理 CH02
University Physics Harris Benson CH02Vectors
Exercise 2-1 (I) The magnitude of the vectors A and B shown in Fig. 2.24 are A = 3 m and B = 2 m . Find graphically: (a) A + B (b) A − B .
2-24 <> A = 3 cos 30 0 i + 3 sin 30 0 j = 2.60 i + 1.5 j
B = 2 cos 450 i − 2 sin 45 0 j = 1.41i −1.41 j
(a) A + B = 4.01i + 0.09 j (b) A − B = 1.19i + 2.91j
2-24-1
2-24-1 Exercise 2-2 (I) The magnitude of the vectors C and D shown in Fig. 2.25 are C = 4 m and D = 2.5m . Find graphically: (a) C + D (b) C − D .
2-25 1
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<> C = 4 sin 600 i − 4 cos 60 0 j = 3.46 i − 2 j D = 2.5 j
(a) C + D = 3.46i + 0.5 j (b) C − D = 3.46i − 0.5 j
Exercise 2-3 (I) For the three vectors shown in Fig. 2.26, take A = 1.5m , B = 2 m , and C = 1m . Find graphically: (a) A + B + C (b) A − B − C .
2-26 <> A = 1.5 cos 20 0 i + 1.5 sin 20 20 0 j = 1.41i + 0.51 j B = 2 cos 200 i − 2 sin 20 0 j = 1.88 i − 0.68 j
C = − cos 60 0 i − sin 60 0 j = −0.5i − 0.87 j
(a) A + B + C = 2. 2.79i − 1.05 j (b) A − B − C = 0.03i + 2.06 j
Exercise 2-4 (I) Consider the three vectors shown in Fig. 2.26. Find the vector D which when added to A + B − C , produces a null vector. <> A + B − C = 3.79i + 0.7 j D + ( A + B − C) = 0 ⇒
3 .79i − 0.7 j D = C − A − B = −3.
Exercise 2-5 (I) Three vectors have equal magnitude of 10m. Draw a vector diagram to illustrate how the magnitude of their resultant can be: (a) 0; (b) 10m; (c) 20m; (d) 30m. <> 2
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Exercise 2-6 (II) Two vectors have equal magnitudes of 2m. Find graphically the angle between them if the magnitude of their resultant r esultant is (a) 3m; (b) 1m. In each case use the law of cosines to confirm your answer. <>(a) 830 (b) 1510
Exercise 2-7 (I) The resultant of two vectors A and B is 40m due north. north. If A is 30m in the direction 300 S of W, find find B graphically. <>60 m at 25° E of N
In the following problems take the x axis to point east, the y axis to point north, and, if needed, the z axis upward. An unknown vector should be expressed in terms of its components: V = V x i + V y j . The magnitude and direction can be found from the components.
Exercise 2-8 (I) A person person undergoes a displacement of 4m in the direction 400 W of N fol follo lowe wed d by by a displacement of 3m at 200 S of W. W. Find the magnitude and direction of the resultant resultant displacement. <>
Exercise 2-9 (I) Three vectors are specified as follows: A is 5m at 450 N of E, B is 7m at 600 E of S, and C is 4m at 300 W of S. Find Find the magnitude magnitude and directio direction n of their their sum. sum. <>
3
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Exercise 2-10 (I) A person walks 5m south and then 12m west. What is the net displacement? <> =
離
12i − 5 j −12
=
122 + 52
= 13 m
tan
−1
5
=
12
22.60
Exercise 2-11 (I) An insect walks 50cm in a straight line along a wall. If its horizontal displacement is 25cm, what is its vertical displacement?
<> 502 − 252
=
43.3 m
Exercise 2-12 (II) An airplane is flown in the direction 300 W of N. N. If the magnitude magnitude of the the westerly westerly component of the displacement is 100km, how far north does it travel?
<> y =
100 tan300
= 173m
Exercise 2-13 (I) Four vectors each of magnitude 2m, are shown in Fig. 2.27. (a) Express each in unit vector notation. (b) Express their sum in unit vector notation. (c) What is the magnitude and direction of the sum?
2-27 <>(a) (b) 4
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(c) (d)
Exercise 2-14 (I) Each of the vectors in Fig. 2.28 has a magnitude of 4m. (a) Express each in unit vector notation. (b) Express their sum in unit vector notation. (c) What is the magnitude and direction of their sum?
2-28 <>
Exercise 2-15 (I) Given two vectors, A = 2i − 3 j + k m and B = −i + 2 j − k m , find: (a) R = A + B ; ˆ . (b) R ; (c) R <>(a) R = ( 2i − 3j + k ) + (− i + 2 j − k ) = i − j m (b) R = 12 + (−1) 1) 2
=
2 = 1.414 m
ˆ = R = [ 1 (i − j)]m = (0.707 i − 0.707 j ) m (c) R R 2
Exercise 2-16 (I) Given two vectors, C = 4i + j − 3k m and D = 2i − 3 j + 5k m , find: (a) S = C - D ; (b) S ; (c) Sˆ . <>(a) S = (4i + j − 3k ) − ( 2i − 3j + 5k ) = 2i + 4 j − 8k m (b) S = 22 + 42 + (−8) 8) 2
=
84 = 9.17 m
S (c) Sˆ = = 0.218i + 0.436 j − 0.873k S
Exercise 2-17 5
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(II) The vector A has a magnitude magnitude of 6m and vector B has a magnitude of 4m. What What is the angle between them if the magnitude of their resultant is (a) the maximum possible; (b) the minimum possible; (c) 3m; and (d) 8m. Do each part graphically and by components. (Let A lie along the x axis.) <>(a) 0 (b) 1800 (c) (6 + 4 cos θ ) 2 + (4 sin θ ) 2 ⇒
co cos θ = −0.896
⇒
θ = 1530
(d) (6 + 4 cos θ ) 2 + (4 sin θ ) 2 ⇒
co cos θ = 0.25
⇒
θ = 75.50
=9
=
64
Exercise 2-18 (I) The resultant R of two displacements is 10m at 370 W of of N. If the the seco second nd displacement was 6m at 530 N of of E, E, what what was was the first? first? <>
Exercise 2-19 (I) In a yacht race the boats sail around three buoys as shown in Fig. 2.29. What is the displacement from the last buoy to the starting point? Express your answer (a) in unit vector notation, and (b) as a magnitude and sirection.
2-29 <>
Exercise 2-20 6
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(II) The displacement A is 6m east. Find displacement B if the magnitude of A − B is half that of A and points points in the direction direction 300 N of E. <> A = 6i 0
0
A − B = r cos 30 i + r sin 30 j A−B
=
1 2
A
=
3m
Exercise 2-21 (II) A ship ship sails from a point a distance of 4km and a bearing of 400 N of E rela relati tive ve to a lighthouse to a point 6km from the lighthouse at 600 N of of W. (a) (a) What is its displacement? (b) What is the least distance between them? <>
Exercise 2-22 (I) A submarine submarine sails 40km north and then 30km west. What third displacement would produce a net displacement displacement of 20km at at 30 0 S of W? <>
Exercise 2-23 (II) In Fig. 2.30, A and B are position vectors. vectors. Show that the position position vector C of the midpoint of the line joining the tips of these vectors is C =
A + B
2
.
2-30 <>
Exercise 2-24 (II) (II) Give Given n two two vecto ectors rs A = 3i − 2 j and B = −i + 5 j , find the vector C in the xy plane
7
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such that C
=
A + B , but whose direction is perpendicular to A + B .
<>
Exercise 2-25 (I) Given two t wo vector, A = 2i + j − k and B = −3i + 2j + k , find the unit vector in the direction of S = 2B − 3A . <>
Exercise 2-26 (I) (I) Give Given n two two vect vector or,, A = 5i + 2 j , and B = −2i − 3 j , find (a) A + B ; (b) A + B ; (c) A − B ; (d) A − B .
<>
Exercise 2-27 (I) If vector A = 6i − 2j + 3k m , find (a) a vector in the same direction as A with magnitude 2 A ; (b) a unit vector in the direction of A ; (c) a vector opposite to A with a magnitude of 4m. <>
Exercise 2-28 (I) (I) Give Given n two two vecto ectors rs,, A = 2i − 3 j + k and B = −4i + j − 5k , find a third vector, C , such that A − 2B + <> A − 2B + ⇒
C
3
C
=
3 =
0.
0
C = −3A + 6B = −3(2i − 3 j + k ) + 6(− 4i + j − 5k ) = −30i + 15 j − 33k
8
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Exercise 2-29 (II) Show that if the sum of three vectors is zero, they must all lie in the same plane. Does the same restriction apply to the zero sum of four vectors? <>
Exercise 2-30 (I) The vectors A and B have the following components: A x B x
= −1.5m ,
and B y
= −2.5m 2.5 m
= 2m ,
A y
= − 3.5m ,
. Find Find the the magn magnit itud udee and and dire direct ctio ion n of C = 3 A − 2 B .
<>
Exercise 2-31 (I) Find the components of the following vectors: (a) P of length 5m directed directed at 1500 counterclockwise from the +x axis; (b) Q of length 3.6m directed directed at 120 1200 cloc clockw kwis isee from the +y axis. <>
Exercise 2-32 (I) A bod body y moves moves from a positi position on with with coordin coordinate atess (3m, 2 m) to ( −4 m,4 m, 4 m) . Find its displacement (a) in unit vector notation, and (b) in terms of its magnitude and direction. <>
Exercise 2-33 (I) Given that vector A is 5m at 300 N of E, find find vector vector B such that their sum is directed along the negative x axis and has a magnitude of 3m. <> 9
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Exercise 2-34 (II) The hour hand of a clock is 6cm along. Take Take 12 noon to lie along the y axis and 3 p.m. to lie along the x axis. Find the displacement displacement (in unit vector vector notation) of the tip between each of the the following times: (a) 1 p.m. p.m. to 4 p.m.; (b) 2 p.m. to 9:30 p.m. <>
Exercise 2-35 (II) Figure 2.31 shows the directions of three vectors whose magnitudes in arbitrary units are W = 20 , F = 10 , and T = 30 . The x and y axes are titled as shown. Find: (a) the components of the vectors; (b) their sum in unit vectornotation.
2-31 <>
Exercise 2-36 (I) The instruction to a treasure hunt state: Start at the oak tree and walk 5m in a straight line at 300 W of N. N. Turn Turn to your your right through through 450 and walk 4m. Dig a hole hole 2m deep. deep. Where is the treasure relative to the base of the tree? <>
Exercise 2-37 (II) Given the vector A = 2i + 3 j m , find the vector B of length 5m that is perpendicular to A and lies in the following planes: planes: (a) the xz plane; (b) the xy plane. <>
Exercise 2-38 (I) A helicopter helicopter rises 100m from fr om its pad and travels a horizontal distance of 200m at 250 S of W. What What is its displa displaceme cement? nt? 10
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<> −181i − 84.5 j + 100k m
Exercise 2-39 (I) What is the angle between the vectors A = i − 2 j and B = 2i + 3 j ?
<> cos θ =
A⋅B AB
θ = cos −1
(i − 2 j) ⋅ ( 2i + 3 j)
=
−4
=
( 12 + (−2) 2) 2 )( 22 + 32 )
−4
64
= 120
(5)(13)
0
Exercise 2-40 (I) Given the vectors A = −2i + j − 3k and B = 5i + 2 j − k , find: (a) A ⋅ B , and (b) ( A + B) ⋅ ( A − B ) <>(a) A ⋅ B = (−2i + j − 3k) ⋅ (5i + 2 j − k ) = − 10 + 2 + 3 = − 5 (b) ( A + B) ⋅ ( A − B) = A2 − B 2 2
2
= [(−2) 2) + 1 + (−3) 3)
2
] − [52 + 22 + (− 1) 1)2 ] = 14 − 30 = − 16
Exercise 2-41 (I) The dot product of two vectors, whose magnitudes are 3m and 5m, is
−4 m
2
. What
is the angle between them?
<> cos θ =
A⋅B
=
−4
15
2
(3 m) m)(5 m) m)
AB
θ = cos −1
−4 m
= 105.5
=
−4
15
0
Exercise 2-42 (I) The components of two vectors are A x B y
= 1.8 ,
B z
= −2.6 .
= 2.4 ,
A y
Find the angle between them. 11
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= −1.2 ,
A z
=
4.0 , and B x
= −3.6 ,
理 CH02
A = (2.4) 2 + (−1.2) 2 + (4)2
<> A = 2.4i − 1.2 j + 4k B = −3.6i + 1.8 j − 2.6k
cos θ =
A⋅B AB
=
B =
−21.2
(4.82)(4.79)
=
4.82
(−3.6) 2 + (1.8) 2 + (−2.6) 2
=
4.79
= −0.918
θ = 1570
Exercise 2-43 (I) The vectors A and B are in the xy plane where A is 3.2m at 450 to the the +x axis, and B is 2.4m at 2900 to +x axis. axis. Find Find A ⋅ B . <> A ⋅ B = (3.2)( 2.4) co cos1150
= −3.25 m
2
Exercise 2-44 (II) The Vector A and B in Fig 2.32 define two sides of a parallelogram. (a) Express the diagonals in terms of A and B . (b) Show that the diagonals are perpendicular if A = B .
2-32 <>(a) A + B A − B or B − A
(b)
(A + B)(A − B ) = 0 A = B
Exercise 2-45 (II) Show that the angles α , β , and γ between between a vector A and the x, y, and z axes, respectively, respecti vely, are given by cos α =
A ⋅ i A
, cos β =
A ⋅ j A
, cos γ =
A = 3i + 2 j + k , find the angle between A and each axis.
<> A = 3i + 2 j + k ⇒ A
=
32 + 22 + 12 12
/
=
14
23
A ⋅ k A
. If
理 CH02
3
cos α =
14 2
cos β = cos γ =
α = 36.70 β = 57.70
14 1
γ = 74.50
14
Exercise 2-46 (II) Given the three vector A = i − 4 j , B = 3i , and C = −2 j evaluate the following expr expres essi sion onss if they they are are allo allowe wed d math mathem emat atic ical ally ly:: (a) (a) C ⋅ ( A + B) ; (b) C ⋅ ( A ⋅ B ) ; (c) C + + A ⋅ B ; (d) C (A ⋅ B ) ; (e) C( A ⋅ B ) .
<>
Exercise 2-47 (II (II) What is the the component of the the vector tor A = i − 2 j + k m in the direction direction of the vector B = −3i + 4k m ?
<>
Exercise 2-48 (I) (I) Give two vectors, rs, A = i + 2 j − 4k and B = 3i − j + 5k , find A × B . <> A × B = (i + 2 j − 4k ) × (3i − j + 5k )
=
=
i
j
k
1
2
−4
3
−1
5
2
−4
−1
5
i−
1
−4
3
5
j+
1
2
3
−1
k
=
(10 − 4)i − (5 + 12) j + (− 1 − 6)k
=
6i − 17 j − 7k
Exercise 2-49 13
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(I) (a) Show, for arbitrary vectors A and B , that A ⋅ ( A × B) = 0 . (b) How could you have arrived at this conclusion without any computation? <>(a) (b) A ⊥ ( A × B)
Exercise 2-50 (I) Vectors A and B are in the xy plane with A = 3.6m at 250 counte countercl rclock ockwis wisee from the +x axis, and B = 4.4m at 1600 countercl counterclockw ockwise ise from +x axis axis . Find A×B.
<> A = 3.6 cco os 25 250 i + 3.6 si sin 25 2 50 j 0
0
cos160 i + 4.4 si sin 16 160 j B = 4.4 co A×B
Exercise 2-51 (II) Show that the t he area of a parallelogram, as in Fig. 2.32, is A × B .
<> = =
( B )( A sin θ )
=
A×B
Exercise 2-52 (II) Given three vectors, A = 2i − 5 j , B = 4 j , and C = 3i , evaluate the following expr expres essi sion onss if they they are are math mathem emat atic ical ally ly allo allowe wed: d: (a) (a) C ( A × B) ; (b) C ⋅ ( A × B) ; (c) C × ( A ⋅ B) ; (d) C × ( A × B) ; (e) C + A × B .
<>(a) (b) (c) (d) (e)
14
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Exercise 2-53 (II) Vector A has a magnitude of 4m and lies in the xy plane directed at 450 counterclockwise from the +x axis, whereas B has a magnitude of 3m and lies in the clockwise from the the +z +z axis; axis; see see Fig. Fig. 2.33. 2.33. Find Find A × B . yz plane directed at 300 clockwise
2-33 <> A = 4 cos 450 i + 4 sin 450 j = 2 2 i + 2 2 j = 2.83 i + 2.83 j B = 3 sin 300 j + 3 cos 30 0 k
3
3 3
k = 1.5 j + 2.6 k 2 2 A × B = 7. 7.36 i − 7.36 j + 4.25 k =
j+
Exercise 2-54 (II) (II) Find ind a vect vector or of len length gth 5m that that is perp erpendi endicu cula larr to both both vect vector orss A = 3i − 2 j + 4 k m and B = 4i − 3 j − k m .
<> nˆ = nˆ
A×B
=
A×B
14i + 19 j − k 142 + 19 2 + 12
=1
5nˆ =
70 558
i+
95 558
j−
5 558
k = 2.96i + 4.02 j − 0.21k
Exercise 2-55 (I) Vector A is 5m directed at 350 above above the +x axis, axis, both both in the the xy plane. Find A − B . Draw a vector diagram.
<> 15
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Exercise 2-56 (I) Vector A is 5m 5m directed directed at 300 W of N and vector vector B is 3m 3m directed directed at at 200 s of W. Find A + B . Draw a vector diagram. <>
Exercise 2-57 (I) Vector A is 4m directed at 350 above above the +x axis axis and B is 2.5m 2.5m directed directed at at 200 above the –x axis, both in the xy plane. Find B − A . Draw a vector diagram. <>
Exercise 2-58 (I) Given the three vector A = 2 m at 280 N of E, B = 1.5m at 250 W of N, and and C = 2.5m due E, find D = A + 2B − C . Draw a vector diagram.
<>
Exercise 2-59 (i) A person starts a two-stage journey at the origin and first walks 6m in the direction 500 N of E. The The final positio position n is 3.5m 3.5m from the the origin in the directio direction n 350 N of W. What was the second stage? Draw a vector diagram. <>
Exercise 2-60 (I) For the vector A we know A x
<> A 42
=
2
A x
2
+
2
Ay
+
= −2 m
, A y
= 1.5m
2
Az 2
= ( −2) 2) + (1.5) + ( A z )
2
16
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and A = 4 m . What is A z ?
理 CH02 ⇒
A z
=
( 4)2 − (−2)2 − (1.5)2
=
9.75 = ± 3.12 m
Exercise 2-61 (I) The vector A = 2i + 3 j m and the sum A + B is 4m long directed directed at 1200 to the the + x axis in the xy plane. What is B ? <>
Exercise 2-62 (I) Vector A is 4m along the + x axis and A + B is 6m along the + y axis. Determine B.
<>
Exercise 2-63 (I) Three vectors in the xy plane have equal magnitude and their sum is zero. If one is 2i m , express the other two in unit vector notation. <>
Exercise 2-64 (II) The vector A is 2m directed at 300 N of E and and B is directed at 500 N of W. The sum A + B is 2.6m long. Find graphically (a) B , and (b) the direction of A + B . (A compass is needed.) <>
Exercise 2-65 (II) Vectors A and B , both in the xy plane, have the same magnitude. Vector A is directed at
300
abov abovee the +x +x axis axis and and
17
/
B
23
isperpendicular to
A , where
理 CH02
A+B
=
2.12m . (a) Find A and B . Determine A + B given that B y is (b) positive,
or (c) negative. <>
Exercise 2-66 (II) Vector A is 2m directed at 300 N of E, the the sum sum A + B is directed at 200 W of N, and B = 3.5m . Graphically find (a) A + B , (b) the direction of B . (A component is needed.) <>
Exercise 2-67 (I) Two vectors are given by P = 3i − j + 2k m and Q = i − 2 j + 4k m . Fins: (a) P + Q ; (b) P + Q ; (c) P + Q .
<>(a) P + Q = (3i − j + 2k ) + (i − 2 j + 4k) = 4i − 3j + 6k m (b) P + Q
=
42 + (−3) 3) 2 + 62
(c) P + Q = 32 + (−1) 2 + 22 =
14
+
+
=
61 = 7.81 m 12 + (− 2)2 + 42
21 = 3.74 + 4.58 = 8.32 m
Exercise 2-68 (I) Two vector are P = 2i − 3 j m and Q = − i + 2 j m . (a) What is Q − P ? (b) What is the distance between the tips of the vectors? <>(a) Q − P = (−i + 2 j) − ( 2i − 3j) = − 3i + 5j m (b) Q − P = 32 + 52
=
5.83 m
18
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Exercise 2-69 (I) Given the vectors A = 4i − j + 2k m and B = −2i + 3 j − k m , find: (a) A − B (b) A − B (c) A − B .
<>(a) A − B = ( 4i − j + 2k ) − (− 2i + 3 j − k ) = 6i − 4 j + 3k m (b) A − B
=
62 + (−4) 2 + 32
(c) A − B = 42 + (−1) 2 + 22 =
21 − 14
=
=
7.81 m (− 2) 2 + 32 + (− 1)2
−
4.583 − 3.742 = 0.841 m
Exercise 2-70 (I) Given R = (3i − j + 2k ) m , what is the vector that is twice as long in the same direction? <> 2R = 2(3i − j + 2k ) = (6i − 2 j + 4k ) m
Problem 2-1 (I) Find a vector of length 5m in the xy plane that is perpendicular to A = 3i + 6 j − 2k m.(Hint: Consider the dot product.)
<>令 B = (b x , by , 0) xy
A ⋅ B = 0 = (3, 6, −2) ⋅ (b , b , 0) = 3b x
3b x + 6by
0
⇒
b x
B = c(2i − j)
c(−2i + j)
=
B = 5 = c
y
x
+ 6by
= −2by
22 + 12
⇒
B = 2 5i − 5 j = 4.46i − 2.23 j
c =
5
−2
5 i + 5 j = −4.46i + 2.23 j
Problem 2-2 19
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(I) The vector A has a magnitude of of 2m and is directed tower the northeast. Find Find the vector B such A + B
=
2 A given the follow: follow: (a) B has the maximum possible
magnitude; (b) B has the minimum possible magnitude; (c) B points northwest; (d) A + B points south.
<>
Problem 2-3 (I) The magnitudes of vector A and B are equal and the angle angle between them is θ . Sho Show tha that (a) (a) A + B
=
2 A cos
θ
2
, and (b) A − B
=
2 A sin
θ
2
.
<>
Problem 2-4 (II) A rectangular coordinate system with axes x′ and y′ is rotated rotated by angle θ from axes x and y as shown in Fig. 2.34. (a) What are the components of the position vector r in the two coordinate systems? (b) Use part (a) to show that the coordinates of a
point P in the two systems are related by x′ = x cos θ
+
y sin θ , y′ = − x sin θ
+
y cos θ
( Hint int: You will ill need to expand cos(φ − θ ) .) These equations show how the coordinates ( x, y ) transform under the rotation of the axes. A vector is defined as a quantity whose components transform in this t his way.
<>(a) r = r cos φ i + r sin φ j
2-34
r = r cos(φ − θ ) i + r sin(φ − θ ) j
(b) x′ = r cos(φ − θ ) = r cos φ cos θ + r sin φ sin θ y′ = r sin(φ − θ ) = r sin φ cos θ
=
− r cos φ sin θ =
x cos θ + y sin θ y cos θ − x sin θ
Problem 2-5 (II) The edges of a cube of side L lie along x, y x, y, and z z axes, respectively. A face diagonal that lies along a face, and a body diagonal that goes through the cube are shown in Fig. 20
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2.35. Find the angle between: (a) the body diagonal shown and the z axis; (b) two face diagonal on adjacent faces; (c) a face diagonal and a body diagonal that share a corner. (Assign vectors to each line and express each in unit vector notation.)
2-35 <>(a) D = (i + j + k ) L D = 3L
cos γ =
D⋅k
=
(i + j + k ) L ⋅ k 1 =
D
3L
⇒
3
γ = cos −1
1 3
= 54.7
0
(b) A = (i + j) L , B = (i + k ) L ⇒
A ⋅ B = L2
cos θ = (c) cos δ =
A⋅B
=
AD
⇒
2
AB A⋅D
1
=
2 6
θ = cos −1
⇒
1 2
δ = cos −1
=
600
2 6
0
= 35.3
Problem 2-6 (I) Personnel at an airport control tower track a UFO. At 11:20 a.m it is located at a horizontal distance of 2km in the direction 300 N of E at an altitud altitudee of 1200 1200m. m. At At 11:15 a.m. the location is 1km at 450 S of E at at an altitude altitude of 800m; 800m; see see Fig. 2.36. What was the displacement of the UFO?
2-36 <> A = 2 cos 30 0 i + 2 sin 30 0 j + 1.2k = 1.73 i + j + 1.2 k km B = cos 450 i − sin 45 0 j + 0.8k = 0.707 i − 0.707 j + 0.8 k km B − A = −1.02i − 1.71j − 0.4k k m 21
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理 CH02
Problem 2-7 (II) Show that the volume of the parallelepiped in Fig. 2.37, whose edges are defined by the vectors A , B , and C , is given by A ⋅ ( B × C) .
2-37 <>
Problem 2-8 (II) show that
A × (B × C) = B (A ⋅ C) − C(A ⋅ B ) . ( Without loss in generality you can
assume that B lies along along one one axis.) axis.) <>令 B = B x i ⇒
A × ( B x i × C) = Bx i ( A ⋅ C) − C( Ax B x )
Problem 2-9 (I) Show that the polar unit vectors r and
θ
unit vectors according to r = cos θ i + sin θ j ,
in Fig. 2.38 are related to the Cartesian θ = − sin θ
i + cos θ j
2-38 <>Place vectors on a circle of unit radius, take components: r = cos θ i + sin θ j θ = − si n θ
i + cos θ j
Problem 2-10 (II) The position vector of a particle is r = xi + yj + zk . The angles between this vector and the x, y, and z axes respectively are 22
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23
α , β , and
γ . Show that
理 CH02
cos 2 α + cos 2 β
<> cos α =
+ cos
r ⋅i
2
γ = 1 x
=
r
r
理 cos β = y cos γ = z r
2
r
2
2
cos α + cos β + cos γ =
( x 2 + y 2 + z 2 ) r 2
=1
Problem 2-11 (II) A three-dimensional vector A has a length of 10m 10m and makes the angle 650 and 400 with with the the
+ x
and
+ z
axes, respectively. Find the magnitudes of its Cartesian
components.
<> A x
0
= 10 co cos 65 65 =
4.23 m
0
A z
= 10 co cos 40 40 = 7.66 m
A y
=
A2 − Ax2 − Az2
=
102 − 4.232 − 7.66 2
=
96/01/03理更
23
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23.43 = 4.84 m
