of
hypotenuse
perpendicular
be
BN
and
actually bisects
of CP
extension
the
their
P
call
medians
that the
13, prove
Example
point. That is, let A11f
of triangle ABC, and
be
ABC
on BC
in a
Consider
direction of
.iE,
IABI
of magnitude
and another
in
acting
\037
the
direction
AG.
IAcl Prove that the resultant is a force of
magnitude
the
3.
Find
(Hint. Use components.) to V = 2i - 3j and whose perpendicular
of AD.
direction a
acting
IADI
\037
in
\037
vector
length is four times the length of V. 4. The coordinatesof two points are Find
the
of the
cosine
originto thesepoints.
5.
Prove
that
the
sum
(3, 1, 2) and (2, -2, 4). between the vectors joining the angle of the
squares
equal to the sum of 6. Prove that the sum of the squares
parallelogram is
the
of
the
squares
diagonals
of the
of a
sides.
of the sides of any (not exceedsthe sum of the squares necessarily plane) quadrilateral of the diagonals by four the square of the line segment times .
that joins the midpoints 7. By
means of
semicircle is a
dot right
of
products angle.
the
diagonals.
prove (If AB
that an angle inscribedin a is a diameter, 0 the center,)))
PRODUCTS)
INNER
75)
--+
--+
on the circle,then point--+ --+ the dot product AP \302\267 BP.)
and P any
8.
the altitudes
that
Prove
(Hint. Use the approach of Theorem 5.)
9.
if
that
Prove
a
of
meet
triangle
13, and
Example
in a
point. on
rely heavily
the line joining their
circles intersect,
two
- OA. Evaluate
=
OB
centers is perpendicular to the line joining their
of
points
in tersection.)
10. Let)
A
B
Find
(a)
- 3j
=
2i
=
- 2i
+
+ j -
4k k.)
A-B.
(b) Find prB A and prA \"B. B. the component of A along Find (e) the work force vector Find done by (d)
ticle from the origin to the work done by Find (e) to (1, 2, -1). origin
11. Let F be the at the
of n
sum
point O. Then
W=F-S=F l .S+F In addition, if
sented by F
F
8 1, 8 2 ,
\302\267 8 =
F
.
\302\267 8
1
.
.
a par-
moving
x-axis.
a particle from
F2 ,
.
.
.
+Fn-S.
, Fn,
(Why?))
consecutive displacements
\302\267 8
2
+
\302\267 \302\267 \302\267
+
F
\302\267
8n.
the
all acting
then)
, Sn,
+ F
along
.S+...
2
the
in moving
forces F 1,
in n
results
(2, 0, 0) A
in
A
(Why?))))
repre-
analytic
geometry)
15.
POINT
OUR
VIEW
OF
should note that the definitions of addition, product of vectors werenot madein termsof-coordinates.At the outset we treated ve.ctors in a coordinate-free fashion. (When such is the case, it is The reader
subtraction,
and inner
often stated that the coordinate
concepts
are
of
\"independent
a
And yet, manyapplications-par-
system.\")
ticularly to geometry-were possible. we proceeded in our development, it is new techniques were employed; this that one should view coordinate systems. light As
in
That is, coordinate
be
should
systems
looked
UpOll as
'another instrument rather than as allother branch of mathematical The philosophy that the authoris study. suggesting
The deals
the
takes
form.
following
branch of mathematics that with, among other things, the
planes,circles, and
The
spheres.
in this field may be
reached 76)))
by
we
call
geometry
propertiesof lines,
solutions various
to problems approaches,
GEOMETRY)
ANALYTIC
some
the
on
depending
learned is
or stronger than othersThe first approach
natural
more
being
77)
at hand.
problem the
usually
which
method,
synthetic
has
some approximation to that exhibited in Euclid's EleThe present work is devotedto a study of the and a Hence vector analytic approach. given geometric one could give it a vector interpretation, or one problem, couldimpose a coordinate on the problem and use system ments.
analytic techniques. course, any three methods is alsopossible.
In
coordinate
summary,
not be
should
systems
of the
combination
Of
viewed as
as another (very powerful) intrinsic geometry but., rather, mathematical with which one attacks geometric problems. those Combining our notionsof vectors-particularly to
tool
scalar
the
of
product-with
tems, geometry.)
shall
we
16.
some
develop
sys-
elements of analytic
LINE
STRAIGHT
THE
We begin
of coordinate
those of the
with the
problem of
the
finding
equatioll
of a
if P = (x, y) straight line in the plane. More precisely, is an arbitrary point ona givel1 line seek a mathewe \302\243, = matical relation that distinguishesP those y) from (x, \302\243. points in the plane that are not 011 of course, There are, many ways to specify a unique line. We begin by specifying two points P1(XI,Yl) and P 2 (X2, Y2),1 and seek the equation of the line determined these The aim is to in arrive at an by points. equation terms of the coordinatesof the specified points. Since P, PI, and P 2 are all one line (see Figure 48), we may employ 4 to write) Theorem
--+--+
\037
OP = 1
This
freely joining
to be
appears
line. It was,
in
translated, two
points.)))
fact,
(1 - t)OPl + tOP
2 -)
the most natural included
way
as Euclid's
states that one and
only
one
to specify first
line
a unique
axiom,
can be
which, drawn
VECTOR
ELEMENTARY
78)
GEOMETRY)
y)
y)
x)
%)
(a))
(b))
48)
FIGURE
in accordance with a common with vectors Or,
this
rewrite
we
the
of
equation
points)
P
Equation
line
our convention of designating origin by their endpointsalone, in terms of the positionvectors
PIP 2.
= (1
- t)P1 +
39 is
often referred
As
takes
t
on
is
the
=
(1
of the
The auxiliary variable t
is also referredto as a of line
is
(39))
For =
midpoint called
parametric
l
+
of
j-P
values, if t =
example, \037P
of
equation
number
real
j-,)
2)
Pt P
2. and (39)
segment
a parameter
representation
(vector)
\302\243.
Rewriting
xi
P.
- j-)P1 + j-P2
vector
position
2 .)
to as a vector
different
we get different vectorsfor P
tP
+
(39)
yj = =
in terms
of the
basis vectors, we
- t)x1i+ (1 - t)Ylj - xI)]i + [YI [Xl + t(X2 (1
However, the representationofa vector
tX2 i +
+ + in
t(Y2 terms
have)
tY2j
- YI)]j.) of basis)))
79)
GEOMETRY)
ANALYTIC
Therefore
is unique.
vectors
X
{Y are the coordinate
Xl +
t(X2
=
YI +
t(Y2
is
case by using
y
(40). Taking two
(0, 1), on the x-axis,we {
y
=
0 +
t(1
=
0 +
t(O
X =
matter
what =
Calling
the value of t, a sufficient
the
(40), yields)
simplified
X =
3 +
=
1 +
Y
If, instead,we would
had
different see
that
(-2,
line.
3),
and applying
2 - 3) t(3 - 1).)
t( -
=
3
-
1 +
with PI
begun
{
we
=
zero.
the
5t
2t. = (-2,3)andP2
= (3,1),
yield:) X
a
is always
y-coordinate
form:)
x =
(40)
O.)
(3, 1) and P 2
{ y
x-axis are)
description of
0 is
PI =
and
t
=
{ Y
Therefore y
(0, 0)
- 0) - 0).)
of the
equations
parametric
say
points,
write) X
In
or
analytic
3).
x-axis
(b)
Yl)
of (a) the parametric representations line joining (3, 1) and (-2, reader of course, that the equation of the knows, = this that 0, but we shall now attempt to verify
is the
No
(40))
-
the
(b)
The
the
Xl)
Find
16.
x-axisand
Thus
-
parametric equations of line \302\243in (in contrast to vector) form.
EXAMPLE
(a)
=
Y
=
- 2
=
3
-
+
5t
2t,)
of the representation the parametric repre8entationof a
parametric
same line. line
i8 not
Thus
unique.)))
choiceof
It depends on the
to derive
used
points
GEOMETRY)
VECTOR
ELEMENTARY
80)
the parametric
equations. it is desirable In instances a in some and to line parameter auxiliaryvariable. In orderto write
xY
-
YI
-
is the
the
form
two-point
for
of
order) of any
two
by the first, getting) -
Y2 X2
YI
the
points
the equation the
difference
,
of a
(taken
-
(41) is called
line. This form
in the
line is
coordi-
of the
terms
reason
x-coordinates
on the
(41))
Xl)
this
For
2 alone.
statesthattheratio to the differenceof
YI).)
of the line,2in
equation
PI and P
-
= t(Y2
YI Xl
-
X
which
- Xl)
t(X2
equation
Y
nates of
=
Xl
second
the
divide
and
elimina-
such
accomplish
(40) as
we rewrite
tion
y-coordinates in the same
the same,namely
. YI - Xl (see Fl gure 49). Not e th at th is isequiva
Y2
X2
the stateme11tthat all the similar.
in
triangles
Ien t t 0
Figure
It should also be notedthatthisratio,
Y2
49 -
X2
the ratio of the
is precisely
metric form parametric
the an
to eliminate form without
many
of
the
form
line.
coefficients
Thus
is independent
of
at least
t in
the
one feature
of the points
-
are YI
,
Xl
paraof
the
used
to)
(41) is not the equation of the line, for it is The equation satisfied by all points of Y2 Yl - Xl), while the this line is Y - Yl = describes (41) - Xl (x X2 in line with the We chooseto use point (Xl, Yl) deleted. of remembered the of because its slight inaccuracy, easily spite 2
Strictly
speaking,
not satisfied by
(Xl,
Yl).
(41),
symmetric
form.)))
81)
GEOMETRY)
ANALYTIC
y)
-
Y2
-
X2 X-Xl
Xl)
Yl)
:>1)
\037) X)
o)
49)
FIGURE
the equations. (See Example ratio of the coefficients of t.) \302\267 \302\267 .. . \302\267
derive the
SIncet e ratIo Y2 h
X2
representation of it light
Figure 49. the
angle
of a
Yl
-
of the
first
by
proceed
of line
of inclination
line is the angle
the
\302\243. The
formed
positivex-axis. Thatis,a is
the
by
\302\243 that
the
a name,
of inclination
angle
line and
given
positive side of the x-axisto the lies above the x-axis. If \302\243is parallel
say that the angle of the angle The tangent slope of the line. Thus) x-axis, we
slope
of
= \302\243
portion
of of
tan
of
inclination is
inclination
a =
the
counterclockwise
measured
from
a in
labeled
angle a
giving
the
in
further
angle
is
.
Xl of
5
Igeb ralC
h tea
to
line, we discuss it
being the tangent We
IntrInSIC
SO
IS
2
16b where
Y2
X2
-
-
YI
of
to \302\243 is
termed
line
the zero.
the
\302\267
(42) Xl)))
VECTOR
ELEMENTARY
82)
Consequently,
a line in effect,that
terms
yield the very
A difficulty
any
GEOMETRY)
(42) states the formula for the slopeof of two and (41) says, in given points; two in the computation used points same
slope.
occurs when
that produced(41)is
X2
not
-
Xl =
0; then for
legitimate,
the division dividing
by
undefined in arithmetic. Analyzing this case we see that the line must be (see Figure 50), separately in order that Xl = X2. vertical The angle of il1clination is 1r /2. The reader is undoubtedly familiar with the is undefined; of 1r/2 fact that the tangent i.e. there is no tan real number that Thus vertical lines 1r/2. equals the concept is merely undefined for such no have slopes; a \037 1r/2 if and 0 < a < 1r, then lines. However, zero is
tan a is have
no
defined,
hence
vertical
lines
are the only
slope.)
Y)
(Xl, Y2))
(Xl, Y1)) x)
X=X1)
FIGURE
60)))
ones
that
GEOMETRY)
ANALYTIC
The x =
equation
Xl
83)
of the vertical line under discussion is to the should be confusing reader, for every value of Y, X is always stating: if
which,
as
thought
of
equal to
Xl.
V
=
X
the reader
and
have the
constal1t;
can easily show
horizontal
that
lines
all
form)
=
y
our analysis
Continuing
constant. the
of
41for line, we solveequatio11 Y =
this
Simplifying
Y2
m=
X2
we
the form)
have
all
erticallines
-
Y2
X2
-
Y2 +
-
Yl
Xl
-
X2
Yl and)
b
=
Y2 X2
Xl)
+
YI.
Xl)
the replacements)
by making
expression
straight
y:)
Yl
- Xl X
of the
equation
-
Yl
Xl +
YI,
Xl)
get)
(43))
y=mx+b)
the line. Note that m, the coefficient of Is the line. therea interx, slope geometric of the constant b? The answer is easily deterpretation mined the Y = b when x = o. Thus the by observing point (0,b) is on the line. This is the point at which the line (43) crosses or intercepts the y-axis, and it 'istherefore called 43 is termed the the y-intercept. Equation of a line, for, from for the slope-intercept equation form this equationwe can immediately off the slope m read
as the
equation
of
is the
of
and the y-intercept,(0,b) EXAMPLE determined pertinent
(1)
What
of
the
line.
17. We return to the line (of the last example) by PI = (3, 1) and P 2 = (-2, 3) and ask several questions is its
it. regarding in equation two-point
form?)))
84
VECTOR
ELEMENTARY
By
(41) we
GEOMETRY)
have)
- 1
y
x-3)
3-
-2 -
1 3)
or)
- 1
y
x-3) (2)
Solving
b =
What
for
11 \302\267 5
- ---2 5)
Observe
that its
(
2
slope is 5 .))
of the line in slope-interceptfornl? 1= and \037x + 15 , so that m \037
is the equation y, we get y =
the line (Figure 51). A simple that procedure form is first to locate the y-intercept slope-intercept of the use (0, 15 1 ) as one point of the graph. Then, making we five to units the two units and slope idea, proceed right downward to locatea secondpoint of the graph. (4) What is the x-intercept? In the process of answering this rather simple question we a bit and reflect on the nature of \"the equation shall of digress the line.\" Logically, the equation y = - ; x + 15 1- is a senis true for only certain choices of x and y. That tence, which the sentence is, some ordered pairs (x, y) render true; these that are said to lie on the pairs are the coordinatesof points line y = - \037x + 151- . All the ordered pairs that render the 1 not on the line y = - \037 statement false represent x + 15-. points (3)
We
graph
applies the
We
therefore
to see
check
\\vhether
a
given
point,
say (3, 1),)
y)
x)
FIGURE
61)))
GEOMETRY)
ANALYTIC
85)
the the line by substituting coordinates in the sentence and determining whether the sentenceis renderedtrue. Here 1 = - \037(3) + 15\037' which we have is true. Hence (3, 1) is
is on
on
actually
a set of pairs.
is (0,b),
the
line.
be said that a line is is a set of ordered points that, expressed analytically, whose is m and whose y-intercept Thus a line \302\243, slope is defined by the statement:
In the
language of
set
= \302\243
the
{(x, y)ly =
-
the
+ b}.) set:
following
\037x
+
line of reasoning, point of intersection
the
Continuing finding
= mx
y)ly
{(x,
The line in Example17is then
of
it would
theory
151_
}.
we considerthe of
two
lines,
problem
and \302\2431
\302\2432,
where) \302\2431: y \302\2432: y
+ b1
=
m1 x
=
m2x + b 2 .)
(44))
= intersection (there is, at most,oneunless\302\2431 \302\2432) statements both and \302\2431 \302\2432 pair rendering true, That is, we seek the ordered pair that satsimultaneously.3 both isfies equations simultaneously. Finding the pair, if it is then a matter of elementary for the exists, algebra-solving solution to a set of two simultaneous linear equations. Thus finding the of our line is the problem of x-intercept for the solution of the pair of equations:) solving
The
is
the
point of ordered
y y
=
-
=
0
\037x
+
J5\037
(x axis).
The x-interceptis therefore C\302\245-,0).) to mean use of the symbols \302\2431 and both lines and sen\302\2432 tencesin one paragraph might be frowned upon by some logicians, but this economy of sYlnbols should not cause the reader any confusion. In the rigorous axiomatic treatment. of analytic a straight line may be defined as a set whose defining geometry sentence is of the form ax + by + c = 0 (see equation (49)). This definition includes vertical lines as well as those that have 3 The
slope defined.)))
GEOMETRY)
VECTOR
ELEMENTARY
8'6.)
A Euclidean theorem states: If two are cut by a lines of transversal so that a pair are corresponding angles if two conversely, equal, the two lines are parallel;and lines are cut by a transversal, parallel corresponding of Applying this to our treatment angles are equal. lines are we may say that two analytic geometry, parallel the same possess if and only if they angle of inclination. For nonverticallines, it may be stated that two lines are parallel if and only if they have the same slope.
EXAMPLE18. To find =
y
-
Since
discussion, it render
-
=
\037x
the
;x +
equation
(For another
2.
+
Exercise8
m =
slope
-
, \037
we
may
(45))
b,)
In accordancewith of (10, -2) in
Thus -2 =
true.
quently, b = 2, and
y
line parallel to
-2).
(10,
to be determined. the substitution
b remaining
foregoing should
-
=
y
,vith
of the
equation
form)
in the
it
\\vrite
the
+ 1;- and passing through have the line we seek must
x \037
-
(10) \037
the (45)
Consedetermined:
+ b.
is completely
approach, see (equation46) in
below.))
EXERCISES
1.
(3, 1)
of the line determinedby
the equation
Derive
(5,
4)
and
in:
(a) the vector form (b) Parametric (c)
2. (a) (b)
as
slope of the line determinedin its x- and y-intercepts?
the
What
is
What
are
through whose
(c)
parallel
4. What is 2x -
is 3y
-5).
slope is to 2x the =
point 12 and
2 and
-
y
=
ExerciseI?
line:
the
4) and
(-5,
through (0, (b)
in (39);
form.
two-point
3. Derivethe equation of (a)
expressed
form;
having
-5;
y-intercept
that
is,
x-intercept is. -3. 3 and
through
(1, 1).
of intersection of the line whose equation the line whose equationis 2x - 5y = 121)))
GEOMETRY)
ANALYTIC
-
(a) 2x
5. Sketch
y
12.
o.
-
2x =
1.
=
4.
(e)
y
(f)
x =
(g)
12.
=
5y
+ 7 =
(c) 3y
(d)
=
3y
-
2x
(b)
87)
X
{Y
?r. =
2
=
-1
(h) \037
{:
- 2t + _
t.
Find the
point of A determined by triangle the use (3, 2). (Don't 6. (a)
t
of
intersection
=
medians
the
1), B = (5, -2),
(-2,
of the
equations
of the C
and
=
medians!)
median emanating from B? of Find the the interior (c) equation angle bisector at A. A from vector which bisects angle A, (Hint. A, emanating (b)
is
What
the
\037
=
is V
of the
equation
\037
AB
+
\037
AC . \037
IACI
IABI
Find the equation of the and r/6 (1, passingthrough -2). (b) What is the equation of the 7. (a)
line
line
an inclination
with through
slopeis undefined? (c)
is the
What
equation of the line through
(1, -2) (1,
of
whose
-2)
whose
with
slope
slope is zero?
8. (a) The equation for m, is
given
Y
x Justify (b)
y-intercept
-
-
Yl -
=
m.
(Xl, Yl),
(point-slope form)
equation of
(0,
b) is
a (47).)))
through
(46))
Xl)
X
Justify
line
(See footnote 2, page 80.) for the line with x-intercept
(46). The
the
by)
+
given Y b)
=
1.
of
(0, a)
and
by)
(intercept-form)
(47))
GEOMETRY)
VECTOR
ELEMENTARY
88)
9. Booby traps:
(a)
x
Sketch
(b) Sketch
. whose 17.
+
y
X
=
=
x +
{ y = t. (c) Find an equation is
x-intercept
y.
1
a line
for
OF THE
GEOMETRY
ANALYTIC
Returning to the useof vectors, is the equation of the line
LINE CONTINUED
we
What
Po = (xo, Yo) = N ai + bj?
and
point
If P
=
y)
(x,
\"is
the
the
pose
to
perpendicular
general
52),
we know
question: fixed the
\302\243 through
point
of the
the
vector
line
\302\243 (see
---?
\037
Figure
is 1r/2and
whose inclination
o.)
that PoP 1..N
or
N
pop.
= o.
Therefore) [(x or)
which is
-
xo)i + (y
a(x -
+
xo)
\302\267
(ai
yo)j]
bey
-
Yo)
the equation of line\302\243. We ax +
by
-
(axo + byo)
bj) =
+ = can
=
0 (48)
0,
(48) as)
rewrite 0)
y)
%)
FIGURE
52)))
GEOMETRY)
ANALYTIC
simply, as)
more
or,
ax + which is
c = 0,)
+
by
the
where c
+
(axo
(49))
byo),
the (x, y)(provided
straight line in be solved for y
can
49
= -
for a
equation
general
Equation
plane.
89)
case we will have the line in slopeform. In the event that b = 0, (49) yields the intercept = and b cannot vertical line x both be zero, for (a the trivial then (49)would become sentence 0 = 0, which is true for all points in the plane). Thus (49) includes b
\037
which
in
0),
-cia
all possible
concept of slope is carried the slopeof N'is bla (if from (49), which can be found
- cia, is
y = -(alb)x
nonvertical.
and
vertical
lines,
straight
If the find that
-alb.
a
\037
its
we vectors, the slope of
to
over
and
0)
form
slope-intercept
conclude
therefore
We
that:
two lines (neither of them vertical) and only if their slopesare negative
are
if
perpendicular
of one
reciprocals
another.
That is,
line
if
is the
m1
of line
slope
and \302\2431,
slope of
the
m2
\302\2432,)
.1 \302\2431
and
if
\302\2432)
only
if)
ml=)
1)
--.)
mt) 4J)
be observed that there are two to this parts but only one has beenjustifiedhere. The reader should the justification.) therefore complete If \302\2431.1 \302\2432and \302\2431is vertical must (no slope), then \302\2432 = be horizo11t.al, O. m'], i.e., should
(It
result
EXAMPLE19. and
Following
the of the equation the vector 3i scheme of reasoning that Find
4j.
to
perpendicular the
line through (2,
climaxed with
\\ve write)
[(x
- 2)i +
or) 3(x
(y
-
(3i -
(-l))j].
- 2) -
4(11
+
1) =
4j)
0,)))
=
0;
-1) (48),
which, simplified
and put in the form
-
3x
EXAMPLE20. and
Find
to
perpendicular the
Writing
given
the
equation
3x line in Y
-
4y
of
10 =
-
4y
O.
of the
10 =
43,..N
reads:
(49),
o.
line through (2,
slope-intercept _-
-
(-1). =
x-2
--,4
_ 5
2')
of the desired N in Example the line (46),
- 5=
- 3y
4x
or)
3
-1)
form)
we see that its slope is I. Therefore the slope this line is -1 /: = -t. with of (Check slope 19.) Therefore, by using the point-slopeform we seek can be written) y
GEOMETRY)
VECTOR
ELEMENTARY
90)
O.)
EXERCISES
1.
line \302\243 is given by ax + by + c = 0, vector N perpendicular to \302\243, in terms
If a
for a
in the 2.
equation of the
through
3. Find
an
of the
expression
coefficients
\302\243.
of the line perpendicularto 2x -
the equation
Find
o and
find
y
origin.
the equation
of
the
bisector
perpendicular
of the
ment joining (1, -3) and (3, 5). 4. Find two vectors of unit length perpendicularto 2x =
1 =
-
y
seg-
1
O.)
DISTANCE
18.
The
problem
A POINT of determining
FROM
betweena pointPo
=
Yo)
(xo,
\302\243:ax
TO A LINE the (minimum) distance
and
by +
+
a line)
c =
0
is easily vectors. We shall actually derive by a formula for this distance,but it is author's sugthat the reader not memorizethis formula. gestion handled
the
One
of
the
as need
stro'ng advantages to pure opposed to be remembered.
of thinking in
analytics-is A
thorough
terms of
that
fewer
familiarity
vectorsformulas
with
the)))
GEOMETRY)
ANALYTIC
91) Po)
\302\243)
53)
FIGURE
often enables one to solve from pattern of reasoning
vector tools a basic
basic
following
as easily ciples-just The present problem
Let P
=
a
applying
first
prin-
formula.
complicated
is a caseill point.
a general point on d from Po to \302\243 could
be
y)
(x,
distance
minimum
as
by
problems
\302\243.
be
Then
thought
the
of
\037
as
the
projection of PPo 011the per-
of the
magnitude \037
uct
will
the
last
N
=
to
PF
pendicular serve
as
section, ai + bj.
\302\243 (see
to finding
an aid
we write
Thlls
53).
Figure
the dot
d. Using the result
Applying Theorem 6, we
have)
\037
IprNPPol =
NI
t.PPo. INI
-
[(
x-
.
Xo
)1 +
(y -
e
Yo
)J J
a.i +
e _
V
-_
taxo
+
- ax
byo
vi a
2
+
of
a vector perpendicular to \302\243ag
\037
d =
prod-
b
2)))
-
byl
.
/
2 a
bj
2 + b
But,
- ax
-
=
by
+ d = laxo 2byo formula coordinatesof the
Po
point
given
c\\
,
(50))
terms of the constants
d in
distance
the
for
+
2 + b
Va
which is a
have
we finally
c, so
GEOMETRY)
VECTOR
ELEMENTARY
92)
and
the in
the given line \302\243. 21.
EXAMPLE
line
\302\243: y
=
jx
formula
Although
the distance from
Find
- 1.
(50)
for the
Po = (1, 3)
distance has
shall illustratethe point made above by abandoning the use of (50) from first principles.
this
section, we
very
We
writing
by
begin
that a vector perpendicular as i
- 2j. In
the
reader doubtlessly INI,
which
jection
by
to
in we were
we were
that on
projecting
fore, be more convenient
the unit if we
2y
immediately
used
that
observed
meant
\302\243 can
computation
the
-
2 = 0, so be written
arriving at d, the forced to divide by
actually computing the provector
convert
\"
I:'
It
might,
the perpendicular
y)
J\\(l,3))
\302\243)
1)
%)
-2
-1
in
suggestion
of working
in favor
-
the
been derived
of
x
form
the
\302\243 in
to
0)
FIGURE
64)))
there-
vector)
to a
93)
GEOMETRY)
ANALYTIC
-+
PoP I on U, choice for PI
where
is the y-intercept -+
d
=
IP\037I. UI
_
(-i _
=
7
V5
o of
projection
\302\243. A
(0,
\302\243, namely
I
(pru
rather
.i -
i
.
- 2j
of simple
Hence,)
-1). \037
=
PoPI)IUII
+ (-1 _ 3)j]
4j)
the
of
\037
=
- l)i
[(0
_
is
PI
- 2j .
i
=
the magnitude any point on
is then
d
distance
present computation.
is)
vector
U
The
of our
outset
the
at
vector
unit
Sucha unit
Ipru PoP
II
2j
o
_ 1-1 +
81
0
o
.
5)
Any
other
say PI = (2,0), should
of PI,
choice
result:)
l)i + (0_ 3)j].
[(2 -
d=)
same
2j
o
70
1+6
5)
o)
Even a generalpoint
i-
the
give
PI
= (Xl,
YI)
l)i +
(Yl
of \302\243 would
the explicit
yield
computation:)
d=
-
[(Xl
-
3)j]
.i \037\037j
-
IXI
V But,
since
(Xl, YI) is
+
2YI
51
5)
on \302\243, we
know
that
Xl
fore)
d =
!2 +
o
51
=
I
\037
0
=
7
0 5)))
.
-
2YI\"
=
2.
There-
vector
.
had
we
Suppose
to
perpendicular
jection
be
would
the
of
value
our
same
method
to a formula.
of. vector
standing of
such
exercises
taking the the value
is grasped, the without directly,
projection and
However,a
-
from (1, 2) to x
the distance
2. Find
the distance
from
(1,
-2)
to x
3. Find
the distance
from
(1,
-2)
to
4. Find the magnitude on the line x - 2y
the
of =
5. Find the magnitude vector 3i - 4j where
5, where of
the
A and
METHOD
ANALYTIC
pro-
computations.
previous
EXERCISES
19.
.
under-
thorough
comes only with the practice it is precisely to promote such Be sure to do a are included.
And
thinking.
Find
umt
thinking
practicethat good shareof them! 1.
of the
sign
projection,
simply
any reference
2j as the
we are
since
but
changed;
the
Then
\302\243.
magnitude,or absolute for d would bethe as in As soon as the of can be made computations
- i+ V5
=
With U
begun
GEOMETRY)
VECTOR
ELEMENTARY
94)
5. =
2y
2y
5. +
and
5 =
O.
segment AB
of the
(1, 1)
projection
B are
-
-x +
projection
A =
=
2y
=
B
(2,
of segment AB
-1).
on the
the sameas in 4.)
OF PROOF
In accordance with the philosophy in Section described shall of we now illustrate the 15, application analytic methods to proving results of geometry. The spirit of will be to divorce our work from the these illustrations
ideas
of
in favor
of
with
working
notions that are analytic two
sides
of a
one half of it. Again,
in
in
tion 15, we
character.
Prove that the line joining the side triangle is parallelto the third
22.
EXAMPLE
similarity) congruence, coordinates, slope, and other (e.g.,
geometry
synthetic
accordance view
this
with as a
midpointsof and
equal
to
the philosophy put forth in Secgeometry problem and impose a)))
GEOMETRY)
ANALYTIC
95) y)
B(b,
C))
%)
A (a, 0))
0(0,0))
66)
FIGURE
coordinate
upon it.
system
nate system in
such a
therefore
vVe
to
as
manner
the coordiour work. For
impose
facilitate
the origin as one vertex of the trichoose x-axis to be along one side of the triangle (see Figure 55). If we do this, and if we desire our proof to hold all triangles, for there is no further choice available. Calling the triangle OAB, we assign coordinates as follows:)
example,we angle
may
and the
= (0,
o
Next, we and
in Example M
midpoint of
M N
slope
=
(a,
0),
OB, by
B = (b,
of M, means
c).)
the midpoint
of the
formula
N =
\037
of
AB,
derived
lOde)
a
=
( The
A
coordinates
the
compute
N, the
0),
of MN
is parallel
b
and) \037
=
,\037))
c/2
- c/2
bj2 _ (a +
to 0 A.)))
b)/2
=
( 2'
.
0, whIch
C
.
2))
proves
that)
secondpart of
As for the of
\037
(;
(\037
\037r
each other. we
Again
elect
the
Let (a, 0),
coordinates,
and C say
a
b
;
=
= \037 \0372
Y
, 1;1
\037IAOI.
that
Prove
23.
-
+
shows that 11IINI =
EXAMPLE
position. = A
-
=
IMNI
bisect
the length
we compute
result,
MN.)
segment
which
the
GEOMETRY)
VECTOR
ELEMENTARY
96)
the diagonals
of
a
parallelogram
the coordinate axesin a convenient be OABC, where 0 = (0,0), parallelogram
to place
= (b, c), asnotedin Figure56. to point B, we find that
If
we
ascribe
d and three other
(d, e),
e are
verdependenton the coordinateschosenfor the the that tices of the paralleiogram. That is, the condition a that its are is sides opposite figure parallelogram, namely,
parallel,forcescertain
coordinates
B.
upon
parallel to OA, the slope of CB is zero; that is, that e = c. And, since (e - c)/(d - b) = 0, which implies OC is parallel to AB, (c - O)/(b 0) = (c - O)/(d a); or B = (a + b, c). d = a + b. Thus, the parallelogram Now that we have imposed condition, we are free to attack our problem-findingthe mid-points of the = \302\253a the The of and OB + b)/2, c/2) diagonals. midpoint of AC = \302\253a + b)/2, c/2), which establishes the result.) midpoint Since
CB is
y)
B(d,e)
C(b,c)
= (a+b,c))
x)
A (a,
0(0,0))
FIGURE
66)))
0))
GEOMETRY)
ANALYTIC
97)
y)
B(b,
C))
%)
A (a, 0))
0(0,0)) M(\037,O))
67)
FIGURE
angle
point.
Let the triangle be
o =
(0,
of OA,
A
0),
N the a
M =
Then
= (a,
ON: y
:
For t\\VO
y
the
in
sho,vn
B=
(b,
AB,
and
Q the
a+ b C ' N= ( 2 2)
0,
lines
median
the
and
57, where
Figure the
CaHill
c).
midpoint
midpoint b
=
Q
tri-
a
of
C
OB.
.
( 2' 2 )
are found to
be:)
x
a+b
BM: y =
AQ
of
midpoint
The equationsof c =
as
OAB,
0) and
2' )
(
mediansof
Prove analytically that the
24. meet in a
EXAMPLE
=
\037a/2)
c/2
y =
or)
;))
(x
(x b/2 _ a
point
equations
-
b_
-
a))
or)
11
=
2cx
2b-a
c b _
of ON and B]1,l,,ve simultaneous pair, getting)
of intersection as a
x=
a+b
3)
and)
y
=
c
-.
3)))
2a
(x solve
-
ac
2b-a)
- a).) the
first
For the
point of
+ \302\253a
point
AQ, we solve the arriving at the same
ON and
of
intersection
third
and
first
at
all meet
GEOMETRY)
VECTOR
ELEMENTARY
98)
equations simultaneously, Thus, the medians of b)/3, c/3). + b)/3, cj3). \302\253a
OAB
triangle
For the purposes of further we use the illustration, vector approach to checkthe coordinatesof the point P of the medians. Referring to Example of intersection the vector of P is written position lOe,
P=
}-A +
+
}-O
= }(Oi +
lB
+
OJ)
= j(a + b)i+ =
P
Thus
-i(ai)
;
+ cj)
+ }(bi
j
(a + b)j3, cj3), which
above
the
with
checks
com pu ta tion.
It
handled
of vectors
use
by
median problem is more than by pure analytics.
the
that
clear
seems
easily
EXERCISES
1. Prove:In
the
triangle
any
sum of
to three-fourths the sum
is equal
of
the squares of the
medians
the
of the
squares
three
sides.
2. Prove:The sum equal to one-half twice
the
median on that
of the
sum
The
lelogram is
the
of the
square
3. Prove:
of two sides of a triangle is squares of the third side, increased by square
the
of
(Comparewith
vector
of the
squares
equal to the sum
of
the
The sum of the squares of quadrilateral is equal to the sum of the points
of
5. Prove
the
four
of a
sides
four
of the
squares
paral-
diagonals.
proof.)
4. Prove:
increasedby
side.
times
the
four
the
sides
of any
of the
diagonals square of the line joining the midsquares
diagonals.
analytically:
(a) the median of a trapezoid.is (b) the lines joining the midpoints rilateral
form
a parallelogram;)))
parallel of
to the the
sides
bases; of a quad-
99)
GEO.METRY)
ANALYTIC
(c) in any
joining the midpoints point that is the midpoint of the segment joining the midpoints of the diagonals. a slope of m1 and an angle of inclination 8 1, 6. If line \302\2431 has a slope of m2 and angle of inclination 8 2, and if line \302\2432 has use the formula for tan (82 - (J1) to determine tan (J, where (J of
is
the'
from
\302\2432).
the
7. Discuss the
lines
of
are \302\2432
(a) (b)
(2,
of
slopesof
the
10.If 0
=
equation
of
lines (0, the
counterclockwise
\302\2432 (measuring
using
the result
of Exercise 6 when
perpendicular. the
triangles
(-4, 13). (-3;
(6, 0),
6),
9. The slopes
in a
58.
Figure
difficulty
and \302\2431
the angles of (5, 0), (8, 4),
8. Find
and \302\2431
See
segments
intersect
sides
between
angle to \302\2431
the
quadrilateral,
opposite
8).
as -2 and 3. Find the the bisect angles betweenthem.
lines
two
that
=
are given
(V3, 0), A line bisecting
1), and B =
(2
V3,
1), find
angle AOB.)
y) \302\2432) \302\2431)
%)
FIGURE
68)))
the
11. If
is given by a1X + b 1 y + C1 = 0 and \302\2432 by = a linear of \302\2431 combination and + Cl 0, we define of the form equation
is \302\2431
alX +
b1y
an
as \302\2432
given
m(alx +
(b)
is the point
If P
P; and
conversely, every
and \302\2431
\302\2432.
The
of all lines through P, can be written
over all
range
(c) If
the
\037y
zero?
and \302\2431
also \302\2432
\302\2432, prove
single
the real numbers(excludingthe casem
is \302\2431
to
parallel
them is also parallel to each of \302\2431 and (51) represents the set of all lines parallel to a pencil of parallel lines. called
\302\2431 (or
12.
line
the
of Exercise
results
11,
the
find
\302\2432.
n =
=
any linear
that
\302\2432, prove
tion of
Using
through
passes
a linear combination the set of this result is that vertex the pencil of lines with 51, where m and n equation
called
often
(51)
P is
through
consequence
P,
of
0
combination
a linear
of \302\2431and
line
=
C2)
are not both
intersection
of
+
b2Y
by
m and n
any linear combination
that
of
described
l\037cus
a line, if
also
lines
n(a2x +
C1) +
+
b1Y
is the
(a) Why
of two
GEOMETRY)
VECTOR
ELEMENTARY
100)
0).
combinacase
this
In
\302\2432),often
the
through
point of intersection of x - 2y = 3 and 4x - 2y = 15, and the point of interpassing through the origin, without finding section of the given lines. follow (Be sure to (51) carefully!) 13. (a) Write an equation that represents the pencil of lines (1, -2). (b) Write an equation that
through
inclination
\\vith
7r
(a) Find the
14.
lineswhose 0, and (b)
through
Find
lines whose
triangle
the
through
are
-
x
(0,
point
the
meet
(;
point
4 =
the point 2y
8 =
+
3y
3 =
-
of
of the
intersection
0 and x
+
8y
+
7 =
, 0).
of Exercise
results
0 and 2x +
\037 ).
the line through the
of the
of intersection
point +
2y
equations are 3x +
0, and through 15. By using
lines
/6.
line
equations the
representsthe pencil of
11, prove the
n1ediansof
a
in a point.)
20. CIRCLES
Let Po =
and
P =
(xo,
Yo)
be
the
center
(x, y) a general point of
of a the
circle of circle
(see
radius
r
Figure)))
GEOMETRY
ANALYTIC
101) Y)
x)
FIGURE
vector
the
Then
59).
59)
of the circle is
equation
easily
\037
found lizing
tor
the
imposing
by the
vectors
position of
equation
condition that IPopl = r. Utiof the points, we write the vec-
circle
the
Ip or
V (P we
form,
- Po) \302\267 (P
as
- Pol = Ir
Po) = as (52)
rewrite
I(x
-
- xo)i +
(52)
r. To get the -
(y
yo)jl
=
r)
=
r.)
analytic
or)
Vex Finally,
by
squaring
(x as
the
whose
of equation radius is r.)))
-
+
XO)2
both XO)2
-
(y
YO)2
members, +
(y
-
YO)2
we get =
r
2)
the circle whosecenteris (xo,
(53))
Yo)
and
the
Concerning
is
of
x
the
coefficients
form
the
in
when
x2
quadratic
2
y2 both
and
- 2xox+
+
X02
-
y2
x
which is, of
We
+ y2
pose
the
C =
+
By
of a
equation
write
2
(x
In.orderto (
+ Ax
do
+
form
of
A
(y2 +
+
Ax)
A 2/4
add
we
so
We
(53).
+
how
y2
+
By
the
in
to paren-
A
2
) =\"\"4
+\"4
members:)
to both
B2 (
we
whether
attempt
By) = -C.
and B 2 /4
2
+\"\"4)
(54)
a circle,
questionsdependon
in the
(54)
form
the
of
represent
accomplish this by completingthe squares theses of the expression
2
a
such
of
and radius?
center
the
find
does
it
If
(2)
The answers to both
x
(54))
0,
circle but
equation
every
can
write this
we may
obscure its essentialcharacteristics, namely, center of the circle. two which can be answered simulquestions,
represent a circle? we
coeffi-
designating
letter,
- Ax +
taneously. (1) Does can
and
= r 2.
yo2
and
radius
the
2
course,
as to
nature
+
2yoY
form
the
in
equal (to unity),
we get
By grouping the constantterms cients of the variables by a single
equation
it should be and y; (2) it has
circle,
x and
in
(5\037).
(53),
Expanding
the
of
equ\037tion
noted that: (1) It
GEOMETRY)
VECTOR
ELEMENTARY
102)
+
B
2
4
-
c)
or)
A x+-
2)
( which
center
is in
2
B
+
the form
of the
(
y+-
2)
2
A
2
=-+--C 4
B
2 ')
4
of (53), enablingus to say
circle is
(- A/2, - B/2)
and
the
that
radius
the is)))
.103)
GEOMETRY)
ANALYTIC
+ B2/4 - C. The overlooked is the questionof real or imaginary. If A 2/4 V A 2/4
For
exists.
circle
example,
one
do
the circle
y2
of x 2
coefficients
the
the equation
2
y2
+
2x +
1) +
(x
the locus
radiusis r =
3
3y +
+
y2
- 1)2 +
(
+
y
is a circlewhose
in terms P =
We
therefore
- 6x +
9y
+
2
rewrite
= o.)
443 ) 2
-3
2)
2 - -,
=-.31 12)
is (1,
center
-j)
and whose
\03703.)
= xi
P
written
zero,
equals
= 1 + -9
-9
consider a circle of radius r in (as Figure 60), the positionvector)
be
3y2
> 0,
/4
get)
If we
can
+
real
cannot repre2 - C B
equal, we may to unity:)
2
+
3y
( or)
+
equal
- 2x +
Completing the squares,we -
3x2
y2 are
and
coefficients
these
with
x
Hence,
-1
expression
the locus
Examine
25.
-_ 0 .
(x2
=
A 2/4
this
if
- C < 0, no
degenerates to a point.)
EXAMPLE
Since
+ if
circle; and
a real
have
B 2/4
+
sent a real locus. However, we
this
whether
x2
has been root is square that
difficulty
conclude
+
the
origin
yj)
of its angle of
r cos 8i
at
centered
inclination
8:)
+ r sin 8j.)
that) X
=
r cos
8 (55))))
{ y=rsin8)
GEOMETRY
VECTOR
ELEMENTARY
104)
Y)
x)
60)
FIGURE
are the the
equations of the givencirclewhere
parametric
parameter.
The single equation of the form (53) sents a circle of radius r, centered at
recovered from (55) by
y2
from
x2 But cos2
8
+
+
equation
=
r
= 1. x
the
r
of the
2
(54)
2 sin 2
2
8
adding
8,)
+
Thus)
+ y2 =
specified
r2)
circle.)))
that
origin,
cos 2 8
r 2(cos 2
y2 =
sin 2 8
=
or
the
that
we deduce
which
alld
squaring
x2
is
8 is
sin 2 8).
repremay be
as follows:)
105)
GEOMETRY)
ANALYTIC
EXERCISES
1. Write the equation (a) center (0, 1) and (b)
center
(0,
(c)
center
(-2,
(a) (b)
passing
2x2
(b)
2 (c) x + 2
(d) 3x
centers and
y2
+
lOx + 7y - 4y 3y2 + 8x
4. Find the points
origin;
and
passing
+
9 = 0; + 15 =
O.
of the line the circleswhose equa-
the equation
and
chord to
common
the following
having
0;
intersection
of
the
contributes
that
=
12y
-
circles
of
radii
- 16x + y2 - 2y 2 = 9x;
+
the
(2,
the origin.
through
(a) x2
circle
-1) and passingthrough through (1, -1), (2, 0), and (0, 3); x-intercept 8, y-intercept -12
(c) having
equations:
2;
the
of
at
center
3. Find the
radius
-1) and radius 2; 3) and radius 3.
2. Find the equation \\vith
with
circle
the
of
are
tions
2 (a) x
x (b)
x
2
2
x
2
y2 + 8y - 6x y2
+
+ y2 +
= 16; 4x 25 = 0
+
8y
+
-
y2
of the
5.
through
Carry
of
equation
+
3 =
o.
(Hint.
interesectionof
meaning
the
= 64
is
What
the
algebraic
loci?)
the following vector line tangent to a
the
approach to determine a given
at
circle
given
point.
Let the circleX be centeredat C = (xo, Yo) (a, b) be a point of X. Call P = (x, y) the general line gJ, which is tangent to X at (a, b). Using radius
gJ
is
vector
represented
(xo 6. Verify
CQ is perpendicular by the equation)
- a)(x -
the equation
point-slopeequation perpendicular
to
the
let
point the
Q = of the
fact
that
that
line
\037
\037
the
and
of
a)
for
the
tangent
+
(Yo
- b)(y
gJ given
line
to QP, prove
and
line at
-
b)
in Exercise
the fact
=.
o.)
5 by using the that a radius is
the point of
contact.)))
7.
of the line tangent to lOx = 60 at (4,2);
y2 + 2 x + y2 x = 2 cos
(b) (c)
+
12y = 36at (6,0); 8
y=2sin8 x
(d)
GEOMETRY)
the equation
Find
(a) x2
=
at CV2, V2);
} 8
2 cos
at CV2, -V2).
y=2sin8) }
8.
Prove
at
(xo, Yo) and
9.
VECTOR
ELEMENTARY
106)
that
the
X
=
Xo +
r cos
8
{y
=
Yo +
r sin
8.)
a parametric
Find
of a
equations
parametric
circlewith
center
radius r is
of the equation
representation
of
the
circle
with center ( -1, - 6x 3x 2
(a)
+ 3y2
(b)
10. Find the points
are
(b)
2
9y
radius
2 =
-
2x 2x -
5;
O. loci whose equations
of the
intersection
of
+ y2 x 2 + y2 -
x
(a)
2) and
+
2y
+
1 =
2y
+
1 =
0 and x + 0 and x2 +
=
y
1;
y2 =
1)
SPHERES
21.
The
which
sphere,
are
three-dimensional
a fixed point. to be precisely that
from
equidistant
easily
be
may
analogue of the defined as the locus
recognized
circle is the of
This of
the
that
points
definition
is
circle
if the
is restricted to a plane. Actually, the very carries over to four-dimensional,fivedefinition
discussion same
n-dimensional
and
dimensional,
spheres
(sometimes
called hyperspheres).
Confiningour discussionto
Po =
(xo, Yo,
zo)
the
generalpoint the (radius the sphere) of
of
fixed
three
sphere,
between
dimensions,
we
call
(center), P(x, y, z) the and r the constant distance Po and P (see Figure 61).)))
point
107)
GEOMETRY)
ANALYTIC
2)
.y)
s)
FIGURE
position Ip
vector
the
Then
-
precisely the
of Po and P, or)
r)
-
vector
the
is
equation
of
have
we
form
(y
the
=r,)
V(P-Po).(P-Po)
coordinate
- xo)i +
of
terms
is)
same as (52),\"\\vhich
the circle. In I(x
of the sphere, in
equation
vectors pol =
61)
(z -
yo)j +
=
zo)kl
r,)
or)
-
Vex
Finally, we
+
XO)2
(y
-
YO)2
(z
get)
-
XO)2
as tl1e equatio11of
+
(y
the
radius
is
sphere
r.)))
(z -
--.:. YO)2 + whose
Po = (xo,Yo, ,vhose
-
=
ZO)2
r.
both membersof this last equation,
by squaring (x
and
+
zo)
ZO)2
center
=
is
r
2
The problem
tion
of
(similar to that form
discussion
in Section
to that
would be quite repetitiousof
20 on
circles.
shall
We
unit
y2 +
+
=
Z2
the components
analyze
matters
whose
radius,
the
for
other
of
the
relegate
an
The unit sphereor sphere at the origin,hasthe equation) 2
therefore
We
discussionas exercise turn attention to
of such a details reader while we now to the sphere. relating the
x
form for the equaof (54)) and of converting
a general
at
arriving
a sphere
equations
of
GEOMETRY)
VECTOR
ELEMENTARY
108)
center is
1.)
of
(56)) the
vector)
position
P=xi+yj+zk)
of point
P on the unit
If a is
the
angle
P makes
with the
62a.
x-axis, (3
the
angle
the
that
z-axis,)
\302\267 =
P
with the
P makes
with the y-axis,and \"I
P makes
that
in Figure
shown
sphere,
that
angle
i
cos a
Ipllil
=
a)
COS
and)
P
\302\267 =
yj +
(xi +
i
zk)
\302\267 =
i
x.)
= COS a; similarly, y = cos (3 and z = cos \"I. cos ai + cos (3j + cos 'Y k . V (see Figure 62b) in three dimensions vector Every to the vector of a point P on the unit parallel position that so the direction of V is completely specified sphere, P. with the vector \"I associated by the allgles a,. (3, and x
Therefore
P
Hence
=
is
For this reason we angles
of V (and
Since Ip!
2
vector
satisfy
the of
angles and P),
of V (alld = p \302\267 P = 1, the the relation
cosines
direction
the
call
also
cos 2
a
+
cos
2
{J +
ex, f3, and \"I the cas a, COS(3,
direction and
of P). direction cosines of cos 2
'Y
=
1.
cos
\"I
every
(57))))
109)
GEOMETRY)
ANALYTIC
z)
y)
z)
x)
y)
x)
(a))
FIGURE
V
Suppose
which
V is
=
mj +
li +
nk.
(b))
62)
Then the
to
vector
unit
parallel is V
=
Ivl
+ mj + nk \ 2 2 2 VZ + m + n li
Thus)
cos a
1
=
V
l2
+
m
2
,
m
cos
2) + n
13
::::a:
,
V
Z
2
2
+\"m
+
n2
(58)
and
COB
'Y
n
=
V
Z2
+
m
2
'
+
2
n)))
GEOMETRY)
VECTOR
ELEMENTARY
110)
which we see that l:m:n = cosa:COS (3:cos 'Y; that is, the numbers l, m, n are proportional tothedirecfor this It is reason that l, m, n, are tion cosines. direction numbers of V. We of the ordered called speak set {l, m, n} as a set of direction numbers. Since the = tli of tV is the same as direction + tmj + tnk(t \037 0) that of V (tV is parallel to V), we also call {tl,tm, tn} a set of direction t \037 0) numbers for V. If (where is a set of direction numbers for a sometripleof numbers nonzero any given multiple of the triple is also a vector, set numbers the direction for of vector, for both triples
from
designatethe
direction
sa\037e
the
(with
of
pOBsibility
condition the opposite sense) and both triplessatisfy 4 to the direction cosines. proportional being 26.
EXAMPLE We
(a)
give
Let V = 2i three equivalent
the given
From
{2, - 3, 6} form of
direction
the
cosa = It should
3t,
2 yields
V.
that
6t}
; e.g.,
t =
-1
12}.
write
2
(-3)2 + 6
2
2 = -,
=
\037i
7
cos{3
7
be noted that the unit
which the
for
immediately
numbers. Two other sets by allowing t to take on two
{4, -6, cosines of V.
we
(58),
!.V 7
from
t =
direction
V 22 +
we see
orderedtriple {2t, -
and
Making use of
V
of
of direction can be found
yields {-2, 3, - 6} Find
sets of directionnumbers
set
a
distinct values in the (b)
- 3j + 6k.
representation
numbers
of
-
=-,
-3 7
cos 'Y =
6 -. 7)
vector)
. 7J
\037
+
\037k
7')
direction cosinescan be read off
directly,
has)
4 it may be advantageous to have In some a developments, V actually set of direction numbers for a vector information impart about the senseof V. If this were the case, we would impose the restriction t > 0; for if t < 0, then {tl, tm, tn} .would imply a sense triple oppositeto the original {l, m, n}. However, our development of direction numbers is principally for applications to lines, in contrast we make to vectors, have no sense. Thus, the which, simple restriction that t \037 o.)))
111)
GEOMETRY
ANALYTIC
the same
sense of
as V.
direction
= -1.V 7777')
However,) . J
\037
+
-\037i
-
\037k
although oppositelysensed,is alsoa unit also yields a set of direction cosines {tor
and
vector
-
, \037 , \037
} for \037
therefore the vec-
V.)
EXERCISES
a set
Find
1.
of direction cosinesfor
the
position
of
vector
the points
(a) (4,3, 5), 12),
-4,
(3,
(b)
(c) (0,
0, 1),
(d) (0,2, 0).
2.
of Example
method
the
Using
of the
sphereswhose
(a)
4x 2 + 4y2
2 (b) x +
y2
+
equations 2 =
+
4z
z2 +
the equation
3. Find
of
4x the
25,
find
the
center
and radius
are
8y,
-
=
6y
sphere
3.
with
at (1, -1, 0) and radius 2, center at (-1,2, -3) and radius V 2, (b) (c) center at the origin and radius 12. 4. Find two sets of direction numbers and two cosines for the vector (a) V = 3i + j - k. V = 2i - j + k.) (b) (a) center
22.
sets
of direction
PLANES
is the locus, or surface, that we discuss in in is t o theline some respects analogous plane, which two dimensions. we have several equivalent Although our of the momellt will best choicesfor a definition, needs
second
The
be served
by
Definition.
locus
the
following:
If Po
of points
is a fixed point and N a fixed
P so
--7
that
called a plane. (In Figure so that its origin at Po.))))
PoP 63,
is we
vector,
the
to N is perpendicular N positioned picture
FIGURE
An
immediate
GEOMETRY)
VECTOR
ELEMENTARY
112)
63)
definition is that
Po P = Po, then PoP is the zero to N (see Theorem5). of the
consequence
--7
is on
itself
that
vector
the plane, for is perpendicular
(P the
vector
in terms
language
of
we have)
vectors,
position
in vector
definition
the
Writing
if
- Po) \302\267 = N
of the plane
equation
(59))
0,)
Po
through
and
perpendicu-
lar to N.
In order to derivean al1alytic expression we follow the usual procedureof calling The POillt of the plane P = (x, y, z). will be Po = (xo, Yo, zo) al1d N = ai + (59) becomes
[(x -
+
xo)i
(y
- yo)j +
-
(z
zo)k]
\302\267
(ai
for the
elements
fixed bj
+
ck.
+
plane,
variable
the
Then
ck) = o.
bj +
Expanding
yields
-
a(x
which Yo,
zo)
is and
the
xo) + bey analytic
perpendicular
-
Yo)
+
c(z
of the
equation
to the
numbers are by character of the equations specified
of
{a, the
-
zo)
=
0,
(60)
plane through (x.o, vector whose direction The analogous b, c}. and the lil1e is))) plane
easily (60)
113)
GEOMETRY)
ANALYTIC
(60)
seen by comparing may be written
ax +
whered = -
+
(axo
by
+
byo +
of (49), which clearly in all three plane is be noted that the equations 49 and 61 and
linear
vectors
(48).
Furthermore,
d =
0, (61) cZo), which is the counterpart shows that the equation of a variables. Finally, it should cz +
of
coefficients are
with
the to
related
variables the
in both
perpendicular
N.)
27.
EXAMPLE
(a)
What are
the equationsof
the
coordinate
planes?
The xy-plane can be describedas the plane perpendicular the origin. Therefore, its equation k and passing through - O)i + (y - O)j + (z - O)k] \302\267 k = is found by simplifying [(x = z the O. reader can show that) O. The result is Similarly, to
z)
z=zO)
(XO,
YO, zo) \302\267)
y)
x)
FIGURE
64)))
the equation is y
xz-plane
(b)
the
for
the
are
What
is x
yz-plane
= o.
= 0, and the equation
for
parallel to the coordi-
of planes
equations
nate planes?
GEOMETRY)
VECTOR
ELEMENTARY
114)
and plane parallelto the xy-plane as shown in Figure 64. Then, as in part (a), we arrive at z = zo, the same reasoning following which states that for all choices of x and y, the z-coordinate of a point on the plane is equal to Zo.
(xo,
through
passing
on a
attention
our
fix
We
to the plane whose
vector
y
perpendicular
the plane the coefficients of the
using
vectors
unit
(b)
What
2i+j-2k
the pointsat which
of the of
the
x-intercept
Intercept by plane.
of the
the
plane of
point
placing y
gives 2x
This is (1,
or
.
equal to zero,
z-coordinates
variables (see(60)
3)
axes? x-intercept. Since every and
to the given plane. can immediately be
plane are then
to the
the intercepts
are
O.
- 2k.
+ j
perpendicular + -
what are
- 2=
to
2i
y-
2z
as)
(61)),
Two
-
vector
unit
perpendicular by
written,
a
find
To
(a)
questionsrelating
is
equation
2x +
A
discuss several
vVe shall
28.
EXAMPLE
zo),
Yo,
=
given
the
0 in the
the
have its
x-coordinate
given equation
- 2 = 0 or x =
0, 0).
is,
the coordinate
x-axis must
we solve for z =
That
locus?
intersects
1.
Thus
the
to verify The reader can carry the through computation that (0, 2, 0) and (0, 0, -1) are the other two intercepts. are the traces of the plane whose equation is (c) What 2x + y - 2z - 2 = 0, in the coordinate We define planes? the trace of a locus oC in a plane II to be the points of oC that lie in II, or simplythe intersectionof oC and II. If we ask specifically for the trace in the xy-plane, we are askingfor the locus satisfying the simultaneous equations - 2z - 2 = 0 2X + y
{z
=
0
(xy-plane)
.)))
115)
GEOMETRY)
ANALYTIC
z)
>
Trace
y)
in yz-plane)
x)
66)
FIGURE
sketch the graph of 2x + y - 2 = 0 in the the xy-plane (z 0) to see the trace in that plane. Again, reader can verify that the other traces areobtained by graphing x - z - 1 = 0 and y - 2z - 2 = 0 in the xz-plane and yzplane, respectively. Graphs of the traces in the coordinate in sketching locus a three-dimensional plane are quite helpful (see Figure65).) Thus,
we would =
23.
A
DETERMINING
BY POINTS
PLANE
ON IT
How many points are actually to determine necessary Geometric intuition indicatesthat three points would be necessary and sufficient, but can this fact be a plane? shown
algebraically?
If
four
the
determined.
pletely
the a, b, c, and d are known, = comis ax d 0 cz + + + by plane that fOllr, rather than This suggests
constants the
of
equation
three, conditions necessary. did not lead us astray, one be chosento be Not are
for
so
unity.
,ve
can
divide
through
by
However, of all
the of a,
one of
four
our
intuition
constants
b, and c are
them,
say
a
can
zero, \037 0,)))
ELEMENTARY VECTOR GEOMETRY)
116
and
get)
x +
Making
-
+
a
- =
+
Z
o.
(62))
a)
(3 =
substitution
the
d
c
-b y a
bfa,
=
'Y
and 0
cja,
= dja,
(62) becomes)
x+
0 = 0,)
'YZ +
+
{3y
three conare from which we see that there essel1tially turn out that stants to determine. In practice,it may = was a performed just 0, in which case the division choose one of the If this be the case,then we could illegal. until we hit upon a legitimate other numbers as a divisor
division (see Example30). and
(2,
the
Find
29.
EXAMPLE
-4).
-3,
=
d
Assuming
1, we
solve for a, b,
ax + By
a +
that (1, -1,
renders
a
and that (2, -3,
this system We can reduce to two equations in two
second equation, which
is
1 =
(63)
true
- 4c
3b
\037
(63))
renders
first by
6a -
....\\ddition
yields
+ 1=
O.)
of three equations
in
most
unknowns
devoid
already
+
12c +
4 = 0
9b
-
12c +
O.)
lOa
-
b
+
3=
7 =
0,)))
three
easily
of c,
8b
yields)
we
0;
the first and third. To this 4 and the tl1ird by 3, getting
4a +
true,
0;
c from
nating
(63)
yields
+
b
renders
4)
2a
the
true
(63)
-
1=
3c +
+
2b
0)
o.)
that (1, 2, 3)
the condition
imposing
get)
1=
0),
the equation)
c in
and
cz +
+
by
3), (1, -1,
(1, 2,
plane through
unknowns
by using
the
and then elimi-
end, we
multiply
GEOMETRY
ANALYTIC
which,
together
117)
- b +
a
with
-- 2
a =
1 = 0, implies b
and)
that)
=-.
3)
these values into the original Therefore the desiredlineis -ix +
Substituting c = -t.
0, or more
-
2x
y
+
z =
2, 4).
Imposing
-
tz +
1=
3.
EXAMPLE 30. Find the plane through (2,
equations gives ty
stated,
simply
and
1 3)
(0,
points on equation
the three
63
1, 2), yields
(1,
-1, -2),
the system
b+2c+l=0
a-b-2c+l first
the
Adding
adding
However,
4a + 3 = 0 or a
card the assumption
a
r!=
0;
particularly
equations gives a + 2 = 0 or a = the second equation to the third gives = \037, an inconsistency. So we must disthat d \037 o. let us assume that Instead, that a = 1. Then we seek b, c, d in)
-2.
two
twice
-
x + the three
Imposing
4c +
2b +
+
2a
=0 1 = o.
+
by
cz + d
=
O.)
given points on relation
(64))
(64), we
get
the
system)
b+2c+d=0
(65))
I-b-c+d=O
2+
+
2b
4c + d
= o.
of (65) gives 1 + 2d = 0 or to the third gives second adding -t; 4 + 3d = 0 or d = -t, which is another inconsistency. So we discard the assumptionthat a r!= 0, upon which the form (64) was based. We now try b \037 0, and use the form of the equaAdding
first
the
d =
two
but
tion with
b =
getting
again the
the
twice
1)
ax Once
equations
+
we impose
y
+
cz + d
the given
= o.)
(66))
points, this time on (66),
system)
1+2c+d=0
a-1-2c+d=0 2a + 2 + 4c +
(67))))
d =
o.)
Adding the first two
equations of
4a +
=
3d
68 and 69 now form two unknowns, the solution of in (67) permits us to back tuting
Equations
we put these values in -
y
the
(Actually,
=
\037z
However,
find
find
or)
in two equations a = d = o. Substithat c = -\037. Therefore the desired plane to be) is
could have
computation
been
have
might
-
2y
had set d
this
(69))
a system of which
(66) and 0)
yields)
O.)
= 0 after the set and then a = 0 after inconsistency if we
what
(68))
to the third
second
the
gives
(67)
2d = 0,
a +
and adding twice
GEOMETRY)
VECTOR
ELEMENTARY
118)
z
=
O.)
been
of the
first
inconsistency.
too far and
jumping
some-
simplified
appearance the second
too fast
for
an illustration.))
These cumbersome methods of finding a plane through three given will points by a far more simpleand elegantvector next chapter.)
24. DISTANCEFROM A The problem of determining
point Po =
(xo,
zo)
Yo,
ax + is
handled
problem Call
=
8 the
(x,y,z)
distance
be an
Figure
superseded in
approach
A PLANE from
distance
the
a given
given plane) cz +
d =
0)
earlier
the
to a linein
from Po to the given plane,and arbitrary point of) ax
(see
+
of
equation
be
in precisely the same manner as was of finding the distance from a point
the plane. P
the
to a by
TO
POINT
the
66).
vector perpendicular
+
Then
by +
8=
cz + d =
let
0)
--7 \\prNPopI,
where
to the plane. Let us-as
N before-)))
is
a
GEOMETRY)
ANALYTIC
119) Po)
FIGURE
use a
unit vector for N.
Then)
V
the
determine
+
bj +
2 a
b2
ai
N= To
66)
+
ck + c
\302\267 2)
projection we resort
desired
to
inner
product)
--7
\037
=
\302\267
IPop
Nt
= 8.)
IprNPoP(
Then)
8=
[(x -
+
xo)i
(y
- yo)j
+ (z -
ai
\302\267 Zo )k]
V
-
But
8
--
ax
-
axo +
-
by
2 V a2 + b
ax + I-axo
by +
cz =
- byo 2 v' a + br
cz -
byo + +
- d. -
cZo
+ c
2
c
bj +
b2
ck
+
2 a
+
2) + c
CZo
2)
Therefore,
dl -- taxo + byo 2 V a +
cZo +
+
b
2
+
c
2)))
dl
\302\267
EXAMPLE 31.
to
the
plane
equation
-
x to our
Conforming
the
-
2y
from Po
distance is)
the
find
We
whose
2z +
1=
policy of minimal
GEOMETRY
VECTOR
ELEMENTARY
120
= (1, -2,
O.)
we ignore
memorization,
derived above and instead reason
formula
-3)
from
first
principles. \037
the projection
seek
we
Since
of PoP on the perpendicular,
P is any point of the given where we shall find some plane, particular P on x - 2y - 2z + 1 = 0, with which to work. in simple result One choice for P that would can computation = = = = z x -1 be found o. P and y Then, by letting ( -1, 0, 0). Therefore) \037
8 =
IprN PoPl
[(-1 -
-
l)i +
(0
-
(-2))j
+ (0 .
(-3))k]) i - - 2k 2j
V1
2
-
- 4
1-2
-
61
3
----12
-3-.)
+
22 +
2 2)
4
EXERCISES
1. and
(a) x + (b) x -
-
(c) 2x =
2y
=
z y
(b) (c)
(d)
(e)
3; +
2z +
9 =
0;
-3.
the equation of the plane (1, 1, 1), (3, -2, 1),and (2, -4, 3); (1, -1,0); parallel to -x + 2y - z = 5 and through perpendicular to 2i - j - k and th\037ough the origin; the origin perpendicular to 2x + 3y - 6z = 12, through and (2, 1, -4). through (-5, 0, 8) and perpendicular to a vector whose directionnumbers are given by the set {4, -3, 12}.)))
2. Find (a)
their intercepts
in the
traces
(d) y
following planes, determining coordinate planes: - z = 3;
the
Sketch
through
121)
GEOMETRY)
ANALYTIC
z)
ce)
y)
\037)
67)
FIGURE
the distance from (3, 0, (a) 3x + 4y - 12z = 52, (b) 2x - y + 2z + 81 = 0, (c) 4x - 3y = 100.)
3. Find
25. THE
IN
LINE
STRAIGHT
Our discussion of
pattern as the
attention
was
Applying collinear
equation
confined
a line
to
THREE
DIMENSIONS
the straight linein
the same
by specifying and seek an P = Let
-2)
\302\243 by
space
shall
discussionof the line
to the two
plane. of its
We
therefore
points PI
=
follow our
when
begin (Xl,
YI,
Zl)
analytic representation of \302\243. (x, y, z) represent the general point of \302\243. 4 to the position vectors of the three Theorem points (see Figure 67), we may write the vector
of
\302\243 as)
P
=
(1
- t)PI + tP2 .)
(70))))
In
xi +
- t)xli +
= (1
zk
+
yj
= [Xl +
Therefore)
= =
Z
is the
txli +
+
-
t(Y2
Zl +
t(Z2
-
t(Y2
YI)]j
- zl)]k.)
t(Z2
- Xl)
t(X2
YI +
tZ2k
+
tY2j
YI)
-
(71))
Zl).)
in of line \302\243
form.
analytic
of the parameter t can be accomplished to that in the discussion the of analogous in We for solve t the plane. parametric
manner in
t)Ylj
xI)]i + [YI + + [Zl +
representation
parametric Elimina tion a
-
Xl
(1 -
- t)zlk +
(1
t(X2
x= Y
line
GEOMETRY)
basis vectors, (70) becomes)
of the
terms
+
in
VECTOR
ELEMENTARY
122)
the
getting)
form,
-
X
-
X2
Thus we
Xl Xl
-
Y
=
YI
-
Y2
Z
=
-
Z2
YI
Zl
-
= t.
Zl
devoid may write the equationof \302\243 X
-
X2
which may
Xl)
-
-
Xl
Y
Y2
be called the
-
-
YI
-
Z
-
Z2
YI
-
Zl
,
(72))
Zl line
the
of
form
two-point
t as)
of
in
three dimensions. a The reader that observe (72), which describes locus in termsof coordinates a11dno variable, is not a single equation. It actually consistsof three will
auxiliary
equations,of X X2
-
Xl
Xl
-
Y
Y2
-
-
are
two
which
YI YI
,
X X2
-
Xl
Z
-
Z2
Xl
and
Y Y2
It
may
necessary
appear strange to describe a
to deduce
sufficient
-
-
Zl
-
Zl
,
YI . YI
the third:)
Z
Z2
-
-
Zl
(73) Zl)
that more than one equatio\037 locus, but a moment'sreflection)))
is
123)
GEOMETRY)
ANALYTIC
it appear quite reasonable. For when we contwo simultaneous linear equations in plane that they defined a point, we stated analytic geometry, two the pointof intersection of lines (if such existed). Each of the equations in (73) is linear and therefore a plane. Two of these considered simultaneously defin\037s consistof the locuscommon to the a line. two, namely, That is, the intersectionof two planes is a line. Each
makes
sidered
plane in (73) is
but
such
to
define
parallelto oneof the coordinate
axes,
In fact,
a line.
be the
need not
planes
special
two
any
only
simultaneous
ones
used
linear
equations)
a1x + { a2X
unless
line
a
define
planes.
have
shall
We
+
sentationsof lines.
b
+
1y
CIZ +
+ C2Z
b 2y
+
d1 = d2
=
0 (74)) 0)
the equations represent parallel more to say later on such repre-
\037
P 1P
P IP
vector
The \037
2 =
(X2
fore has a
-
2
is parallel
that
x1)i +
(Y2
-
Yl)j +
set of direction
numbers
to (or (Z2
{X2
is along) line \302\243
zl)k and thereYl, Xl, Y2
If the concept of di,.ectionnumbers is applied Zl}. that to lines (see Figure 68), we see the denominators the form (72) are precisely the directionnumtwo-point bers Given a set of directionnumbersfor of the line. direction for the line can be found by equations cosines
Z2
-
in
\302\243,
58.
with direction l1l1mbers {ll, m1,nl} is direction 11l1mbers {l2, m2, n2}. \302\2431 = k if if vectors VI to \302\2432 and only lli + m1j parallel = and .This V 2 latter l2i + m2j + n2\037 are parallel. if a condition is satisfied if and V is 1 (nonzero) only Let
and
be \302\2431
a line \302\2432
a
line
with
+ n1
multiple
of
V 2,
say)
V1 =
tV
2,)))
GEOMETRY)
VECTOR
ELEMENTARY
124) z)
,\302\243)
Pl(Xl,Y1,Zl)) Y)
x)
68)
FIGURE
case II
in which
=
=
ml
tl2,
II _-_
m1
l2
m2
-
Hence and
a necessary be
to \302\2432
and
parallel
tm2,
and n1
=
or)
tn2
--_en1
(75))
n2) condition
sufficient
for
their direction
is that
lines
\302\2431
numbers
be
proportional.)
26.
BETWEEN
ANGLE
TWO LINES
The angle betweentwo lines between two vectors, one parallel lines. As a consequence of this of the angle betweentwo lines
is
as the angle of the given we may speak
defi11ed each
to
definition, even
if the
lines do
not
intersect.
Consideringthe lines
and \302\2431
we shall
derive a formula for
between
and \302\2431
the
lines.)))
in \302\2432
terms
as \302\2432
.specified
the cosine
of the
of
direction
the
earlier, (J
angle
numbers
of
ANALYTIC
GEOMETRY)
125)
The angle (J is the angle between V I and V lli + mIj + n1kand V 2 = l2 i + m2j + n2k. scalar products, we have)
where
2,
VI =
COB0
=
V I
.V 2
\302\267
V2
VI
=
Iv11lv
VV
2 1)
,
\302\267
I
Using
VI
V' V 2
\302\267 V 2)
or)
+
lIl2
(J =
COB
Vl
I
2
+ mI
2
mlm2
+ nI
2
V
+ nIn2 12
2
+
as well,
given direction numbersare (76) simplifies to become
cos
cos al
If the
(J
=
cos a2
+
COS
2
+
direction
n2
(76)
2
cosines
{3I COS {32
+
As a
\302\267
m2
COS'Y
corollary, we may concludethat
I COS 'Y2.
the
two
(77) are)
lines
z)
\302\2431) , .)
VI) y)
x)
FIGURE
69)))
if and
perpendicular
PI
=
+ nln2
o.)
(78))
the equation on the line
Find
32.
-1,3) tOP 2
= (2,
if
mlm2
+
lll2
EXAMPLE
only
GEOMETRY)
VECTOR
ELEMENTARY
126)
=
\302\243 joining
-5).
(1,0,
Form. From (71) '\037le have)
(a) Parametric
X = \302\243:
(1 -
2 +
2)t
y=-l+(O-(-l))t
{ z
= 3
+ (-5
-
3)t.)
+
t
yields)
Simplifying
X=2-t \302\243:
=
y
-1
{ z = that
We
see
A
set 0f
3
..
..
d lrectlon COSInes
_
x-2
V
- y+1.
-1
.
.
(a) which
line
x-I We
set 2
can
be restated
oC
=
of (72)'we
of
Example
y+3
= z
-1
as a
_
V
8 /-
66)
\302\243.)
\302\267
}
have)
z-3
-8)
-
1.
2
of \302\243';(b)
representation
and \302\243'
between
V 66
66
-
X
I-='
_
1.)
01 EXAMPLE 33\302\267a 1ven 1Ine d...J: a parametric
1
/_'
for
numbers
direction
1
-
IS
By means
form.
Two-point
8t.)
1, 8} is a set of
{-I,
{
(b)
-
=
the
y
+ 3 _ 1 = z,
cosine
fi
of the
n d
( a) )
angle
32.)
=
t ')
parametricrepresentationof
x=I+2t
y =
-3 -
z =
t.)))
t
\302\243')
(J
(b)
127)
GEOMETRY)
ANALYTIC set
A
therefore
of
apply
cos O =
numbers for which (76), gives)
direction
\302\243' is
Y(-1)2 + 1 + 2
Since
\302\243\" is
to
parallel
Z
zt,
where
\037
is given
as
a point
x-I are
-1,
+ 1
2)
I} .
-1(2)+1(-1)+8.1 82
2
y2
+
(-1)2
=y=
-1
(Xl, YI,
of
Zl) is
\302\243\".
to
of the
\302\243\" is \302\243,
We
5
EXAMPLE 34. Find the line\302\243\" parallel and passing through (1,'0, -2).
=
{ 2,
the
6Y
=t
\302\243\".
32)
Example
x
form
a point of
Therefore
\302\243 (of
.
ll)
But
l=Y
YI \037
(1,
0,
of the
equations
-2)
line
z+2 8
Thus far we have ignoredthe possibility one of the of twothe case numbers being zero, in which point form (72) could not apply becauseof the illegality of division Before proceeding with the algebraic by zero. us determine let the geometric meanproblem involved, direction
This number zero. direction ing of, say, the being = = a would mean that cosa 1r/2. implies 0, The line in question wo1ildthereforebe to the be or the line x-axis; equivalently stated, a to (or actually in) the yz-plane. In parallel line is parallel with a set of direction numbers m, n} a that to the yz-plane. Similarly, the can show line with a set of direction is parallel {I, 0, n} to the xz-pla11e; a set of direction numbers and a line of the direcIf two {I, m, O} is parallel to the to one of tion numbers are zero, line the is parallel the axes.) first
which
perpendicular
would
summary,
{O,
reader
numbers
with
xy-plane.
then
35.
EXAMPLE
and (3, 2, A
set
of direction
that the
Find
equations
-2). two-point
numbers form
of the
line through
is easily found (72)
cannot
to
be
be used.
(1,
2,
-1)
-1}, so However, a)))
{2, 0,
GEOMETRY)
VECTOR
ELEMENTARY
128)
be written
parametric form of the line can x=I+2t
y=2
-1 -
z = the
of
Elimination
yields
parameter
= Z+l
X-l
-1
2 { is a
which we
nonparametric
to writing the these representations z are. That x and
both
2,
line
the
of
form
come
may
from
=
y
t.
two-point that
and is It
form.
=
y
2 no
as closeas to see
is easy
matter
what the
is, the line must be in the plane = 2 and y is, therefore, parallel to the xz-plane. EXAMPLE 36. the equations Find for the line through with direction numbers given by the set to,4,OJ. (1, -1,2), An set of direction numbers is to, 1,O};indeed, equivalent to, m, O} (when m \037 0) is an equivalent set of directionnumbers. We may thus write a parametric form as: values of
x= =
y
Actually,
the line
the equation
is
value,
z =
is what
which
-1
z =
2.
=
-1
y
+
+
t
t
is
superfluous,
quite
the
as
determined
completely
planes x = 1 and
1)
intersection
for
of the
2. The y-coordinate may take on may happen in the foregoing parametric
of the line. We now return to the
any
form
section
of
course,
been
two
35, (79)is the it.
Let us
EXAMPLE
.
a line as the interequation 74). This idea has, of for instance, in Example throughout; of a line in terms of two planes through
representationof
(see
planes
employed equation
examine a more generalcase. 37.
a line
Given
\302\243 whose
IS)
2x
+ 3y x
-
y
algebraic
-
z
+
z =
=
1 2,)))
representation)
129)
GEOMETRY)
ANALYTIC
exhibit a method finding
we shall
for the
forms
other
equations
of aC. adding the two
z by
Eliminating
3x + the
x from
Eliminating
y
-
=
3z
3z-3
-2 =y=
which
yields
-3.
3x-3
and
3.
allows us to write
of these
each
in
=
2y
equations
given
5y
Solving for
equations gives
5)
may now be put in the form (72) by dividing denominator of the first and third membersby
x-I
=
-(2/3) a set
Therefore
y
3:
z-1 (80)) 5/3.)
of directionnumbers
By setting) x-I
=
-(2/3) we arrive at a
=
numerator
=
y
is
{
z-1
= t,
5/3
parametricrepresentationof x = 1 - it I
=
Y
z =
1, i}.
-i,
line
the
t
1 + it.)
EXERCISES
1.
as
equations, follows:
through (b) through (c) through (d) through
(a)
= 1;
(e)
through \037(two
(f)
1, -2)
(-4,
-2) (-4, 1, -2)
(-4,1, (3, 2, (3,
form, of the
in two-point
the
Find
mined
-
5)
2, -5)
and
and
parallel
and
(5,3,
and
(3,
lines deter-
to z-axis;
-1); 1, 2);
perpendicular
and having
to 3x
cos a =
-
\037and
2y
-
cos {3 =
possibilities);
through (-4, 3, 4, 12;
(g) parallelto 5i -
1, -2) 4j
+
and
having
6k and through
direction
the
origin.)))
6z
numbers
VECTOR
ELEMENTARY
130)
2. Find
of parametric representations
lines
the
GEOMETRY)
1.
in Exercise
Prove that 3. Let AB, AC, and AD be three edges of a cube. plane through D and the line joining the midpointsof AB and AC is tangent to the sphereinscribedin the cube. 4. (a) Find a set of direction numbers for the line) -
X
x-
{
= 0
3y + 7 -
2z
4 = o.)
(b) Find a parametricrepresentationfor the line of part (a). of the form 5. (a) Find equations (72) that represent
3x -
6z =
-
2y
-9
x+4y-8z=-16.) (b)
6.
a parametric
Find
Prove that the
(a) (0, 7, 6), (b)
7. Three (0,6, 0), 8.
Find
a
of
the direction
taneously,
triangle:
Find the
cosines of
a
vertex.
fourth
line
simul-
perpendicular,
to)
--
x-I
y-l
--
z-l
9. Find the cosine of
and
6
3
2
Exercise
(a).
part
are (4, 3, 5),
ill order
parallelogram
1, 4).
(-8,
of a right
-2), (2,4,1).
-4,
(-3,
of the line of
(6, 4, 8);
1, 3),
vertices
and
vertices
are
following
(-2,
-2,4),
(0,
representation
the
angles
made
x- -----. Y
3
4
z
12)
by the two
lines
of
8.)
27. INTERSECTION
OF
A
LINE
WITH
A PLANE
of line may be representedby the equations there two planes through it, the point of intersection (if a plane is one) of the line with may be found by solving
Since a
the three
linear
illustrate utilizing
equations
simultaneously.
However,
we
another method for solving this problemby of the line.))) the parametric representation
AN AL
YTIC
131)
GEOMETRY)
EXAMPLE 38. Find the point of 2x +
intersection
z =
-
3y
line
of the
1
\302\243:
{ and the
plane)
(j-, 1,I) native
II: x + of point we use
the
as
approach,
2y
-
2z =
-3.
linear
three
these
Solving
x-y+z=2)
equations simultaneously intersection. However, for the the parametric form of \302\243:
yields
alter-
x=l-jt
y=t .z which
found
was
= 1+
in Example
-it,
37.
in the equation Substitutingtheseequations (1 - it) + 2t - 2(1 + it) = -3, the solution
give the that
value of
is also
the point we
on II seek
the
parameter
of
by x =
This
j-, y
FIGURE
70)))
will
yields the point of \302\243 value is t = 1. Thus, = -i.) = z 1,
y)
%)
II gives
which
t, which
(see Figure70). is given
of plane
VECTOR
ELEMENTARY
132)
GEOMETRY)
AND 28. ANGLE BETWEEN A LINE A PLANE How can we extend our understandingof the angle of a pair of lines to define the concept of intersection
angle
a
between
and
line
If
a plane?
\302\243 is
the
line
of
the
and II plane II
in the of lines \302\24311\" the plane, there are an infinity that might conceivably be usedto find an angle between II (see Figure and \302\243 of choices 71). Since such an infinity for the angle between a line and a plane is undesirable, the one usually selected as the of intersection angle II is the minimum and between \302\243 of the {3 infinity angle
formed
by
the
and \302\243
lines
a defini-
such
\302\24311'. However,
difficultthe problemof finding {3 extremely to turn at this We therefore point. impossible indeed, between the more convenient definition: The angle {3 II is defined to be the complement line \302\243 and of the plane a line m perpendicular \302\243 and to II.) between angle
tion renders
EXAMPLE
line \302\243 and A
vector
pendicular
plane
parallel
to plane
angle of intersection between the Example 38. is L = -fi \302\243 + j + j-k. A vector perthe
Find
39.
the
II of to
II is N
=
i +
2j
- 2k.
ffi,)
FIGURE
71)))
Thus
the
comple-)
GEOMETRY
ANALYTIC
be'found by dot products)
of (j may
ment
r _
cos
is
+ 2
= \\-2/3
(3
( 2 which
133)
(j =
to sin
equivalent
=
\037O/3\\
2_
,
y39)
y9
09/9
)
-
2/09.)
EXERCISES
1. In
the
to take of Example 39, we were careful value of the numerator in the determination of
computation
absolute
the
-
COB
Why?
(3).
(;
coming, which the tion of the definition. 2.
a general
Find
our definition
that
indicates
This
reader
3.
two
Using
different
of intersection
4. Find the anglebetween 3 above.
5.
Find
of the
the
a representation)
y+2
z-3
-1)
the
6.
lOx
(a)
planes.)))
Give
line
z =
-
2y
and plane
a suitable
+ 2y
3)
6)
- l1z =
3.)
definition
of
the
5.
given in Exercise intersection
z+3
y+3
2)
and)
is 3x +
of intersection
x+3
angle
3)
and the angle of whose equationsare) plane
point
line and
the
coordinates
with
whose equation
plane
{3 is
where
find the
methods,
2)
the
shorta slight modificaa small by has
plane.
of line, x-3
and
{3
formula for the sin {3,
formed by a line meeting a point
of
eliminate
should
for
the
angle
between
two
(b) Find a generalformula whose equationsare)
and)
(Hint. Use two planes.
alx
+ b1y
a2X
+
b 2y
+
CIZ
+ C2Z
each
vectors,
angle
between two planes,
+
d1 =
0
+
d2
the
for
GEOMETRY)
VECTOR
ELEMENTARY\"
134)
= o.) to one
perpendicular
of the given
)
of the planes (c) Find the angle made by the intersection whose equationsare2x + 2y + z = 1 and 2x + 10y - llz = 3. the two planes whose equations (d) Find the anglebetween are x - 4y + 8z = 2 and 2x + y - 2z = 18.
7.
Show
that
line given
the
X
-
{x (a)
is parallel
= 0,
fundamental
Apply
3y + 7 2z
=
plane whose vector
= 0
4,)
to the plane whose
(b) lies in the
8.
by
equation
is 3x
equation
is 12x
techniques
+ 4y +
to
problems.
(a) Find the angle between
of
its
the
diagonal
-
of a
6y
-
2z
6z = 1.
the following cube and one
edges.
the angle between the diagonal of (b) Find diagonal of one of its faces. of a (c) Find the angle betweenthe diagonal
of its faces.)))
a
cube
cube
and
a
and one
cross
products)
29.
CROSS
PRODUCTS
As we
have seen,
the inner product of
vectors
two
yields
a scalar quantity, hence the synonym scalar product. In the present section we introduce another kind which leads to (1) a vector of vectors, multiplication and a relation of the angle quantity (2) involving the si11e vectors
two
between
(as opposed
the cosine dot products). has gainedsome the liberty of a longer in
security
Since, in
dealing
excursion
of
to the appearance of by now, the reader with vectors, we take
through
the
algebra
of
before illustrating the theory. two Let A and B be any product. and call 8 the smaller angle between them. A X B, is a cross product of A and B, denoted
cross products vectors, Then
cross
of
Definition
the
vector)
IAIIBI
where U is
a
unit
vector
sin
8 U,)
perpendicular 135)))
(81)) to
the
plane of
GEOMETRY)
VECTOR
ELEMENTARY
136)
AxB)
B\037
U
\037\037 A
B
U \037\037 A
\037)
\037)
BxA)
FIGURE
72)
as to
and pointed in such direction a triple. That is, if {A, B, U} right-handed who walks from the terminus of A to the and
A
B,
B through the emanate from a
then he is always B,
{A,
being
X B}
vectors
three
the
A
upright
standing
and lengthsof Because
are
vectors
the cross
right-handed triple if as
<
vector
popular-terminologies.
immediate product
corollary defines
of the
X B (if
A
scalars. a
vector
We often termed the vector product. our earlier policy of utilizing both names reader will become familiar with all the
An
of
none
nonThis assertion U. of U in (81) can1 when 0 < 0 < 1r,
non-negative
product results in
it is
equally
left,
with his the direction of U.
are the zero vector,for
direction null) points in the same follows from the factthat the coefficient not be negative because0 < sin0
of both
to U,
parallel
precisely
form a
to his
0
has
always
terminus vectors
two
the
(when
direction\"
\"\037pward
Thus,
angle 0 singlepoint 0)
make
observer
the
quantity,
shall so
continue that
the
standard-and
definition is that the
a noncommutative
multiplication.)))
CROSSPRODUCTS) In
137)
the
we have
fact,
Theorem 8.
A
This result
B
X
follows
- B X A.
=
the
from
simply
quite
observation
that B X A has the same magnitude as A X B but points in the opposite direction (seeFigure 72). the word \"smaller\" were left out of Query: Suppose the definition of vector Would the resulting product. to the be definition equivalent original one?
Attempting to gain further physical or geometric of cross product, we observethat) into the notion insight A
for
if
A
and
is parallel
B are
(Is the converse
of
to B
implies
B
X
=
(82)
0;
then 0 = 0 and sin (J
parallel,
A geometric
true?)
(82)
A
=
o.
interpreta-
tion of the magnitude of the crossproductis given by Theorem 9. The area of the parallelogram by generated vectors A and B is equalto IA X B \\. The area of the paral73.) Proof. (See Figure A B and is equal to) lelogramgenerated by IAlh
30. TRIPLE
=
IAIIBI
=
sinO
IA
X
BI.)
SCALAR PRODUCT
The product A. B X C is called the triple scalar product of A, B, and C. N.B. (1) Although parenthesesmight appearneces-
sary-at leastto assist the
operatioll it should
beginner-to
is performed (the \"cross\"or the \"dot\") be noted that such parelltheseswould
\037h)
)
first,
)) A)
73)))
com-)
be
hl)r\\
A)
FIGURE
which
indicate
GEOMETRY)
VECTOR
ELEMENTARY
138)
I I I
\037-----------------
\"
\"
/ //B
C)
FIGURE
order for the
two
\"Cross\"
operations.
we would have
\"dot\"; otherwise
one possiblelogical
is only
there
for
superfluous,
pletely
74)
a
vector, which has no meaningfor A X B \302\267 C also The product (2)
must
precede with a
scalar
crossed
yields
a scalar
scalar
product
us.
quan-
of tity and thus is alsocalledthe triplescalar product A, B, and C (in that order). Callingboth quantities, A
\302\267
B
X
B
A X
C and
\302\267
C,
the
triple
be reasonable and warrantedonly so they are, as we shall equal-and 10. The magnitude of A Theorem volume V of A,
B,
the
parallelepiped
(See Figure
Proof.
IB X ci = IAI
V
IA.
B
=
a
cos
Therefore + =
were
they
always
see.
generated
the vectors the by
(Theorem
9)
\302\267 B
X
C represents
C.
and
and
if
soon
would
V
xci.
=
74.))
area
of
base
+ altitude IAIIB
X CI
of parallelepiped. cos a, which
states that
The reader should investigatethe questionof sign by the following answering question: What kind of orientaof A, B, and C leads to a positivetriple scalar tion product? kind to a l1egative triple scalar product?))) What
CROSS
PRODUCTS
139
In which case does the triple scalar - V? and in which case does it equal
V
equal
product
A. B X C = A X B.C. Since both triple scalar products represent Proof. the same and since the sign dependsonly on volume, of the the two orien\037ation triple {A, B, C}, products Corollary.
the
be equal.
must
-Our final result on the algebra of embodied in the following theorem.
Theorem 11.
vector
The
(i)
(ii) Proof of D Our
proof
A
X
C) =
+
(B
+
(A
distributive
is
product
is
products with
That is,)
to addition.
respect
vector
X C
B)
=
A
X
B
+
A
X
C
+
A X B X
C, and C.
A
X
B
-
X
Let
(i).
-
= A X
(B + C)
will be
completed
C.)
D is
that
show
can
we
if
A
necessarily the zerovector. We
v
scalar
the
take
with an
of D
product
arbitrary
V.)
vector
\302\267 =
V
\302\267
= V
\302\267
D
X
[A
A
X
C)
-
C)
-
(B + (B +
A
V
X
B
\302\267
A
X
-
A
X
B
-
V
C] \302\267
A
X
C
(Theorem
=
V
X
A
\302\267
(B
+
C)
-
V
X
\302\267 B
A
(Corollary of Now, using the distributivity first term of the right member,we
V.D
dot
-
C V X A \302\267 to Theorem 10).
on the
product!
get)
-VXA.B
=VXA.B+VXA.C
7)
-
V
X
A
\302\267
C.)
Hence)
v
\302\267
D
=
O.)
(83))))
VECTOR
ELEMENTARY
140)
V is
that
Recalling
statesthat the
Consequently, D = 0 Part (ii) is left as
an
Theorem8.
from (i) and
Weare
in
now
A
in terms
our
of
i i
X
j
and imposing
=
are
we
First,
k}-basis.
kXk=
j =
(81), we
O.
(84)
further see that
k = i, and k
j X
k,
=
i
X
Theorem 8 on the relationsof k X j = -i,
= -k,
j X i
j X
definition
the
Applying
=
i
X
develop that
vectors
two
{i, j,
note that
readily
tools to
of sufficient of
represented
follows
it
for
a formula for the crossproduct
C.)
A X
B +
X
exercise,
possession
vector.
have)
we
and
that (83)
to any
is perpendicular
C) =
(B +
A X
we see then
arbitrary, D
vector
GEOMETRY)
i
and
yields
(85)
k =
X
(85)
j,
-j. (86)
Now let) A
=
a1i +
a2j +
ask
B =
and)
b1i +
b
2j
+
bsk.)
Then A
X
B
=
a2j + ask) X
(a1i +
= (a1i +
a2j
+
ask)
+
(b1i
b 2j +
bsk)
X b1i
+ (a1i + a2j + +
X
ask)
b 2j
(a1i + a2j
+ ask) X
b 3k)
= a1i
X b1i + a2j X b1i + ask X b1i + a1i X b 2j + a2j X b2 j + ask X b 2j + a1i X bsk + a2j X bsk + ask X
By using to obtain)
A X
B =
(84),
(a2bS
(85), and -
a Sb 2)i
(86), we simplify + (aSbl
this
bsk.
expansion
- a1bs)j
+
(a
1b 2
- a 2 b 1 )k.
(87))))
141)
PRODUCTS)
CROSS
reader who is familiar with determinants to it is see how convenient pleased express for A X B in the language of determinants. The
(87)
be
will
to
formula
the 1
Equation
states)
A. X or, more
B = i a2 a3 b2
-j
b3
simply
ple 28,
b3
of
k
a2
a3 .
b
b2
b 3)
a vector
finding
b1
a2 b2
(88)
of the use
this conceptto the
to apply
namely, that
J
i
al
+k
al
illustration
first
Our
40.
will be
products
a3
b1
1. . AXB=
EXAMPLE
al
problem
of
cross
of Exam-
to the
perpendicular
- 2 = O. plane equation y the cross product of two vectors is a vector perpenSince dicular to the determine plane of the given two, we need only in the plane 2x + y - 2z - 2 = 0 and take their two vectors cross product. We therefore select three points, arbitrarily in the given plane: A = (0, 0, -1), B = (0,2,0), C = (1, 0, 0). \037 \037 AB = 2j + k and AC = i + k. Then is 2x
whose
-
+
2z
Hence) \037
AB
\037
X AC
i
j
k
= 0 2 1 = 2i +
j-
2k
101)
vector perpendicularto the plane of reassuringto note that this answer agrees
is a
in our
first
Although
solution
to
the area
this of
B,
with
C. (It is the one found
and
problem).) a
simple theoretical task to compute cumbersome when given the coordinates the power of vector products
triangle
a problem, it is quite often an area by the usual formulas of the vertices. We exhibit in attacking such a problem 1
A,
is a
as)
For the reader who with determinants, is unfamiliar brief word of assistance in the appendix.)))
\\ve
have
a
ELEMENTARY VECTOR GEOMETRY)
142)
A)
FIGURE
A
-5, 2), B =
= (1,
---7
=
AC
X
3k
-3 4
---7
---7
Since
lAB X ACI
Figure 75),
three
the shaded Therefore)
of
our
earlier
\037
AC
5
C
and
=
(5, 0,
2).)
whose
i-115i-
= 15i-
desire is
12j+
the
of
are A,
vertices
work we
+ 5j.)
4i
12j
+
17k.
0)
equals the area
triangle we
K = In
of
and
j k 8 -3
i
---7
---7 AB
(-2, 3, -1),
by)
---7
8j -
Then AB = 3i +
be given
vertices of a triangle
Let the
41.
EXAMPLE
75)
precisely
=
17kl
(see
parallelogram
B, and
i V
the areaK C,---7 ---7
\037 IAB
X
ACI.
658 .)
methods, solved, by analytic the equation of a plane The vector approachto
of determining problem three through given points. the
thisproblemis
demonstrated
in
the
next
example.)
A = (1, 2,3), EXAMPLE 42. Find the plane through B = (1, -1, 0), and C = (2, -3, -4). (The analytic solution The vector to this problem was given in Example 29.) ---7 ---7 AB X AC is perpendicularto the desired plane and is therefore to every vector in the desired plane (see))) perpendicular
143)
PRODUCTS)
CROSS
Hence P =
Figure 76). if and
\037
\037
Writing
- l)i + In terms of [(x
(y
condition +
explicitly
+ zk] \302\267 (-3j
l)j
x-I
y+1
-5)
1
dot
of
31.
we
product,
the
determined
plane
DISTANCE The
-2x
distance
by
FROM
from
+
y
A,
5j
-
7k) =
o.
becomes) = O.,
-3
-7)
formulas for cross product - z + 3 = 0 as the equation
of the
means
get
X (i -
z
-3
o
gives
- 3k)
relation
this
determinants,
or, expandingby
point in the plane
X AC = o.
only if BP. AB this
a general
z) is
y, \037
(x,
B,
and
and
C.)
A POINT TO A PLANE a point to a planewas found
by
pro-
unit perpendicularto the plane. But now this for by means of cross may be facilitated, procedure a planemay be found to the unit perpendicular products, of the from three directly points plane without any need for determining the equation of the plane. EXAMPLE a plane be determined by the points 43. Let B = (-1,1, -1), and C = (2,1, i); and let A = (-3,0,1), P = (1,- 2, - 3) . We shall determine the distance d from jecting
Po
on a
to the
plane of
Example31,which
A,
is
and
B,
the
\037
C.
present
AB x
(The reader is referred problem in disguise).)
\037
AC)
P(x,y,z)) A)
B)
FIGURE
76)))
to
VECTOR
ELEMENTARY
144)
GEOMETRY)
\037\037 X AC)
AB
----
D
i d
---:-\\
1 ;:\\\\\037)
B)
FIGURE
A
vector
77)
to the
given plane may be found by \037 AB X AC (see Figure 77, which
perpendicular
\037
cross product the geometry)
the
taking
symbolizes
\037
\037
AB X AC a unit
Then
i
j
5
1
X
lAB
with
(Compare
=
Example
=
3(ti - j
- k).
is)
.!i-j-k v i + 1+
] 3 1 =!(i-2.-2k)
31.))
= IPoB.
+ 3j
NI
+ 2k) --l(i
- 2j -
2k)1
=
t(2
+ 6
+ 4)
4.)
DISTANCE
32.
3k =
\037
IprN PoBI 1(-2i
= ACI
\037
=
-
3j
\037
ABXAC
N=
-
\037
perpendicular \037
d
k
= 2 1 0 = ji
BETWEEN
TWO
LINES
We now direct attentiontooneofthenastiestproblems
of
elementary
analytic
geometry,
namely,
the problem
of)))
145)
PRODUCTS)
CROSS
space.
to this
refer
We
is usually
student
beginning
means
is the
geometry
analytic
pure
between two nonintersectinglinesof as a nasty problem because,if
the distance
finding
plagued
by
of
the
analysis, the
of
problem
is required to \"see\"the derivation the and then he is plagued by another complifQrmula, formula to memorize. Once again, the vector cated that
visualization
both
with
assists
approach
and If two lines \302\2431
of
difficulties.
are \302\2432
the
given,
minimum
distance
and d between them is the distancebetween \302\2432 \302\2431 along in mutual the perpendicular (QR Figure 78). Thus, A and B points of \302\2431 and we call \302\2432, respectively,)
if
\037
=
d
(Readers who are encouraged The
jection.)
EXAMPLE =
(1, 0,
point of
proin
the
44. Let \302\2431= AC, where A = (2, -1, 3) and and let \302\2432= BD, where B = (1, 3, 0) and
-5);
D = (3, -4, and \302\2431
this have difficulty understanding to check back on the definition procedure for computing d is given
example.
following
C
IprQRABI.)
1).
We
shall
determine
\302\2432.)
FIGURE
78)))
the
distance
between
that is simultaneously
A vector
\037
= AC
N
to
perpendicular
i
\037
X BD
j
= -1
2
1
GEOMETRY)
VECTOR
ELEMENTARY
146)
k
= 55i -
- 8
and \302\2431
is) \302\2432
15j+ 5k.
1)
-7)
Then)
---7
d =
=)
ABI
\\prlf
N
ABo
1N !
- 3k). 5(11i- 3j 5 v'121 + 9 + 1
+ 4j
I(-i
-_
1-11 -.12 -
=
3\\
131
v'
+'k)1
26 .)
v' 131
EXERCISES
1.
the following
Evaluate
(a) i X (j
+ k),
+ j)
X k,
(i
(b)
(c) (i + j + k) X (d) (2i - 3j + k)
a tetrahedron, (b l , b 2 , b a ),
B=
k), - k),
(2j
(e) (-i + 2j + 4k) X 2. Using the fact that a tetrahedra of equal volume, of
j +
(i + X
(i +
2j +
parallelepiped determine
4k). can be sliced into a formula for the volume
six
the four vertices of which are A = (aI, a2 aa), = = C (dl, d2 , d a ). (CI, C2, ca), and D
the volume of a regular pyramid whosebaseis a square whose height is h. (A pyramid is called regular when its base is a regular polygon and the altitude from when
3.
Find
of
side
a and
the apex meetsthe base in its center.) 4. Prove that the answerto Exercise if the sec3 is unchanged ond condition of \"regular\" That is, a change in is omitted. the positionof the apex does not alter the volume of the pyramid as long as the apexremains at a fixed height h above the base.
5. Determine and
parallel
= 0,
the equation of to
V
=
li +
where P = (x,
y,
the
line
.\302\243 through
Po
mj + nk by observing that z) is any point of cC.)))
= (xo,
Yo,
zo)
\037
PoP
X
V
CROSS
PRODUCTS)
6. (a)
If P = (x,y, the
justify
point of
plane ABC, AB X AC = 0 as being a vector
an
is
z)
---?
\037
AP.
equation
arbitrary \037
(See Example 42.) of part (a), q.erive the equation = = (0, 1, 0), and A (1, 0, 0), B by
the
of
equation
147)
plane.
(b) Usingthe method determined
plane
(0, 0, 7.
distance
the
for
be
V
and
to plane
parallel
II2 .
to plane
parallel
and x =
x =
1
x-I (b)
=
Y
t,
-Y = z
=
2
2
x- = y+2 2 -1)
and
(c) line AB
=
Y
2t,
-
X
z =
-t
2t,
z =
assumption
4),
B)
\302\267
(C
= z.
CD, where
=
A
(0, 1,
2), B
D)
11. Find a set of of
the
and
a2X
cross
products.)
33. TRIPLE
b 2y
+
= (1, -1,
B, prove
that
= O.
direction
whose
planes +
0), and
(0, 2,
X
that
- t.
D ;= (-1, 2, 2). 10. If C is perpendicularto the plane of A and C =
Yare
Wand
while
+ 1
line
and
t, 1
III,
further Y) = o.
the
Under
III 1. 112 , prove (U X V) \302\267 (W 9. Find the distancebetween)
X
=
of this chapter, determine a vector from the origin to plane ABC.
methods
the
Using
8. Let U
(A
C
\037).
formula
(a)
the
of
C2Z
numbers
for the
equations are alX + d 2 = 0, by a
+
line of intersection + dl = 0 CIZ
bly +
direct
application
of
CROSS PRODUCTS
to more For the application of complex algebra problems of geometry and, particularly,to the developthe reader will see in the next of trigonometry-as vector
ment
chapter-we
shall
find
of three and morevectors.)))
\"it useful
to expand
cross products
VECTOR
ELEMENTARY
148)
GEOMETRY)
by obtaining an expansionfor the triple A X (B X C). Unfortunately, there is product and well-motivated development for su.cha no simple A of the efforts of several matheformula. summary We begin
vector
maticians tions
the
of
in a
appears
problem
VectorTriple
S. Klamkin
can Mathematical Monthly, December an approach, the virtue of which If we are given the three vectors
sider the three emanating
from
to bethe originofa Figure (i)
79)
(ii) the (iii)
the
system, It
with
follows
C =
adopt
A,
C, we conwhich is taken 0,
B,
and
coordinate
We
system. way
(see
:)
is taken
x-axis
the
1954).
system in the following
the coordinate
impose
(AmeriWe
is simplicity.
a point
rectangular
simplifica-
short paper, \"On the
Murrey
by
Product,\"
various
to
contributed
have
who
is
y-axis
from
c1i + C2j.
plane of Band C; so that the xyz-coordinate
in the
taken k} as a is
z-axis {i,
along B;
taken
j,
(i)
basis, is right-handed.
Since A
b1i, and from
B =
that is
free
from
C =
FIGURE
79)))
any
cli +
(ii)
stipulation,
C\037)
that it)
149
CROSS
PRODUCTS
must
be written
=
A
+ a2j
ali
1. . B
C =
X
+ a3k. Then)
J
k
bl
0)
o =
CI
C2
0)
blC2k,
and)
(B X C) =
A X
observe that C, that is,
We B
X
i
j
k)
al
Q.2
Q.:\037
o
0)
b
A
coordinate X
(B
must be
plane.
of be a linearcombination fore simplify (89) by factoring the it is exhibitedas a linear combination
X
Band
right of li
C. We thereso that member Band C. Thus)
- a l b1(cli
+
C2j).)
\\ve have)
Finally,
(B X C)
A X an
+ a2c2)b
= (alcl
X C)
(B
\037o
to (B X C) is perpendicular the cross Thus triple product, of Band C. in the Hence,
it must
A
perpendicular
the
must be
X C),
- alb1c2j. (89)
a2blC2i
1c2)
X C)
(B
X k.
vector
A
X
A
=
C)B
-
(A
\302\267
(90))
B)C,
triple cross product in
of the
expansion
= (A.
elementary
tern1S.
By using sho,v that
Theorem the triple
8 and (90), the reader cross product (A X B)
ca11
X C
easily
has the
expanSIon) X
(A
EXAMPLE C = -4i A
X
45.
+ k, (B X
X C
B)
Given
= (A. A
=
X
B)
- j,
and
,ve computeA X (B X C) and (A X B) X C. C) = (A. C)B - (A.B)C - 3)(i - j) - (2 + 3)(-4i + k) = (-8 = - IIi + IIj + 20i - 5k
= 9i + lIj -
(A
2i
C)B - (B \302\267 C)A. - 3j + k, B = i
X C = = =
5k
(A. C)B - (B \302\267 C)A -IIi + IIj - (-4)(2i - 3i - j + 4k.)))
- 3j +
k)
VECTOR
ELEMENTARY
150)
EXERCISES
1.
Verify
2.
If A,
B) X
(A X
that
B, and
C = (A. C)B
C are given
(B X C) X
in
as
(B
\302\267
C)A.
45, compute
Example B X
and
A
-
GEOMETRY)
(C X
A).
3. Prove the identities:) X
(A
B) X
D) =
(C X
(A
B
X
-
\302\267
D)C
(A X
B \302\267 C)D
(91))
and)
(A X 4. By
(A
(C X B) \302\267
D)
making use
of
\302\267 B
X
= (A. (91),
- (A. C)(B \302\267 D)(B \302\267 C) D)
show
that
(A X
B) X
(A
X
(92) C)
=
C)A.
5. Prove
Lagrange's identity: (a2 b s - a Sb2)2 + (aSb l - a 1bs)2 + (a1b 2 - a 2b1)2 = (a12 + a2 2 + as 2 )(b 1 2 + b 2 2 + b S2) - (a1b1 + a2 b 2 Use (92), which is, indeed,sometimesreferred (Hint. the Generalized Identity of Lagrange.)
6. If A,
B,
C,
and
D are (A X
coplanar\"
B) X
prove
that)
(C X D) = o.)))
+ a sb s )2.) to
as
trigonometry)
short
This
of
application
standard
the chapter is devoted to illustrating notions to the development of vector of
formulas
and
plane
trigonometry.
spherical
We shall see that spherical trigonometry, admits to a simple analysis in termsof vectors.) 34.
PLANE
Law where
TRIGONOMETRY
of cosines. = a, IAI IBI
Consider the triangle of Figure = b, and lei = c. It isclear C =
and
e
that)
\302\267 e
.
c2
gIves)
c2 of
sines.
=
(B 2 a
-
B
-
\302\267
(B
A)
2 + b
=
=
a
2
2 + b
Here we
-
2A
-
A)
\302\267 B.)
get
the
- 2ab cosI')
seek a relation involving 151)))
80,
that
A,)
the last dot product,we
When we expand law of cosines)
Law
particularly,
familiar
(93))
the
VECTOR
ELEMENTARY
152)
FIGURE
of
sides
angles.
the
triangle
We
therefore
80, and the sinesof its the
c X C = CX which implies)
80)
of Figure employ
o =C
(B
-
B
X
GEOMETRY)
cross product) -
A),)
C X
A,)
or)
A =
C X Equating (94)
the
CX
of left
magnitudes
B.)
(94))
and right membersof
yields)
sin {3 =
ca
which is
law
the
to
equivalent
a sin
By repeated the variables,
cb sin
a,
of sines b)
sin {3
a
application, or simply by the we have the completerelationship) b)
c)
sIn
c)
'
-.
. a)
sin
\037)))
SIn
'Y)
symmetry
of
(95))
153)
TRIGONOMETRY)
states that the sides of a triangleare proportional sines of their respective opposite angles.
which the
to Sum
and
difference
formulas.
sum
and
difference
formulas
requires somewhat
The usual treatment of in high school textbooks
are
fU:fther complicated
eral
cases, lie.
angles
vectors,
by
Once by
by the need to
again means
consider
sev-
quadrants in which the matters are greatly simplified of which all cases are treated the
on
depending
that
arguments
geometric
messy
simultaneously.
Let
Q and
R be
points on the unit
circle
origin)
as
of radius one centeredat 81, so
on a
=
R =
Q
\302\267 = R
cos
{3i
sin {3j
+
cos ai + sin aj. IQIIRI
cos (a
-
(3).)
:y)
R)
x)
FIGURE
81)))
circle
in Figure
that)
Q
Then,)
the
(i.e., shown
GEOMETRY)
VECTOR
ELEMENTARY
154)
Hence)
-
cos (a
=
(3)
COS{3
COS a
Once again, referringto Figure
i cos
j
sin
Thus.
which
{3
sin a
cos a
= (sin
0
sin
-
(a
(3)k.)
a cos {3
-
sin
{3
cos
a)k.
0)
- (3)k = .)
(a
write)
may
k
sin
{3
=
(3)k
(96)
this cross product becomes
of coordinates
terms
In
we
81,
= IQ11RIsin (a .-
Q X R
sin {3.
sin a
+
a
(sin
cos
- sin
{3
cos
{3
a)k,
implies sin
- (3) =
(a
sin a cos{3
-
sin
a.
cos
{3
(97)
EXERCISES
1. Using sin
(a
(96)
+ (3),
2.
By
as
requested
and
cos 2a,
methods, in Exercise 1.
3. Observewhat (96) and
cos ( -
0)
derive direc.tly
vector
using
(97). and
Area. The
for
formulas
deduce
(97),
sin 2a.
if {3
happens
- a
is chosenfor
the
area
the
for
formula
familiar
two
terms
+
(3),
the sameformulas expansions
Is the result consistentwith the facts sin 8 = - sin ( - 8) for all angles8?
of triangle of Figure 80 in included angle is an immediate
sides
cos 8
=
K of
the
and
the
of
consequence
cross
for
products,
K =
X
\037 \\A
K
Hence)
Since
cos (a
a triangle
sides,its
area
must
B\\
=
= lab
sin
is completely therefore
\037 \\A\\
\\Bl
sin
\"I.)
'Y.)
determined by its
be completely
terms of the sides. We shall now apply to the determinationof such a formula.)))
three
expressible in vector products
155)
TRIGONOMETRY)
K
=
2K
X
1A
BI.
product to expressthe length
the scalar
Using
Its
the triangle of Figure 80. K = ilA X BI, which implies
once more may be written
Consider area
A
of
X B,
we have)
4K 2
=
IA X
= (A.
=
B\\2
A)(B
B) \302\267 (A (A. B)2
(A X
-
\302\267
B)
= a b 2 -. (A . B)2 - A = + B)(ab
X
B)
(by (92))
2
derivation
in the
But,
A
\302\267
(ab
\037f
a
2
(98))
B).
b
+
2
- c2
that
we found
of cosines,
law
the
=
A.B
\302\267
,
2)
ill (98) gives) 2 - c2 - a2 + b
and substituting
4K 2
=
ab
2
(
= =
Calling - 2a t t
-
2c
ble
a 2 - b 2 + c 2)(2ab + a 2 + b + c) (a - b + c) (a + b
\037 (
-a
t
=
a +
- a + = a + b 4K 2
=
c, we make
b +
=
b
+
c,
t
b
+
2
half
then)
-
- 2b)(t -
2c)t,)
b
+
c,
c).)
and
c.) -
\037 (t
2a)(t
that a greater simplificationwould were clever enough to introducea of t. Because, if 28 = t, that is,
4K 2 =
+
b
the substitutions
- 2b = a
if we that
- c2 )
- c)(a +
we observe
resulted
\302\267
)
2)
-
\037 (2ab
Thus) and
c2
-
a2 + b2
we get
Simplifying,
4K 2
(
)
ab +
(28 \037
8 =
i(a +
-
2a)(28
b
+
c),)
- 2b)(28-
2c)28.)))
(99)) have varia-
VECTOR
ELEMENTARY
156)
GEOMETRY)
A)
82)
FIGURE
K
Hence)
and we
a)(s -
S(S -
2 =
arrive at Hero'sFormul
s(s-
-
b)(s
the
a for
c),) area of
a triangle
- c), wheres = j-(a+ b + c). A theorem in Euclidean geometry states that the area a is equal to one-half the product rhombus of of its a Before ofthis diagonals. giving simpleproof theorem, we first a preliminary result: The diagonalsof a prove
K=v
are
rhombus
b)(s
perpendicular.
sides of the as shown in Figure
Let B,
-
a)(s
are
the
by
=
have
we
equal,
rhombus be designated 82. Sinceall sidesofa IBI.
IAI
pendicularityof the diagonals, (A +
(A B) \302\267
B) =
which shows the diagonal
A
+
-
the
K
the
quantity
K of
area
= inside
the
rhombus
the perproduct
= 0,
is. given by
B) X leA
absolute
inner
and
be perpendicular to
B to
the rhombus
21i-(A +
IBI2
IAI2
the diagonalA-B. Now,
the
take
we
-
determine
To
A
value
- B)I, signs being
twice)))
TRIGONOMETRY
K
=
B) X
i-ICA + K =
Hence
shaded region in Figure82.
of the
area
the
157
+
\037 IA
and
center
is
form
the
Call a, {3, and
O.
.
2
result.)
triangle.
involving the We therefore
on a unit
sphere
'Y
the
whose
arcs
circle
great
2
the spherical triangle ABC;a
sides of
B, and
opposite
{3 being
A,
opposite
and
B,
A,
1r sin
B(
desired
the
is
basic relations
a spherical C as points
of
angles
consider
- B(,which the
derive
to
wish
We
B(IA
BIIA -
= i-IA +
B)(
TRIGONOMETRY
35. SPHERICAL
sides.
-
(A
Then
Since
C.
opposite
\"I
that
being
the sphere is of unit radius, the arcs also \"I are {3, the radian measures of the central angles formed by A A Band and an and We further C, B, respectively. C, stipulate that a, {3, and 'Y be less than 1r. in terms information of the Amassing all the given 0 of the sphere, we vectors emanating from the center
a,
and
have
\302\267 A B = cos \"I, B \302\267 C = cos a, and A \302\267 C = cos {3. As we saw earlier, the angle is two between planes determined most easily by finding the between angle A to Thus the interior the angle perpendiculars planes. as is the and OAC same between OAB the plalle plane A X B and A X C. Since by equation 92 allgle between
X
(A
B)
sin 2A
\"I
circle
great
a plane through globe
(A
on a
are all great
shortest
and B
X
not
distance
C)
antipodal,
\302\267 C
-
cos a
(A. B) (A
-
sphere is the intersection of the sphere.
of
great
paths
on the
is along
the arc
A and great
with
sphere
of
path,
circle arcs.)))
if
Thus,
(unless
points) we
see
that
A the
on
lying
unique
Finally,
the
the so-called
sphere. the
the
than
(other
opposite,
B.
(101))
{3,
the
circles are
the shortest
diametrically are
latitude
surface of a
a sphere,
or
circle that passes through sides of a sphericaltriangle
Great
(100)
C)
Meridian circleson
circles, but parallels of circles.
\302\267
cos
cOS'Y
center
are two points of that joins A to B
and B are
= B A =
{3 cos
sin
the
equator) are
sphere
\302\267
A
great the
158)
FIGURE
is often termed
which
the
GEOMETRY)
VECTOR
ELEMENTARY
83)
law
cosines
of
for
spherical
triangles.
EXERCISE
The perpendicularvectors to angle equal or supplementary planes. Verify, by considering
lars,
that
A
B and A
X
the
angle
supplement
of
triangles
by first (A
(See
X C
equation the
Taking
does, in
to
angle the direction
in an the given
intersect
may
planes
the
between
of the perpendicuthe scalar by product (100)of A and not the fact, yield angle
A.
now
shall
We
determined
two
X
B)
derive
the law
of
sines
for
B
\302\267
spherical
considering the expansion) X (A
91 or
X C) =
(A
X
(102)
C)A.
Exercise 4 of Section33.)
magnitudes
of both
members
of (102), we
get)
IA
X
BIIA
X ci
sin
A
=
IA X
BI.lcos
01,)))
159
TRIGONOMETRY
the angle between A
0 is
where
C.
Band
X
Thus,
we
have) X
IA
to
is equivalent
which
sin ,8 sin
=
A
X
X (B
A)
yields
IB X
X C) =
(103))
01.)
both membersof
and
Simplifying, we
(103)
of
X
BllCllcos
(91)))
(see
of,)
Icos
81.)
of the angles of the
have
sin B
(104) we
and
= Icos01
(104)
deduce)
sin,8sin
A,)
law of sines for sphericaltriangles. Again, of letters, we may write the complete interchange sines in the more usual form) is
an
C)B)
in terms
rewriting
sin a sin B = which
IA
cl sin B =
sin a From
\302\267
B
X
(A
to)
IB X
triangle,
-
A((B X ci sin B =
is equivalent
which
law
cos
expression)
(B
by
I
the magnitudes of
evaluating
Now,
the
= Icos OJ,)
sin A
cl
the
.
.
SIn
a
sin
A)
-
sin
,8
sin
B)
-
SIn
'Y
sin
C)))
more
this
devote
We
geometry)
chapter
to a
BY
INEQUALITIES
considerations.)
36.
DEFINED
LOCI
potpourri
of geometric
we earlier work with lines, planes,and spheres, discussed loci that were defined by algebraic vector) (and that It has no doubt occurredto the reader equations. relations other than equality may be utilizedto specify a locus. the conditions For example, if P is the defining a in vector of three-dimensional position point space, < r defines a a that of then locus consistsof sphere Ipl radius r and all points interior to the sphere,whereas > r defines the locus that consistsof the sphere and all I pi relations exterior to it. Analytically, these two points
In our
be expressed
would
x2 +
+
y2
r
x2
and
+
46.
yv e
y2 +
Z2
>
r,
examples of loci so definedfollow. sketch the locus defined by
Further
respectively. EXAMPLE
Z2 <
-2x - 2 <
y
160)))
<
x +
1.
MORE
161
GEOMETRY .\\\\\\ \037\\
y I
\302\2431 \\ .'
1 Iv \037
\\II i\037
I
r-
,- x)
1
0
1
(a)
y
y \302\2432
2
2
1 1
x
0
0
1
(b)
x
1
(c)
FIGURE84)
We
first,
consider,
that
represents the familiar Figure 84a). Sincey desired
locus
line itself
The
locus
consists
y
is
(but
x +
larger
of points
(see Figure 84a). - 2x + 2 < y,
the points below
<
straight
not
by
1. line,
than
The equation which
y
or equal
to x +
above the line as well similar
including)
= x
+ 1
in
is sketched
reasoning,
the line y
as
1,
the
on the
consists
= -2x +
of 2.)))
84b.
(See Figure is \302\2432
a broken
used
have
We
in the
included
not
the
both
\302\2432.
the
includes the darkened not include any point of
does
is indicated by the fact is drawn as a broken line.)
latter
The
but \302\2431
boundary
of the plane that This region is
locus
desired
The
of line
portion
the region
simultaneously.
inequalities
shadedin Figure84c. lower left
may
determine
to
sketches
two
satisfies
indicatethat
two simulinequalities be found by combining
the
solution
final
line to
locus.)
The original problem imposes
taneously, so the
GEOMETRY)
VECTOR
ELEMENTARY
162)
that the
\302\2432-portion
of
EXERCISES
1.
(a) x > 0
and
(b) x < y
< 1.
(c) -2 < < - 1<
y.
x
the
is
What
3. Let 0
the
be
x +
2.
of the region 1 < Ipl < 2 if P is a planepositionvector? P is a position vector in three dimensions?
shape
vector
the
<
2y
(a) the vector (b)
o.
<
y
(d) 4x
2.
defined by the relations
the regions
Sketch
of a
origin
rectangular coordinate
system and
\037
P,
-4
IPQI
= 1.
radius is 3
lie
inside
and
C
triangle (0, 1).
=
6.
x
(b)
x
- a = b; - a < b. locus
the
is
What
(a) x < I? (b)
x
(c) x
(d)
2xl
=
that the points (x, y) region whoseinner radius
<
Iy
-
ABC, where
of points
II?)))
is
2 and
(take the centerat (0,0));
(Assume
y?
< y?
and
of Q?
=
A
5. Confining attention to the x-axis, give pretation of the points (x,0) such that (a)
2
fact
the
an annular
lie inside
outer (b)
locus
the
is
What
4. Expressanalytically (a)
dimensions such that lopi =
in three
points
Q any
b >
0 in
(1, 0),
a
B =
geometric
both parts.)
(x, y) suchthat
(1, 2), inter-
(For assistance
to Section37
5 and
Exercises
with
reader is referred
6, the
c.))
BOOBY TRAPS
A FEW
37.
163)
GEOMETRY)
MORE
(a) How shall
the
sketch
we
locus
plane
defined by
the equation)
x+
=
y
x +
y?)
because the line glance it appearsto bea straight is But linear. look once of the appearance equation like terms, that is, by subtracting By again! grouping x + y from both members, we get)
At first
0=0.)
(105))
If you suggest that there isno are you us consider the problem in the light of primary
locus,
ordered
What
ordered
any
(x, y) satisfy the be substituted may
pairs pair
statement
choice values for the graph consistsof the of
entire
is the
What
(b)
Y
t ranges
which Adding
the
line
straight
locus,
for
x=
two
the
- 1!
there The
is an
every the
xy-plane.
locus of the parametric eqllations) x
in
question:
relation? Clearly, in (105), and the
identity, true for variables. Consequently, It
true.
remains
Let
wrong!
=
cos
=
sin
2
t
2 t,)
the real numbers? is yields x + y = 1, which equations in Figure 85a. But this cannotbe the x = 2 or of t that is no value yields over all
occurred
slip-up
when
2 sin t
we
blithely
are ignored the fact that cos t alld non-negative. The locus (see Figure 85b) is, therefore,the segment of x + y = 1, with the further restrictionsx > 0 and y > 0, for the given parametricequationsplace such 2
restriction
on
x and
y.)))
GEOMETRY)
VECTOR
ELEMENTARY
164) y)
y)
1)
x)
%)
o)
1)
(a))
(b))
85)
FIGURE
is the graph of the relation
(c) What
-
It <
Ix
In order to
insight
geometric
gain
2?)
we begin with an algebraicanalysis. it true that Ix - 11= 2? The reader question in several ways:
(i) Using the
that x-I
or x =
-1. he
Or,
(ii)
and only if
a
may 2
=
b
to the solution of
2
the
aware of the
become
point
x
point 1. the
the
(on
distance
In
-2,
axis)
general,
between
fact that is at the
this
value,
results
he reasons in x = 3 =
Ibl
if
problem
equation
- 1)2 = 3 or x
x is
approach
may
which
polynomial
x =
yields
again
=
relation,
which
For
be aware of the fact that lal , in which case he reduces the
(x which
absolute
of
definition
= 2 or x-I
this
into
4,)
= -1.
In
case
either
we
states that the Ix 11 a distance of two units from -
quantity
x and a.
...:..-2
Ix
-
Thus it is
al
symbolizes
reasoned
that)))
165)
GEOMETRY)
MORE
-'
2
x r
2
A.)
-1)
,r
locus x
and
I
I
1)
2)
86)
the segment between x = - 1 the endpoints. The bobby trap the question, \"What is overlooks
2 is
including
3,
arises
the
when
reader
the dimensionin which Hence
setofall is that
if
our
(x,
y)
the
< 3,1
Ix
and
- 11<
any real number,
which impliesa locusconsisting not only of the on the x axis but alsoof the entire strip between including
lines x
the
= -1 and x
now
answer
The
be
may
y
2.
stated?\"
we seek the
the plane,
is of
that
such
been
has
problem
discussion
-1 < x
L3)
I
<
I x-II
=
X \\
o)
FIGURE
the
A.)
segment
= 3 (see
Figure
and 87).)
)') I, \"':I \037/, \037\037 \037/
Y
\037
I
-2
3
-1
.... \". x)
\037 I'
//,
i%:'l)
FIGURE
1 In the
terms of set theory,
sanle set
we
seek
as {(x,y)ll < x < 3}.)))
87
{(x, y) Ilx
- 11 < 2},which
is
VECTOR
ELEMENTARY
166)
(d) What
is the graph of) (x
Again ordered note
that
+
y) (x
=
y)
O?)
(106))
to the fundamental question:What (x, y) render the given relation true? We pairs (106) is true if and only if)
that is, if
and
satisfying the
or)
only given
origin. If point
(x,
from
-
we resort
x+y=O)
relation
GEOMETRY)
is b\037th
satisfied. relations
if y =
-x or y
relation
is a
x. Thus, the locus of lines through the pair one of these lines, the y) is on either that this is quite different (Observe then satisfied simultaneously; being
we would have had an (107) and the locus would
the two lines in
the
(107))
x-y=O)
\"and\"
be
=
between
the
point
the
equations
of intersection
figure.))
y)
x)
FIGURE
88)))
in
of
167)
GEOMETRY)
MORE
to
The language of set theory helps clarify sort. Since the locus is defined with must have the union of two Therefore sets. this
of
matters an
we
\"or,\"
locus
the
may be written as the set) s=
= -x}.)
{(x,y)ly=x}U{(x,y)ly
This description of the locus and clarity that often resultfrom tion of the ideas of set theory.) 38.
the
illustrates an
simplicity applica-
elementary
AND CONVEXITY
SEGMENTS
and sufficient Theorem 4 providedus with a necessary condition for a vector to have its endpoint on a line. We now turn to confining a the vector to specific endpoint segment. Let A and B be two points and C some point of the AB. can be described by stating This situation segment -4 -4
that BC =
< 1.
< t
0
where
tBA,
In the languageof positionvectors fixed point 0, we write) =
C
C=
tA
+
teA (1
t =
<
t
That
1.
The
1.
tains of the
A
set
all
<
t
1.)
(108))
form
+
tA
is
{pIp =
called
tA
+
convex if and
only any.
be
points in
\302\243.)))
rep- t)B,
0
- t)B}.)
(1
entire line segment that joins set. That is, a set S issaidto convex A
(1
occur when t =
endpoints
the
any two
can be
that
points
is,
segment AB =
Definition.
o<
of the
vectors
position
by
where 0 <
contained
- B), - t)B.)
the segment AB consistsof
Thus
resented and
B +
some
to
relative
and
B of
S, the segment
it
if
two
con-
points if,
given
AB is entirely
VECTOR
ELEMENTARY
168
EXAMPLE 47.
We
Let the sphereof A and
B be any
set.
<
sphereand
its
and
let
origin,
of the
points
IAI
centered at the set. Then,)
r be
radius
two
of a
the points
that
prove
a convex
form
interior
GEOMETRY
IBI <
and)
r)
r.)
AB is also a We must show that any point of the segment distanceof less than or equal to r from the origin. The series of
inequalities)
ItA +
(1
-
t)BI
<
ItAr + =
the
establishes
desired
EXAMPLE 48.
We
1(1 -
tlAI
t)BI
+
(1
-
t)IBI
<
tr +
- t)r =
(1
result.) continue
the
of our ability
consequences
an algebraic to write segmentsin vector languageby finding condition that a point lie interior to, or on the boundary of, triangle.
Let A, B, and We assume, for the
Figure where
89)
D
consists
may
be
r)
C be three distinct purposes of the
any point
points
of illustration,
points
not
of segment AB.
(see segmentsCD, Again,
c)
A)
B)
FIGURE
89)))
line.
on one
that a triangle
of all the
a
using)
vectors to
position
=
D
tA +
(1
-
x =
+
mC
= (1 that (1
Noting
the
ABC,
triangle
sB +
.
. whIch
In
tC,
o<
{
<
t
=
t
m < 1.)
- t)B] mC.
+
t)B
+
t)
m =
r, s, and t
1, we draw
or boundary
interior
of the
point
r + s+ 0 < r < 0 < s <
(1
- m)(1 -
are scalars
there
0 <
- m)(1 -
(1
Conclusion. If X is ariy of
+
(1
+
m)t
write)
may
where
m)[tA
+
m)tA
-
we
m)D,)
-
(1
< 1.)
the triangle, there is someD so that
mC + (1 -
x =
V\\Te
consideration,
0 < t
where
t)B,)
X is on the segmentCD, and Then)
under
segment AB by the equation)
point of
is any
X
if
represent points
D of
point
any
represent
Thus,
169)
GEOMETRY)
MORE
X =
that
so
rA
+
1,
1,
1, 1.)
EXERCISES
1. Justify
2.
Graph
(a) (b)
each step
+
y2
=
4. Representthe
5.
1)(x
points
to the
similar
What
2
+
y2
-
condition
on
2) = 0 2) = 0 those
the
of
coefficients be
on
the
ABC?
and
prove
the
at the origin
centered
of a tetrahedron and of a square in one exhibited in Example 48.
guarantee that a point would
6. State
47.
0
-
(c) (x + 2y + 1)( -x + y + 3. Prove that all spheres-not only -are convex.) manner
in Example
of inequalities
series
the
loci
the
x 2 --- y2 2 (x
of
converse
of the
A,
B,
and
boundary
conclusion of
a
C would
of triangle Example
48.
might expect that the relation betweenX, A, B, and C of Example 48 would somehow on the location of the depend of the auxiliary (the origin position vectors). However,))) point 7. One
the result the
Can you explain is independentof Consider the meaning of r + 8 +
otherwise. the four vectors
indicates
example
between
(Hint.
point?
auxiliary
=
t
the
of
relation
the
why
1.)
8. What Example
That
is,
modification, if any, shouldbe made to the result of A? the position vectors all emanate from 48 when when A = o.
9. Let triangle ABC =
A
B =
(1, 0),
medians), and write
its
(b) Let r = j-, \"8
(1, 1),
point M rA
+
and
=!,
the
=
follows:
(0, 1)
8B + t = t.
tC.
Check the result
associatedwith
i\037tersection
is inside,
of the
in the form
these
of
values
of ABC, the outside, or on boundary of the
altitudes
triangle.
tetrahedron ABCD be defined 0), B = (1, 0,0), C =
10. Let A
=
of intersection
(point
46 by locating the point Example of the coefficients.
(c) Find the point of and determinewhether it
C
and
vector
position
M =
vertices as
by its
defined
be
(a) Find the median
of
GEOMETRY)
VECTOR
ELEMENTARY
170)
(0, 0,
by
its vertices
(0,1,0)
as follows:
and
D
=
(0, 0,
1).
an necessary conditions, (a) Write stipulating expression, for point X to be on the face ABC of the tetrahedron. (b) Do likewise for X to be on the face BCD. (c) How would you guarantee that X be inside (not on the of
boundary)
39. LINEAR
the
meat
(beef
tetrahedron?)
PROGRAMMING
Considerthe A hospital
given
following
problem.
2
its
is concerned with minimizing the costof and diet. The average hospital diet pork)
in suggested by a Jack Spratt problem Problems for First-Year Algebra\" by in The Donovan and Marilyn Zweng, published is recomfine article Mathematics Teacher, March 1960. This in the details of a successful mended to the readerwho is interested experiencein the teaching of linear programming to high school 2
This
\"Linear
students.)))
problem
was
Programming Lichtenberg
MORE
171)
GEOMET:ay)
and of fat 1.5 pounds requires 2 pounds of lean meat which meat per person per week. costs The $1.00 beef, The pork,which is 0.2 fat and 0.8 lean. costs pe\037 pound is 0.6 fat and 0.4 lean. If the hospital $.75 per pound has200 patients on this diet, and if it cannot purchase more than 900 pounds of meat per week because of refrigerator space,find out how many pounds of beef
and how many pounds of that the cost is at a minimum. This
programming. involving
program
we
that
fact
will
programming
determine a
the
Here
operations.
optimum
hospital
of seeks a program(for the purchasing meats) in that it would the sense the satisfy optimum the hospital while minimizingthe cost.
Let's
the
extract
which needs
the
from
data
pertinent
so
is used because be linear rela-
The word are trying to
see.
soon
we
linear
variables
the
purchased
elementary problemsof
word
The
the relations shall tions as stems from the for
of the
is typical
problem
linear
be
should
pork
is of
hospital
problem. Let
B = poundsof
to
beef
purchase
each
week,
each
week.
P = pounds of pork to purchase I t is clear that) B
The restriction
>
of
O)
rest
of the
space
refrigerator
B + P The
P
and)
<
and
> o.)
(a))
tells
us
that)
900.)
given data can
(b))
be
in
summarized
table)
Fat
,.,' \"'.
Lean
CostjIb
$1.00
Beef
0.2
0.8
Pork
0.6
0.4
0.75)))
the
172)
200
a total.of
Since
required,
0.8B
is required,
0.2B
of
we
the
tions
meat
(d))
for the cost C(in dollars)
(a)
yields
(d) is
then,
problem, a
minimum
(a)-(d) hold. Thus graphically, the set the restrictions.
(e))
turn
of
ordered
0, P >
B
0,
2B + P
(B, P) that satisfy
pairs
is)
set
This
given
to finding,
attention
our
we
R
{(B, p)IB >
+ :P .)
prediction, all the relaP are linear relations. The on Band P. restrictions place to select the ordered pair (B, P) for C(B, P), provided value
Band
-
= B
P)
the earlier
with
involving
inequalities
R =
of fat
pounds
> 300.)
0.6P
an expression
write
accordance
that
300
(c))
purchase)
C(B,
Our
=
\302\267 1.5
+
.
In
> 400;)
have)
we
Finally,
meat is
of lean
pounds
+ JP
of 200
a total
since
and
400
2
that)
see
we
\302\267=
GEOMETRY)
VECTOR
ELEMENTARY
+
P <
900,
> 1000, B
+
3P
> 1500}.)
in Figure 90. The set R is the darkest shown We region have thus narrowedthe choiceof orderedpairs to those number of of the triangle R, but we still have an infinite too many for a trial and error prochoices-certainly cedure unless we were willing to settle for an approximate
answer.
however,
Fortunately,
assist us in narrowing Theorem
12.
search
the
Let f
be
a
linear
of definition is a polygon. of f(x, y) is attained at a
to
considerably. function The
convex
vertex
a theorem
is
there
of
the
whose domain value
minimum polygon.
(The)))
173)
MORE GEOMETRY P)
1000)
500)
B)
1000)
(0,0))
90)
FIGURE
word \"maximum\" without
disturbing to
Returning
vertices of
R
the as:
U = (300,400).
may be substitutedfor \"minimum,\" the validity of the theorem.) we problem, T = (100,800),
original S =
Theorem
Applying
narrowed
1500)
our search to
determine the
(600,300), and
12,
the pointsS,
T,
has
which and
U,
we
evaluate)
giving pounds '.gram
3
C(100,800)
= 1 \302\267 100
+
C(600,300)
= 1 \302\267 600
+
C(300,400)
= 1 \302\267 300
\302\267
800
=
$700
\302\267
=
$825
\302\267
=
$600,)
4 300
\037
3 +
4
400
400 the conclusion that 300 pounds of beef and of pork make up the optimum purchasing profor
the
hospital.)))
174)
P 2 (X2,
show
We
Y2).
the end-points
the
of
is con-
P
when
y)
= (Xl, YI)
where and PI under this restriction, the values of f(x, y) are attained at 2
,
segment.
leari1ed that
38 we
Section
c.)
that,
minimum
and
maximum
In
P 1P
to the segment
.fined
+
by
of f(x,
values
the
consider
first
3
linear
the
of
by)
given
= ax +
f(x, y) We
Let the value
of Theorem 12. at P = (x, y) be
Proof function
GEOMETRY)
VECTOR
ELEMENTARY
the segmentP IP
=
P
if
belongs to
(x, y)
2 , then)
X =
(1
=
lYl +
(1
0 <
l <
y in
which
= aX2 + = (ax2 + = f(X2, Hence
values the
l)X2
-
l)Y2
=
X2 +
l(XI
-
=
Y2 +
l(Yl
-
1. Then the value
j(x, y) = a[x2+
o < l
-
+
lXI
-
l(XI
+
bY2
+
-
l(axi. +
bY2
Y2)
X2)] +
+
b[Y2
+
aX2
l[(axi +
c) +
YI)
l[f(XI
of
-
Y2),
f at
l(Yl bYl bYI
+
P is
f(X2,
+
bY2)
c
+
Y2)]
c
c)
- (ax2 +
bY2
+
c)]
Y2)].
and f(x, y) ranges between f(Xl, YI) f(X2, Y2) as < 1. The extreme and (maximum minimum) of the function are attainedat the endpointsof
segment.
What
Query. amine the
an exercise
for
tionsof
the
polygon
and
Since the term \"linear\" of the form
expressions
to
exhibit
of
\"linear.\)
a
proof
of
your
proof
reader, convexity
with
answer
=
f(X2,
12 is Theorem the hint that the are essential. for
Ex-
Y2)?
is substantianted.
left as
defini-
function is linear.if and only if c = o. to include is often used in a loose manner ax + by + c, when c F- 0, we have chosen 12 for such a loose interpretation Theorem
this
speaking,
Strictly
if f(Xl, YI)
happens
proof to see if The remainder of the
3
X2),
175)
GEOMETRY)
MORE
ADDITIONAL EXERCISE the
Make
in the
replacements
following
statement
Theorem
of
12:)
for
\"polyhedron\" \"f(x,
y, z)\"
and
\"polygon,\"
for\" f(x,
y).\
Prove the resultingstatement.) van company charges25 cents per from N ew York to San Francisco and 15 for pound moving crates on the same crossIf at least one quarter of each load is furniture country trip. and at least one quarter of each load consists of crates, find and maximum cost per pound for a load. the minimum of a load that is furniture, Let F = that fraction of a load that consists of crates.) C = that fraction 49.
EXAMPLE
for
moving cents per
pound
A moving furniture
3
1
-
Then)
-4
4-
1 3 -
and)
since
Furthermore,
we have
crates,
a load
consists entirely
furniture
of
and
the additionalrelation) F
+ C
= 1.)
a segment, These three relations define as shown in Figure 91. of cost on a one-dimenOur problemreducesto a consideration are (-1-, : ) and (i, \037 sional set, a segment whose ). endpoints the relation The cost K(F, C) is given by K(F, C) = .25F +
.15C.
K 1' 4 3
(
K
( Hence,
pound,
4
= 0.25 )
3' 1 4
4
)
= 0.25
1 + 4 ( )
3 0.15 4 ( ) 1
3
+ 0.15 4 ( 4) ( )
the minimum cost while the maximum
=
=
for a load would would
be
22!
0.70= 0.175
4
0.90
\037
4 be
17
cents per
0.225.)
\037 cents pound.)))
per
ELEMENTARY
176)
VECTOR
GEOMETRY)
c) 1)
(i,t))
3 4\
1 '2)
1 4\"
(f,t)
1
(0,0)
1
4
3
'2
91)
FIGURE
bats
A sporting
50.
EXAMPLE
that cuts the wood paints,
goods company makes baseball processed first by a lathe that machine by a finishing
which are bats, to size and then
softball
and
F
1
4)
and labels the bats. The lathe produce a baseball bat and 3 minutes bat. The finishing machine operates
dries, polishes,
6 minutes to a softball produce ates
operto
for because
on a bat, no matter which kind. However, and maintenance problems,the lathe can operate can operate 4 hours machine only per day and the finishing bat is $1.00 a baseball on If the only 3 hours per day. profit if the the on a softball bat 75 cents, and and company profit can sell all the bats it makes, find how many of each kind it should producedaily in order to realize maximum profit. to be a mass of detail, the problem can there seems Although be summarized easilyin the following Let manner.
3 minutes of loading
The
b =
the number
of baseballbats produceddaily,
8 =
the number
of softball
restriction
(4
\302\267 minutes
60
glves) 6b
+ 38
bats produceddaily.
per day)
< 240 or
2b +
on the lathe 8
<
80.)
operation (a))))
177)
GEOMETRY)
MORE
(3. 60 minutes per
The restriction machine gives)
3b +
180
60.)
(b))
data are the relation
the
all
and
o.
(c)
8 >
o.
(d))
8)
PCb,
Thus
finishing
b >
day would
8) per
PCb,
profit
(a)-(d)
8 <
b +
or)
the
also that)
We know
The
<
38
on
day)
=
be b +
:8 .)
(e))
summarized in terms of (e). We seek to maximize
the convex polygon det\037rmined the by The polygon is exhibitedin Figure92.
inequalities
restrictions
PCb, 8) on (a)-(d). of
vertices
The
the
polygon are
o = (0,0),
A
Evaluating would
maximum
bats
=
(40, 0),
the
profit
at
B = these
achieved
be
(20,40), four
by the
and
points
C
=
(0, 60).
shows that a
production
softball
of
only.) s)
80)
40
.
B)
20)
b)
0(0,0)
20)
60)
FIGURE
92)))
80)
EXAMPLE 51. A chocolate manufacturing milk chocolate, semisweet chocolate,and but demand It can produce 500 poundsper day,
types is such that the maximum
GEOMETRY
VECTOR
ELEMENTARY
178
amounts
makes
concern
chocolate.
bitter
that
for the can be
various sold are
as follows)
milk-400
per day
pounds
semisweet-300 bitter-200
If the for
is 80
profit
semi-sweet,
the
per day.
pounds
cents per pound for bitter, 75cents per pound milk and 60 cents per pound for chocolate,
for production that
the program
determine
pounds per day
maximize
would
profit.
Although this problem couldbe handled by only problem ables, we shall treat it as a three-dimensional Let poses of illustration.
two for
8 =
pounds
chocolate produced per day of semisweet chocolate producedper day
b =
pounds
of bitter
=
m
pounds
varipur-
of milk
chocolate producedper day.
Then)
< 500
m+8+b
0<
The m \037
profit :8 +
+
function b \037 (in
The graph An algebraic A
=
P is
400
o < 8
< 300
o<
<
b
(b)
(c)
200.)
(d))
defined by the relationP(m,8,
b)
=
dollars).
in 93. of restrictions (a)-(d) is shown Figure solution for the vertices of the polyhedron shows
B =
(400,0,100),
D = (200,300,0), E and
G
=
(400,0,0),
=
(0,
(300,
We may concludethat the profit by
<
m
(a)
producing
only
is The latter alternative 500 pounds would always concludethat two kinds So the searchis confined
300,
C = (400,100,0), F =
(0, 0, 200),
achieve
a maximum
200),
0, 200).
concernwould
or one type of types because excluded producing two
in a loss. of chocolate should result to
an
investigation
chocolate.
less than We may therefore be manufactured. of the profit at)))
179)
GEOMETRY)
MORE
b)
E)
t
I
400
:> 8)
500)
m)
FIGURE
points m
A, C, D, E, and G-the b = 500.
93)
a plane
of
points
8 +
+
The computation
cise
for
the
of
these
at
profit
vertices is
represented
by
left as an exer-
reader.)
EXERCISES
Carry out Example51as a two out Example 51 with Carry maximum amount of bitter chocolate 1. (a) (b)
problem. that restriction
that
can be
the
sold is 100
day.
pounds
per
2. Find - 1
the minimum on the set defined
4y
variable the
and
values
maximum by
the
- 2x +
inequalities
3y
6
< 2
x+y
x+
<
2y
>
3.)))
of I(x, y)
= 2x +
VECTOR
ELEMENTARY
180)
3. The delivery
trucks
capacityof 1000 least 400 gallons
On each trip a gallons. of regular gasoline, at
owned
by
GEOMETRY)
company have a truck must carry at least 200 gallons of
oil
an
and, at most, 300 gallonsof white gasoline, gasoline. 4 cents for high profit per gallon is 3 cents for regular, test, and 5 cents for white gas, find the program for loading the trucks that would yield a maximum profit. (This can be test
high
If the
done with
4. Assume What kind
5.
two
only
variables.)
the trucks in Exercise3 areall loaded of loadingwould the least profit? yield
mountain
climbing
and B-rations for
an
A
per
unit
expedition.
of carbohydrate
Units of
protein
Units of
fat
B
units,
of the
ARISING IN
40. THEOREMS The
the
study same
is not a closed polygon you have
measurements,
the
minimum?))
MORE
GEOMETRIES
geometry deals with someof
of projective
Euclidean
in
discussed
entities
namely points, li11es,and planes. geometry
climbers.
mountain
that the convex region (Observe in this case. How do you know that
GENERAL
1 $1)
carbohydrate units, 18 the minimum cost diet
and
the requirements
satisfying
3 4
3
are 10 requirements 6 fat units. Find
minimum
The
1 3
$2
Cost
protein
A-rations and costs
wishes to purchase The food values
party
A
Units
capacity.
as follows:)
classified
are
to
discards and
the focuses
notions
However,
geometry, projective
of distance and
attention
on incidence
angle prop-
erties: points lying on \037ines, lines passing through points, For eleintersections of various sorts, and so forth. is the reader mentary discussions of projectivegeometry, referred to Chapter IV of What is Mathematics? by)))
181)
GEOMETRY)
MORE
and Courant H. S. M. by
Robbins, and to
Coxeter.
We
the
Real
principal results of this field of study: the Desargues
Plane
Proiective
into
look
shall
of the
two theorems
of
and Pappus.
Desargues' theorem.
If
two
have
triangles
corresponding
vertic\037s by concurrent lines, then the intersections of joined are collinear sides (see Figure 94). correspond\037ng that the theorem The readershouldobserve is devoid
with of metric concepts. It is concerned neither lengths of of sides nor with the sizeof the angles the triangle. One of the axioms of projective geometrystatesthat that is, parallelism is ruled out. two lines every intersect, Consequently,
the
intersections
to in Desargues' exist in pro-
referred
theorem, namely points P, Q, and R, must the jective geometry. But, if we consider in Euclideangeometry, Desargues' theorem interfere
may
with
the existence
therefore precede the
P, projective
generalized
s)
FIGURE
of
94)))
hypothesis
of
parallelism
Q, and
R.
version
We of)
GEOMETRY)
VECTOR
ELEMENTARY.
182)
o)
/)
//
./
/ //
/
A') B')
96)
FIGURE
theorem with a
Desargues' Euclidean
geometry.
Let ABC and A'B'C' to A' B' and BC parallel
D.
Theorem
AB parallel
AA', BB', A' C' .
and CC' meetin O.
In terms of
have
the
to
triangles
AC
A' B'
to
parallel =
is
vectors position collinearities: following of
These imply 0, B, and B'. = = B sB' and C. tC'. And AB implies B - A = l(B' - A'). But
and
- rA'.
sB' C
-
B =
with
lines to parallel
while
B'G'
A';
A = rA',
respectively:
A
we
two
be
Then
use
standard
our
emanating from 0, 0, C, and G'; 0, A,
B-
specialcase that arIses111
Hence, r = s =
m(C' -
l.
B') = tC' -
Similarly,
sB',)))
GEOMETRY)
MORE
implies 8 = = m = r =
which l
have
183)
C -
in
(Note that same plane.
to prove. be in the
out
set
not
may
both
we
result the given trianglesmayor The vector proofis to A'G', the
is parallel
AG
that
states
which
A
t = m. Combinillg these results, s = t.) Now,) = tC' - rA' = r(C' - A'),
we
valid
cases.)
that every two lines meet, we of general form of Desargues'
the assumption
Under
vector
a
present
proof
theorem.
94, let, the
in Figure
As
two
with the correspondingverticesjoinedby at S. Call P, Q, and R the meeting of corresponding sides, as follows:
A'B'G'
lines
current
sections
P
of
intersection
the
use
we
Again
point 0
not
(which
rA +
This triple
Noting
members
-
(1
appear
that
=
r)A'
sB +
(1
the
implies
equality
=
sB =
(1 -
s)B' -
sB
-
tC =
(1 -
t)C' -
tC
-
the
sum
in each of
(1
- r)A'
of the
three
(1(1
+
tC
following
-
=
of cor-
pair
every
- s)B'
rA
rA
A'.)
in the
The fact that S is collinear with allowsus to write) vertices responding =
C'
emanating from some figure).
vectors
position does
con-
inter-
and
B'G',
intersection of CA and
R the
and
A' B'
and
AB
Q the intersectionof BC and
S
ABC
be
triangles
-
- (1 -
coefficients of
(1
- t)C'.)
relations:)
r)A'
s)B'
t)C'. left
and
right
these three equationsis equal,we)))
opportunity to use Theorem
have the
having the sum
r
8
_
A
B =
r-s
r-s
8 B_
s-t
t C
s-t r
t
C-
t-r
A
t-r)
=
=
to 1.
equal
1 - 8
1B' _
1C' _
- t
1
- r
1A' _ t-r)
t-r
Each memberof (109)representsa 0, with its .endpoint on A'B' This vector is P. That taneously.
from
vector
(109)
B'.
(110)
C'.
(111))
t
emanating on
and
A'.
8
s-t)
s-t 1
r
r-s)
r-s
and
dividing
by
4,
coefficients
the
of
GEOMETRY)
VECTOR
ELEMENTARY
184)
AB,
simul-
is,)
s
r)
p=)
-
r
r -
s)
s)P =
(r -
or
A-
-
rA
B s)
sB.
(112)
Similar reasoning with (110)and (111)yields t)Q = sB - tC (s
(113))
and)
(t
(r
the
(114) = - (s -
- s)P
the
establishes coefficients
t
-
t
-
r-s
8
Q +
collinearity
us
-
r r-s)
sir - sand
(114))
rA.
give
t)Q
p =
or)
-
tC
(112), (113), and
Finally,
which
- r)R =
(t
- r)R,
t
R,
of P,
Q, and R, for
r - tlr - s
sum
to
unity.
EXERCISE tIle divisions What justification is there in performing by r - 8, 8 - t, and t - r to obtain (109), (110),and (Ill)? of these denominators What would be implied by any being zero?)))
MORE
185)
GEOMET.RY)
theorem. Pappus' was the last A.D.) of antiquity. ticians
a
that
theorem
devoid categ9ry
metric
of
of
Pappus
Alexandria
(3rd
century
theorems.
projective
If alternate
of
of the remarkableGreek mathemaAmong his original contributionsis can be stated as a pure incidence theorem, and therefore into the concepts, falling
vertices of a plane hexagon
on
lie
two
lines,
the three pairs of oppositesidesmeet in three collinear points. We may restate the theorem in specific terms as follows: and points of line \302\2431 If A, B, G are distinct A', B', C' are distinct points of another line \302\2432,the three points of AB' and A' B, and BC' intersection of the pairs of lines
and B'G,GA',
and
G' A
are
collinear.
Before proceeding with the proof,we again that every two lines are assumed
out
point to
intersect
once in
If one considersPappus' theorem geometry. in the strict realm of Euclidean geometry, one must conarise and intersiderspecialcaseswhere parallelism may
projective
fere with the existenceof Exercise 4, page 191).
Proof Pappus' we call U a
Theorem.
of
unit
vector
along
certain
intersections
Referring and \302\2431
FIGURE 96)))
V
(see
to Figure 96, a unit vector)
GEOMETRY
ELEMENTARY'VECTOR
186
along
=
A
a'V
is
collinear
4 to
write)
L
Since
Theorem
,)
0', we
with Band
=
+
lbU
invoke
may
- l)e'V.)
(1
B' and C, we
collinear with
is also
L
since
eU
b'V,) C' = e'V.
B' =
,
=
C
bU,)
(1 - l)C'
L = lB + And
B =
aU,
=
A'
and)
write)
we may
Thus
\302\2432.
may
also
write)
(1 -
L = XC + two
These
=
X)B'
XeU +
-
(1
X)b'V.)
imply)
representations
lb=Xe
l)e' = (1 -
(1 which
-
bb'
Thus) L =
and)
To seek
it
can
M
=
N
r,
=
X,
- ee'
b'e
l=
Similarly,
l and
for
solving
yields)
and)
ee')
bb' -
be'
-
ee')
X=
bb'
be(b' - e') U + bb' - ce , ae(e' - a') U ee' - aa' - b')
aa ,
bb'
determine
whether
8, and
t, such that)
rL +
+
-
(b
bb'-
e) ee)
,
V.
U +
+
a'e'(eee' -
a)
V,
aa')
L, M,
8M
b'e'
\302\267
that
be shown
ab(a'
X)b',)
tN
a'b'(a - b) V. , -) bb' aa and N are collinear,we =
O.)))
187)
GEOMETRY)
MORE
relation must r, 8, and
other algebraic
(What
t
satisfy
conclude collinearity?)
we may
before
Perseverance
with
the
yields
algebra
elementary
solution)
r = aa'(bb' -
ee'), s
=
-
bb'(ec'
aa'), =
t
from
linear. zero
we
which
do
know
we
do
(Now,
that that
conclude
ee'(aa'
L, M, and r, 8, and t
- bb'),
N are colare not all
?)
just proved possessesmathematical far' beyond significance any dreams that Pappus had in the third century. could have back Approxia one hundred method of mately years ago building up number systemsfrom geometry was first discovered by a we have
theorem
The
Since German mathematician, Karl G. C.von Staudt. time much creative work has been done on the
that
of
foundations
and
geometry
the algebraic structureof number
with
its interrelationship that
systems
can
built up from geometry. One of the great achievements of this study is a remarkabletheorem, first proved of around turn David the Hilbert (1862-1943) which states:) century, A
law valid
number
a. in
related
system
b = b the
\302\267
a
and
if
geometry.
be by the
to a
geometry satisfies the only if Pappus' theorem is (The
here
\"dot\"
stands
for
multiplication.))
The geometry
we have beenworking
with
is Euclidean,
number system for our analytic geometryisthe in which multiplication real number is commutasystem, \302\267= \302\267 tive, that is, a b b a. Thus we were able to prove and the
Pappus' theorem;and,
conversely,
if
we
had
number system with the aid of Pappus' would necessarily be commutative.))) multiplication
developed theorem,
a its
VECTOR
ELEMENTARY
188)
Since it is hardly possibleto enter cussion of the foundationsof geometry
we shall have to content ourselves remark that it is a rich subject,
GEOMETRY)
in this
short
closing
by
work, the
with
hfts resulted
which
dis-
a detailed
into
in the
as as well discovery of many strange numbersystems There are geometries for which the coorgeometries.
dinate
number
has
system
1 + 1 = O. There are theorem nor Pappus' And
stranger
their
accompanying
variety
of
in
other
and
\037
2;
in
geometries
number
applicat\037ons
agricultural
1
in
fact, which
where neither
Desargues' theorem holdstrue. the fact that such geometriesand
is
yet
1 +
fou11d a wide as the designof military logistics,
have
systems
as diverse
fields
experiments,
of mathematical machines. psychology, and the study The theorem of Menelaus. Menelaus of Alexandria, a treatise on spheres and actuallymade who wrote some in spherical in the 1st century discoveries trigonometry is also noted for having discovered an interesting A.D.,
theorem
transversals.)
concerning
ABC
triangle
AB, BC, and CA points L, M, and N, respectively,)
cuts the sides
a transversal
If
in the AL
LB
.
BM
.
MC
CN
=
of
(115))
-1,
NA
where all segmentsreferredto are directed segments. Conversely, if points L, M, and N are on the respective sides AB, BC, and CA or triangle and) ABC, AL BM CN _e_e_ LB
then
We
shall
L,
]f, prove
ing the converse
and N only
MC
-1
NA)
,)
are collinear.) the first
part
as an exercisefor
of
the
the
theorem,
reader.)))
leav-
189)
GEOMETRY)
MORE
A)
B)
M)
FIGURE
all
Consider
point
A,
emanating from
of points
vectors
position
and
L
These
97)
ratio l: 1 -
AB in the
divides
- m 1 - n.)
M dividesBC
in
the
ratio
m: 1
N divides AC
in
the
ratio
n:
state)
conditio11s
CN
AL BM _e_e_
LB
l
N A)
]lifC
-
m
l
1-
l-ll-m)
n)
n)
Then) \037
L
=
lB,
N =
nC,
\037
and BM = mBC=
m(C
-
B).
(116)) l < 1, n < 1, and (Figure 97 illlistrates the casewhere - B = mC - 11\037B a11d) M m > 1.) Tllerefore M = (1 - m)B + mC.) ( 117))
Now,
since L,
exists
a real
we It.1,alld N are collinear,
know
that
there
number r such that) M
=
rL +
(1
- r)N.)
(118))))
(117), and (118),we
From (116), (1 Hence)
.
wh
IC
deduce)
- m)B + mC = rlB + (1 - r)nC. - r)n and rl = 1 - m,) = m (1
\302\267
h Imp
1
l =
Iy
- m
-
1
-
in a
now
Weare
r)
position to examine the productof of rand n alone:)
terms
in
ratios
three
1l
1-
I-n
m
.
l 1-
(1r
-
n
m
- r)n .)
(1
r)
the
GEOMETRY)
VECTOR
ELEMENTARY
190)
I -
I-
r)n
(1 -
r)n
r
- r)n
(1
1
-
r
- (1
- 1+
1 - (1 -
-
r)n
-
-
which
result
is the
(1
(r - 1)(1-
we wished
n)
I-n
r)n
1 - (1 -
-
n
r)n
r)n
(1 -
(1
- r)(l
I-n
-
n
r)n
-1
,)
n)
to prove.)
EXERCISES
1.
a + b
=
aL
= 1, by
(You might
of the Menelaus theorem. bN for some scalarsa and b. +
converse
the
Prove
Hint: M
using
try
solving for it in
to
the
relation
eliminate
terms of
m
l
one of and
m
1 _ l.
n.))))
1
\037
.
Prove that
I-n
m
the scalars, say
=
-1.
by
first
n l,
GEOMETRY
MORE
191)
A'
A) \"
\"
\" \"
,
\"
,
\" \"
,
,
,
\" \"
\"C)
B)
C')
B')
FIGURE
98)
2. Prove the special Euclideancaseof Desargues' which reads: Let ABC and A'B'C' be two triangles parallel to A'B', and BC parallel to and CC' are parallel. Then AC BB',
B'C'
lines
while
be
must
theorem, with AB
parallel
AA',
to
A'C' (see Figure 98).
3.
case of Pappus' theorem. If and A', B', C' on two coplanar A, B, C lines \302\243 and are so related that the lines AA', \302\243', respectively, of the pairs CC' meet in a point, the points of intersection BB', of lines AB' and BA', BC' and CB', CA', and AC' are collinear with the point of intersection of \302\243 and \302\243'. 4. Prove Euclidean by vector methods the following special case of Pappus' theorem: If A, B, C are distinct points of line AB' is that \302\243 and such A', B', C' are distinct points of \302\243' parallel to BC' and A'B is parallel to B'C, then AA' must be two
Prove
sets
the following of three points
parallel to CC'. Why
41.
is
this
a special
APPLICATIONS
special
case of Pappus' theorem?) OF
PARAMETRIC
EQUATIONS TO LOCUSPROBLEMS
The cycloid. Considera mitted to roll slippage. without
wheel
set
An
on
a line
interesti11g
and perproblem,)))
192)
FIGURE
and ing
VECTOR
ELEMENTARY
point
Figure
P on
is the
the rim
99)
are applicable to certain engineerWhat is the locusof a fixed
results
whose
one problems,
GEOMETRY)
following:
as the wheelrollsalong
the
line?
(See
99.)
To simplify matters allow the line to be the x-axis and let P beginat the of the coordinate system. r the We call C the center radius of the wheel. The which the the parameter 8 isusedto denote through the) radius CP has rotated. Referring to 100, we
origin
and
angle
Figure
y)
x)
o
F)
M(x,O))
FIGURE
100)))
193)
GEOMET.RY)
MORE
......-...)
OF
distance
horizontal
assumption
that
the
trigonometry
we
know
in
......-...
fact in determining x = from
r8.
=
PQ
= r x
sin 8, =
to
referring
Again,
y
Q
Hence)
y
=
r
=
=
PQ -.
PC
r)
r(8 - sin8).
100, we
= CF -
CP
cos 8
- r cos8 =
note =
QC
that)
- QCj
r
= r cos 8.
r(l -
cos
8),)
we have) =
X
{Y as
-
PQ
- r sin 8 = Figure
C
PQ.)
of P is and the x-coordinate
= FQ
= MP
but)
and
r8
P.)
see that)
PCQ we .
important of
- MF = r8 -
SIn 8
Thus,
this
use
shall
We
the x- alld y-coordinates OF
triangle
r8, where8 is measured
FP =
that
OF =
Thus
radians.
But
the arc FP, by the not slip; and from
is equal to wheel does
the
parametric is
which
called
=
r(8 r(l
- Sill 8) - cos 8))
representation
of the
desired locus,
a cycloid.
We shall confineour attention to involutes the for these curves are easily treated without circles, of calculus. The more general theory is part of the alld requires-as the name of differential geometry
Involutes. of
use field
suggests-differential calculllSas its analysis.
Consider zero
thickness)
of
instrument
idealsort,
a string (of a mathematicalor to be tightly wound about a
with
circular
post,)))
VECTOR
ELEMENTARY
194)
GEOMETRY
Y)
x)
(b))
(a))
FIGURE
101)
P the endpointof the string. The involute of a P as the curve described by point string is unwound while beingheldtaut (seeFigure 101). We shall now derive parametric equations for the the fact that the portion of the string involute, using to the circle (because it is is tangential alreadyunwound and call
circle
is the
being heldtaut).
Letthecircleof
radius
r be
origin, and (r, 0) on to take unwinding at the
centered
suppose the initial positionof P to
be
at
A
=
the x-axis. We arbitrarily choose the in a counter-clockwise the place direction,calling unwound of isa the where OT radius portion string PT, an angle 8 with the horizontal radius OA. making TGP in Figure 101b, we observe to triangle Referring that angle GTP is equal to 8, which allowsus to write) GP
=
Since the
TP
sin
unwound
string
......-...
arc
AT,
we
have
GT
and)
8)
TP =
TP
= .TP
is equal
cos 8.
in length
(119)) to the
......-...
AT = r8, which
permits
us
to)))
(119) in
rewrite
are
the
=
GP
We-
195)
GEOMETRY)
MORE
now
form)
8
r8 sin
GT = r8 cos8.)
and
expressions for the
to determine
ready
coordinatesof P in
terms
of
the
OF + GP = r cos8 + r8 sin and y = FT - GT = r sin8 x =
8 =
a parametric
Thus
X = =
{ y
Vector
8 + 8 sin
resin
8
the
- 8 cos 8).)
involute
of
8)
the
(120))
- 8 cos 8).)
of assistance in determining therefore
We
considered.
just
type
r(cos
are often
methods
lociof the
resin 8
of
representation .)
8),
8
cos
r.8
8 + 8 sin
r(cos
=
circle is
8.)
parameter
a derivation of (120)fromthe vector to hope that it may be instructive \037 OP = xi + The vector position
attempt in the
of view
point
reader.
the yj
\037
\037
=
OT
=
r cos
TP)
+
\037
8i +
r sin 8j +
TP.)
\037
We
call
= ai
TP
+ bj and determinea and \037
= Va 2
ITPI \037
TP
and
Since
r
r!= 0,
=
\037 \302\267 OT
we may
=
ar cos
8 + br
divide by r,
a cos or)
+ b2
8 + b
a = -b
r8,)
sin 8 ::cO.)
getting)
sin 8 = 0 tan
(J.)))
b.)
(121))
VECTOR
ELEMENTARY
196)
this va lue for a
Substituting
in
(1 21)
2 2 V b tan 8 b
or which
+ b2 = sec 8 = r8,
b =
implies
r8 cos
GEOMETRY)
yields r8)
8
a = -r8 sin
and
8.)
\037
now
are
We
an
\037
getting
for OPe)
\037
OP
to the problemof
to return
equipped
formulation
explicit
sin 8i + r8 cos8j.
= -r8
TP
Thus
=
r cos
8i +
=
r(cos
8
r sin 8j -
- 8 sin 8)i
+
r8sin8i+ r cos
8j)
resin
8 +
8 cos 8)j.
Therefore)
X = { y differs
which
from
See Exercise 1 below.)
=
the
- 8 sin 8)
r(cos
8
resin
8 +
parametric
(122))
8 cos 8), form
(120)!
Why?
EXERCISES
1. The careless
error of
representation (122) is in error, for some was put forth in the argument. Find the derivation of the involute valid vector-type
parametric algebra
and
give
a
a circle.
r rolls along a line without of radius slipping. described by a point P on a spokeof the wheel, of the wheel. This the center where P is at a distancea from curve is known as a trochoid or prolate (see Figure l02a). cycloid P is on locus if a > r, as is the \037ase when is the (b) What the rim or flange of a locomotive wheel (see figure l02b). that a descripThis curve is called a curate (Observe cycloid. tion of this locus shows that some part of a locomotive is movis moving forward!)))) ing backward, no matter how fast the train
2. (a) Find
A
wheel
the locus
MORE
197)
GEOMETRY)
(a))
(b))
102)
FIGURE
42. RIGID
MOTIONS
Euclid's theory of congruence is predicatedon the to a figure from one position to another move without its metric properties (e.g. length of disturbing ability
motions are generally edges, size of angles). Such or comtermed rigid motions and consist of two types called binations of these two types. The first, transB= a a to fixed refers figure by allo,ving displacing lation, B=. vector T to act on each point of the given figure P is translated That point is, if P is a point of B=, the to Q, where) \037
\037
=
OQ
OP
1'hesecond
of
type
figure
is
104
an
shows
ellipse
is the rotation, in which a motion a fixed point in the plane. Figure and a triangle rotated simultaneously
rigid
about
rotated
(See Figure 103.))
+ T.
about the centerofthe ellipse. We
shall
obtain
for these
expressions
analytic
rigid
motions.
Supposeit isdesiredthat
the
of the translation to the origin To keep matters straight we Ilates of a point with primes.
translation
(h, k)
point
(x, designate
nl0ved by y)-coordillate system. (h,
k) be
the.
Thus)
) (h',
k') = (0,0),)))
new
coordi-
GEOMETRY
VECTOR
ELEMENTARY
198)
Y)
x)
Q)
103)
FIGURE
in terms
which,
lation
T
vector
Hence)
of the definition,states that has
the trans-
the effect)
hi + kj +
T = O.
=
- kj.)
T
In general, the point(x,
-hi
moves
y)
to
point
(x', y'),
and)
y)
y)
x)
FIGURE
x)
104)))
MORE
GEOMETRY)
their
relationship
199)
follows from the xi
or)
(x
-
h)i + (y
x'i + y'j,
- k)j =
x'i +
yj.
X'=X-h
Thus)
y'=y-k)
{ are
T =
yj +
+
equation)
the
equations
of translation.
a rotation In order to discoverthe equationsdescribing the origin of the coordinatesystem,let P = (x, y) clockwise through the angle 8 to its new rotate position = p' For purposes of our analysis,let a be the (x', y'). that vector P makes with the positive the position angle
about
x-axis
x =
(see
p
\302\267 i =
Then)
105).
Figure
Ipl cos
a =
Ip/l cos a (since
y = P
\302\267 =
j
Ipl sin a
Ipl
=
(123)
'p/l)
= Ip/l sin a,)
y)
P(x,y))
P'(x',y')) x)
FIGURE
105)))
GEOMETRY)
ELEMENTARY.VECTOR
200)
and) =
X'
\302\267 i =
P'
-
cos(a
Ip/t
=
(J)
\\p/\\ COS
+
y' =
P'
\302\267 =
-
sill(a
Ip/l
j
=
(J)
a COS(J
\\p/l sin
a cos -
Ip/l cos
a sin 8.) (124),
at)
arrive
X' = { y' = which
(J
right members of
in the
(123)
equations
Imposing
we
(124)
(J
a sin
sin
\\p/l
x cos (J
y sin (J
+
- x sin
(J,)
rotation
the
describes
analytically
y cos
+
(J
(125))
stipulated
above.) EXERCISE
are
Rotations
usually
from (125) by
follow
by a
described
solving
x and
for
pair y.
of
that
equations
(a) Carry out this
algebraicprocedure. (b) Derive these equations directly by vector methodsanalogousto thoseusedto determine(125).) Take
52.
EXAMPLE
We discuss the
circle
the
is at the origin. The
the origin
under the
{
Thus,
the
x-I
y' =
y
circle is
translated
unal
all,
the
circle's
- 1)2+ so that
-2).
2)2 =
+
(y
4.
its new center
The
(1,
point
-2)
translation)
X' =
the algebraictransformation after
at (1,
is centered
circle
given
becomes
(x circle
the
of
translation
did
metric
+ X'2
not
2.)
y'2 = 4. Observe that affect the radius, which is,
+
property
that was to
remain
tered.)
EXAMPLE 30\302\260)clock\\vise,
(1, -2)'?)))
53. If
the points of
what
will
the
be the
plane
are rotated
new location
of
7r/6 (or the
point
MORE The
201)
GEOMETRY
rotation through ?r/6 are:)
of the
equations
-
x'
=
x
-x-
1 2
-
y
2
=
y'
1
-v - 3+
2 \037
+
13 \037
y
2)
Therefore)
X' =
- 1 = approximately
V3
-0.13
2)
y' =
which states
-
1
2
- v/\037
=
.3
-2.24,)
approximately
that)
rotation
) (-0.13,
-2)
(1,
-2.24).
?r 16)
thru
EXERCISES
1. Find the coordinatesof the point moves the origin to (a) (4, 5); (b) (-3, 3); (c) (-5,
2.
the
Find
coordinates
(2,
-3) if the translation
4).
of
(a) (2, 4); (b) (-3, 6); (c) (-2, 0) when the axes are rotated counterclockwise arcsm
through
\302\267 4
the
angle
'5.
the sameas Exercise 2 when clockwisethrough the ?r 1 4. angle 3. Do
4. Show that the equation x 2 of
rotation
the
+
the
y2 =
axes
are rotated
counter-
r 2 is unchanged by any
axes.
surface is generatedby a line moving in such parallel to its original position. If, as the generating line moves, it remains perpendic\037lar to a fixed is the cylinder plane and traces out a circlein the fixed plane,
5. A cylindrical
a
way
termed (a) circular
as
to
right
be always
circular.
Provide cylinder.)))
a reasonable
definition for
the
radius
of a
right
ELEMENTARY VECTOR GEOMETRY)
202) (b)
the cone, define
loose-with
analogy-perhaps
By
(Seethe first
right circular cylinder. Exercises that follow.) for a
the
of
axis
Miscellaneous
the equation of the right methods, (c) Find, by vector in the eircle (x - 1)2+ circular cylinderthat cutsthe xz-plane (z + 1)2 = 1 and whoseaxisis parallelto the y-axis. 6. Let Po = (xo, Yo, zo) be the center of a sphere that has = PI (Xl, YI, Zl) as one of its points. Using vector techniques,
that
prove
(x
-
- ZO)(ZI.-ZO) equation of the planetangent to the sphereat Pl.) -
XO)(XI
is the
XO)
+
(y
a line line
fixed
is
said
at
Yo) + (z
vector
is represented
=0
EXERCISES
always intersectsa angle j-a( < 1r/2), then X cone with axis cC,vertex V,
a way that it
a fixed
angle the
in such
varies
to generate
Prove by vertex
X
cC in
vertex
and
-
YO)(YI
GEOMETRIC
MISCELLANEOUS'
1. If
-
point V a right circular a (see Figure 106). at
methods
origin,
that
axis the
by the equation
an
the right circular
z-axis, and X2
+
y2
-
3z
vertex 2 =
cone, with angle
21r/3,
o.)
.\302\243)
FIGURE
106)))
MORE
203)
GEOMETRY)
(Hint. Let P bethe position Then p. k = (pI cos ?r/3
= (x, y,
of P
vector
vector equation
is the
the cone.
z) on
the
of
cone.)
2. Find the equation the x-axis, vertex is at
that represents the cone whose axis the origin, and vertex angle is ?r 12.
3. Find the equation the z-axis, vertex is at
that represents the cone,whose
(0,0, 2),
vertex
and
4. Given points A, B, and R in space, distance d from R to line AB. \037
\037
and AB make an angle 8, then d
If AR
(Hint.
\037
sin
but
be determined from that (i) the equation
8 may
Observe
and (ii) interchanging
5.
Prove,
by
vector
of
trapezoidto the endpoints
the
of
one of
the
sin
IARl
AB.) is never
AB
and B
that the
methods,
by joining the midpoint
X
line
of A
the'roles
\037
=
8;
\037
AR of
for the
formula
a
is
is 2?r/3.
angle find
axis
is
needed;
yields a check.
area of
formed
a triangle
sides of a the nonparallel half that of side is opposite
trapezoid.
6. Prove inclined
to
by vector methods: If a straight line is equally three coplanar lines, it is perpendicularto their
plane.
7.
Prove:
sum
The
hedron is equal of the midpoints
8.
Find
regular
the
to
of the four
times
squares of the the
sum
edges of
of the
tetra-
any
squares
of
joins
of opposite edges.
angle
tetrahedron.)))
between
two nonintersecting
edges
of
a
appendix-
expansion
of
determinants)
the
For
we
determinants,
method
is unfamiliar a brief provide
\\vho
reader
of determinant
not attempt to
expansion.
with the theory of description of the This discussion does
mathematical
communicate
insight
device theory but merely provides a mnemollic of a compact ing with cross products by means
The
two by two determinant,
for
notation.
written :\037
sents
The
a quantity
three
according to the following al
a2
b1
b 2)
by three
into deal-
::
repre-
expansion:)
.
= a 1b2
determinant 204)))
-
a2b 1.
is expanded as fo110v,\"s:
OF
APPENDIX-EXPANSION
al
a2 ag
b l
b2
bg
Cl
C2
Cg
+ ag
=
al
b2
b g
C2
Cg
b 1 b2
=
- a2
a 1b 2C3
C2
Cl
b1
bg
Cl
Cg
- a 1b 3 C 2 - a 2 b 1c3
+ a2bgCl that
(Observe
205)
DETERMINANTS)
expression
-B X A.)
There are
various
to
\037pproaches
and there are
minants,
of
equivalent
algebraic
B =
2 Cl.)
aab
the second and third rows of the one we negative
interchanging
leadstoal1 which is the have computed. This is the A X
-
aab 1C2
+
several
deter-
expanding
devices
by
one
which
is not the expansion. However, the readerwho familiar with determinants would be wise to stickto the in this volume. of cross products particular expansion In accordance we view the with the expansion, given
recalls
cross
product)
(ali
as equal i
j
k
a2
aa
b2
a2j +
aak ) X (bli +
b
+
2j
bak))
to)
al b1
+
=
i \\
b a) -
a2
a3
b2
b
(a2 b 3
_
j
a
- a3b 2)i
+
al
a3
b l
bg
(agb l
al + k b \\
i
- a1ba)j
+
(a
1b 2
a2 b2
- a2b l)k.)))
answers)
4.
6.
2
l =
(a)
SECTION
3)
SECTION
4
6. 1000.)
.y 89 .
m
=
'3 5.
\"5'
4
(b) l =
m
=
SECTION 6.
3
.
7.
\"7'
(c) l =
5-3 2'
m
=
2.)
6)
are equal.)
They
SECTION 8
- 10k, C 4i 10j + 4k. (c) 4i + 12j- 6k. \037 = 6i + 6j + 14k. BD (d) A (f) (-18, 4, -30). (g) (-18, 4, -30). 2. (a) 4i - 8j. (b) 3i + 6j. (c) -7i + 2j. (e) Yes. (f) Yes.
1.
A
(b)
(2,
+ 4j
+ 10k,B
=
-6i
- 10k,D = B = 12i + 20k,
- 6j
=
(g)
= 6i
3).
-3,
(a) No.
3. No.4.
4 /-
\037 \037
.
\302\267
(31 + 2J) or 13
- 4
/-
\037 \037
.
(b)
V
=
158 A +
3
3
B + C.
14
SECTION
3 \302\267
+ 4j
(31 +
. 2J.)
13 206)))
.y2i 4\302\267 7
0 (a-II. )
\302\267 1.
3
207)
ANSWERS)
(b)
-11 V6
6
=
y
(c)
16
(c) 5x +
=
2y
33.
5-
(b) x =
+j).
+t(3i
+4j)
t)(5i
_ . 2
x
= 1.
2t,
33 33
-5
2. (a)
\037
5t.
4 +
.
SECTION
? = (1 -
1. (a)
V29
29
11.
4 units.
(d)
(2i + j - k). 29 ;;\037 Because the dot product is distributive.) -11
and
-.
-,
(b)
5 2 3. (a) 9x + 5y + 25 = o. (b) y = 2x + 6. (c) 2x - y = 1. 4. (6, 0). 6. (a) (2, -1) (b) x + 3y + 1 = o. (c) x = -2 + (7 V 34 + 5v'58)t, y = 1 - 3(V 34 + V 58)t. (a) 3y = V3 9. (c) x = o.)
x
7.
- (6 + V3).
(b)
4
.
bj. 2. (a) . . (21 - J ).)
ai +
VS
+
5
2y =
x +
o. 3. x
8
0
. 2.
6.
4. o.
O.
3.
O.
9. -7
=
14.
5x
-
ml
\302\267 8.
(a)
mlm2
\037\037 +V 50.
m(x
(a)
m2
-
1 +
13. (a)
+
4y
=
6.
11 5\"\
19
SECTION
6. tan 8
-2.
18
SECTION
1.
=
y
17
SECTION
1. N =
(c)
10.
y
1) + n(y
+ 4y
- 2=
5V3
.
=
9 1r 25 -, arc tan -,-. 9 25 2
arc tan
+9V2
x.
12. x
-
8y
=
O.
9
- 2) = o. O.
(b)
SECTION
7x
(b)
y
- 32y
= V3
3
x + b.
- 3 = O.
20
- 1)2= 4. (b) x 2 + (y + 1)2 = 4. (c) (x + 2)2 + (y - 3)2 = 9. 2. (a) (x - 2)2 + (y 18 = 2 o. x - :y 5x + 5y2 - \037 (b) 5 (c) x2 + y2 - 8x + 12y = o. 3. (a) (8, -6).)))
1. (a) x 2 + (y
+
1)2 =
5.
208
10.
4 2 3' 3')
-
(d)
(
7. (a) 9z + 2y 9. (a) x = -1 x
= 1
10.
(1,
(b)
(a)
+
V65 3
=
40.
+
5
_/-
\302\267 4.
(c)
141 9,
\037V
(0, 1).
0) and
(a)
_
5V2
/_'
0) and
(1,
/_
_
5V2
and
2. (a) 6x +
4y
+ 7z
(e) 4z -
+
12z
1)2 +
26
case. The line is is nonAgain, two-point form
(c)
- z+2 2+2
x+4
LIne
y
b
3+4
2+5
-5
and 9
+
=
4.
1 +
-5 (a)
=
t.
6_
-4 _ 0
- 0
5
z =
z
1.
x-O = y-O = z-O
(g)
y
=
2t,
Z
=
+ t.
-2
- 6t. (f) x = -4
6, 2, 7.
:1. 6.
(b)
x
in this
non-existent
is
x-3 = y-2 = 3 - 2 2 - 10
(e)
=
.
2.
-1).
an d
x-3 3
- 2
(b)
x
=
=
=
y
(c) x = 3
+
3t,
- 34 +
8.
+
y-2
-4
20t,
Y
/_
_
/_
=
14
2t,
-2
/-
181'V
+ 12t.
+ 18t,
1
6
V 181' V
-
== 39
_
z+5
+ 9t, Z
-
=
10 -5 + 1
= 2
+ 4t,
.
-12 _
1i
3t,
= 1 y
1.
2-
O
7\"
(-12,4,
4.
SECTION 24 (c) 2x - y - z = O. 19 88 = 76. 3. . (a) -.13 (c)-. 5)
... IS gIven
eXIStent.
=
Z2
= 17.
Two-point form by x = -4, y
.
1)2 +
+
(y
(m ;:C0).)
-m
m,
3m,
-
(x
SECTION
given
1. (d) 0, 1, O.
0, 0,
(c)
\"'\" j_.
5V2
4. (a) 3, 1, -1
1. (a)
(0, 1).)
SECTION 21
2. (a) (0, 1,0), 1. 3. (a) 2 (c) x + y2 + z2 = 144.
3y
-
(b)
431
1.
-x
= 2
=
Y
+ 2 V2. 5 + COB 8. 31.../+ V 1418. 2 6
=
y
COB 8, Y
= 24.
3x + 4y
(a)
GEOMETRY
VECTOR
ELEMENTARY
181.)))
209)
ANSWERS)
28)
SECTION 1.
In order
3.
(9,
2
-
10
(
Use vectors
first
be
(3
- 33'
6.
12).
-5,
6. (a)
1r
to have
69'
-
20
to the
normal
angle.
quadrant
- 39 SIn. '
)
20
8.
=
(3
21
(c) cos 9
planes.
13
_ .
=
V45)
8. (a) cos9
1
=
'
(c)
. 8m
1
=
fJ
V3
1.
(a) k - j.
V3)
32
SECTION O.
(c)
- 4k.
(e) -8j
2. Usetriple \037
6. (b) x
+
y
+ z
7. d =
= 1.
A
AB
-
(c)
11.
3.
-
blC2
b2 C l,
cross
their
is the
product
product.
X
(a) o.
9.
Acl
a2bl-)
33
SECTION X
.
\037
- alC2, alb2 +
a2Cl
- 7j + 5k. 6. A
2. -13i
X AC
\037
lAB 2
scalar
\037
Band
C X D
zero vector.)
are parallel.
Hence
SECTION36
2.
the bounding Annular centered at origin (including (a) region circles. (b) Spherical shell including the spheres. bounding 3. A torus whose circular cross-section has a radius of (doughnut) 4. (a) 4 < x 2 + y2 < 9. one. (b) 0 < x < 1 and 1 - x < y < x + 1.)
4. If
T4D,
TaC +
6. At
is
ABCD
with
SECTION 38 a tetrahedron or a square then 0 < Ti < 1 and Tl + T2 + Ta
least one coefficient
equals
zero.
X
+
T4
9. (a) (\037,
M
=
j-A +
10. (a)
X
and Tl +
= T2
iB +
-lC.
TlA + +
Ta
=
(b)
(1,
= TlA = 1.
\037).
1).
T2B + TaC,with 0 < Ti < 1 1. (c) Insist that 0 < Ti <
1.)))
+
T2
B
+
VECTOR
ELEMENTARY
210)
GEOMETRY)
SECTION 39
1. (a) Note
500 for maximum profit. 100 bitter. (b) 100milk, 300 semi-sweet, 2. max = 7, min = 5. 3. 400 gal regular, 300 gal high 200 gal high test. 300 gal white. 4. 800 gal regular, 6. i A-ration, 4 B-ration.) m + 8
that
x
= r8
b
=
sin
a
- a sin
8,
8, y
- a cos8. - a cos 8.)
y = r = r
42
SECTION
1. (a)
6.
(c)
(c) (:-3, 1).
(6, 2). -
(x
1)2 + (z +
2.
(c)
(a)
e51
1)2= 1.)
y2
z2.
+
\037
\037
4.
d =
\\AR
3.
3(z -
X
ARI .
\037. 8 \302\267 2)))
IABI
,\037}
EXERCISES
MISCELLANEOUS
2. x 2 =
test,
41
SECTION
2. (a) x = r8 (b)
+
2)2
=
x
2
+
y2.
6.
( 5
58}
index)
Absolute
value,
Addition
of
14, 57-9 9, 11-13
vectors,
Altitudes, 75 technique
')
30
Axioms, 1
196
Cylinder,
Bound vector,
9
201)
9,
Commutativity,
Dimension,
187
63,
Direction
73
68-70,
Distance,60-1
Coxeter,
76
methods,
vectors, 27, R.,
Courant,
34
181
H. S.
product,
147-50)
M., 181 135
108-10,
cosines,
Direction numbers,
Convexity, 167-9 Coordinate-free
appendix
141,
49
Direction angles, 108
Cone, 202
Coplanar
21 fi.
191
54-9
Complex plane,
1
science,
Determinants,
Circle, 100 fi.
Components,
Deductive
Dependence,
Desargues'theorem,181fI.,188,
19)
lemma,
Bypass
66-7
perpendicular,
Bisectors,
triple,
196
Prolate,
46-9
Basis,
Cross
191-3
Cycloid,
Curate,
point
Auxiliary
Curate cycloid, 196
Distributivity,
Division
125
110-1,123-5 139
65,
of segments,
Dodecahedron,54 Dot product, 62 fi.')
27-9, 34
ff.
Equilibrium,
10,
20)
211)))
212)
INDEX)
7-8, 10,
Force,
Freevector,
20, 73
Parallel
6-7,
Parametric
11
Galileo,
(= spans), 48 Great circle, 157)
Pencil of
146, 191 100
lines,
63, 89
Perpendicularity,
Planes, 111ff.
19
D. W.,
87
Hero's formula, 156
Point-slope
Hilbert, David, 187)
Positive triple, 43
\"If
Projective geometry, 180fi. Prolate
16
146)
Radial vectors,
Law of
Left-handed
Linear
triple, 43 fi. Rigid motions, 197-201
Right-handed
43 fi.
Donovan,
170 170 fi.
programming,
Scalar, 4-5 20-21,
combination,
Linear dependence, 7-8,
Mechanics,
Menelaustheorem, Model,
Negative Newman,
21 fi.)
10, 68-9,
73
43
40 theorem,
Parallelism,86)
12 fI.)
fI.
Segment division, 27-9, 34
Set theory,
21,
Slope,
165, 167
35, 85, 19
81
Slope-intercept
form, 83
Space,49 Span,
Sphere, unit,
Statics,
ff.
48
106 fi. 108-10
10
Staudt, K. 185,
63
137-43
Shortcut lemma,
43 ff. triple, J. R., 53)
Orientation,
Scalar,product, triple,
188-90
by scalar,
Ordered set, Pappus'
35
3
Multiplication
197, 199-201)
Rotation,
Length, 57-9 Linear
181
H.,
Robbins,
7-8
action,
156-7
Rhombus,
150-1, 158-9
triple,
Lichtenberg,
Resultant, 10, 20
158
150,
Law of sines,
Line of
of, 150
identity cosines,
56
52-4,
43
Radians, Lagrange,
196
cycloid,
Pyramid,
S., 148)
Murray
170 fi.
64-6
Projection,
Involute, 193-6) Klamkin,
2
linear,
Programming,
81
87
additive,
fi.
parallel,
Postulates, 1
Inclination, angle of, 60 fi. Inner product, Inverse,
46, 49
vector,
Postulate,
only if\", 6 Incidence, 180, 185
Intercept form,
form\037
Position
and
78-
representation,
80, 122,
Generates
Hall,
2
postulate,
Parameter, 78
9)
188, 191
Stevin,
G. C. von,
Simon,
Straight line, 77
187
10, 11 fi.,
121
fi.)))
213)
INDEX)
16-17
Subtraction,\"
Terminus (= endpoint), 5, 9
Traces,114-15
147-50
Triple scalar product, 137-43 196
Trochoid,
80, 126)
form,
Two-point
Uniqueness of 34 Unit
representation,
magnitude
dependence of, of, 4, 6
origin
of, 6
108
Unit vector,
radial, 52-3
subtraction of,
Vector(s),
of,
9, 11-13
5
zero,
definition
of,
5, 12,
of, 4,
6,..13)
14, 22 136 fi.
product,
Vector
product, triple,
Weyl,
Hermann,
Work,
73)
Zero vector,
4
16
Vector
bound, 9 direction
7,
9
unit,
5)
addition
21 \302\243f.
multiplication by scalar, 12ff.
ff.
sphere,
20 ff.,
35
sum of,
Union, 167
of, 7-8
action
linear combinations of, linear
197-9
cross product,
Triple
10
free, 6-7, 9
line of
Translation,
of, 6,
equality
Vec.tor(s),
53)
Symmetry,
Zweng,
147-50)
53
5, 12,14,22
Marilyn,
170)))
schLlster)
seymour primarily
matical
tool
trigonometry, both plane and
school students and theorems
by employing
spherical,
and
vector and
features
text
it offers
greater
to
for high into
insights
than synthetic
rather
proofs,
analytic
mathe-
applications
Appropriate
algebra.
undergraduates,
college
algebra as a
of vector
with the development in geometry, this elementary
Dealing
proofs.) with an overview of Starting tors, explores their fundamental
discusses
Subsequentchaptersexamine
vectors
vec-
defines
and
linear combinations,
and
properties
technique
point
auxiliary
elementary operations, the text
and uniqueness of representations. In coordinate systems, properties
and elements of analytic geometry-the methods of proof, and circles, spheres, and planes. The text concludes with surveys of cross products, plane and spherical
and
of inner
formulas
products,
line, analytic
straight
and additional geometric convexity and linear programming. book, with solutions at the end.)
trigonometry,
and
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