Vectors in 2-Space and 3-Space
Chapter Contents
Introduction to Vectors (Geometric) Norm of a Vector; Vector Arithmetic Dot Product; Projections Cross Product Lines and Planes in 3-Space
1 Introduction to Vectors (Geometric)
Geometric Vectors
Symbolically, we shall denote vectors in lowercase boldface type. All our scalars will be real numbers and will be denoted in lowercase italic type
terminal point initial point
• The vector of length zero is called the zero vector and is denoted by 0. • Since there is no natural direction for the zero vector • the negative of v, is defined to be the vector having the same magnitude as v, but oppositely directed.
Definition
If v and w are any two vectors, then the sum v+w is the vector determined as follows: Position the vector w so that its initial point coincides with the terminal point of v. The vector v+w is represented by the arrow from the initial point of v to the terminal point of w.
Definition
If v and w are any two vectors, then the difference of w from v is defined by v – w = v + (-w)
Definition
If v is a nonzero vector and k is nonzero real number (scalar), then the product kv is defined to be the vector whose length is |k| times the length of v and whose direction is the same as that of v if k > 0 and opposite to that of v if k < 0. We define kv =0 if k = 0 or v = 0.
A vector of the form kv is called a scalar multiple.
Vectors in coordinate Systems(1/2)
In Figure 3.1.6, that v has been positioned so its initial point is at the origin of a rectangular coordinate system. The v1 , v2 of coordinates the terminal point of v are called the components of v, and we write
v = (v1 , v2 )
Vectors in coordinate Systems(2/2)
If v = (v1 , v2 ) and w = ( w1 , w2 ) two vectors are equivalent if and only if v1 = w1 and v2 = w2 and
Vectors in 3-Space (1/4) coordinate axes
origin
rectangular coordinate system
• Each pair of coordinate axes determines a plane called a coordinate plane. These are referred to as the xyplane, the xz-plane, and the yz-plane. • To each point P in 3space we assign a triple of numbers (x, y, z), called the coordinates of P.
Vectors in 3-Space (2/4)
Rectangular coordinate systems in 3-space fall into two categories, left-handed and righthanded. In this book we shall use only right-handed coordinate systems.
Vectors in 3-Space (3/4)
A vector v in 3-space is positioned so its initial point is at the origin of a rectangular coordinate system. The coordinates of the terminal point of v are called the components of v, and we write v = (v1 , v2 , v3 )
If v = (v1 , v2 , v3 ) and w = ( w1 , w2 , w3 ) are two vectors in 3-space, then
v and w are equivalent if and only if v1 = w1 , v2 = w2 , v3 = w3 v + w = (v1 + w1 , v2 + w2 , v3 + w3 ) kv = (kv1 , kv2 , kv3 ), where k is any scalar
Vectors in 3-Space (4/4) Sometimes a vector is positioned so that its initial point is not at the origin. If the vector P1 P2 has initial point P1 = ( x1 , y1 , z1 ) and terminal point P2 ( x2 , y2 , z 2 ), then P1 P2 = ( x2 , y2 , z 2 ) − ( x1 , y1 , z1 ) = ( x2 − x1 , y2 − y1 , z 2 − z1 )
In 2 - space the vector with initial point P1 ( x1 y1 ) and terminal point P1 ( x1 y1 ) is P1 P2 = ( x2 − x1 , y2 − y1 )
Example 1 Vector Computations with Components If v=(1,-3,2) and w=(4,2,1),then v + w=(5,-1,3), 2v=(2,-6,4) -w=(-4,-2,-1), v – w=v + (-w)=(-3,-5,1)
Example 2 Finding the components of a Vector The components of the vector v = P1 P2 with initial point P1 = (2,−1,4) and terminal point P2 (7,5,−8) are v = (7 − 2,5 − (−1), (−8) − 4) = (5,6,−12)
Translation of Axes
In Figure 3.1.14a we have translated the axes of an xy-coordinate system to obtain an x’y’-coordinate system whose O’ is at point (x ,y)=(k ,l ). A point P in 2-space now has both (x ,y) coordinates and (x’ ,y’) coordinates. x’= x – k , y’= y – l , these formulas are called the translation equations. In 3-space the translation equations are x’= x – k , y’= y – l , z’= z – m where ( k, l, m ) are the xyzcoordinates of the x’y’z’-origin.
Example 3 Using the Translation Equations (1/2)
Suppose that an xy-coordinate system is translated to obtain an x’y’-coordante system whose origin has xy-coordinates (k ,l )=(4,1). (a) Find the x’y’-coordinate of the point with the xy-coordinates P(2,0) (b) Find the xy-coordinate of the point with the x’y’-coordinates Q(-1,5)
Example 3 Using the Translation Equations (2/2)
Solution (a). The translation equations are x’=x-4, y’=y-1 so the x’y’-coordinate of P(2,0) are x’=2-4=-2 and y’=0-1=-1. Solution (b). The translation equations in (a) can be written as x=x’+4, y=y’+1 so the xy-coordinate of Q are x=-1+4=3 and y=5+1=6.
2 Norm of a Vector; Vector Arithmetic
Theorem 2.1 Properties of Vector Arithmetic
If u, v and w are vectors in 2- or 3-space and k and l are scalars, then the following relationship
Norm of a Vector (1/2)
The length of a vector u is often called the norm of u and is denoted by u . Figure (a): it follows from the Theorem of Pythagoras that the norm of a vector u = (u1 , u2 ) in 2-space is u = u12 + u22 Figure (b): Let u = (u1 , u2 , u3 ) vector in 3-space.
be a
u = u12 + u 22 + u32
A vector of norm 1 is called a unit vector.
Norm of a Vector (2/2)
If P1 = ( x1 , y1 , z1 ) and P2 ( x2 , y2 , z2 ) are two points in 3-space, then the distance s between them is the norm of vector P1 P2 P1 P2 = ( x2 − x1 , y2 − y1 , z 2 − z1 )
Similarly in 2-space:
the length of the vector ku :
ku = k u
Example 1 Finding Norm and Distance The norm of the vector u = (-3,2,1 ) is u = (−3) 2 + (2) 2 + (1) 2 = 14 The distance d betwwen the points P1 (2,-1,5) and P2 = (4,−3,1) is d = (4 − 2) 2 + (−3 + 1) 2 + (1 + 5) 2 = 44 = 2 11
3 Dot Product; Projections
The Angle Between Vectors
Let u and v be two nonzero vectors in 2-space or 3-space, and assume these vectors have been positioned so their initial points coincided. By the angle between u and v, we shall mean the angleθ determined by u and v that satisfies 0 ≤ θ ≤ π.
Definition
Example 1 Dot Product
Component Form of the Dot Product (1/2) Let u = (u1 , u 2 , u3 ) and v = (v1 , v2 , v3 ) be two nonzero vectors. If as shown in figure 3.3.3, θ is the angle between u and v, then the law of cosines yields 2
2
2
PQ = u + v - 2 u v cosθ
(2)
Since PQ = v − u , we can rewrite (2) as 2
1 2
2
2
u v cosθ = ( u + v − v − u ) or
1 2
2
2
2
u ⋅v = ( u + v − v − u )
Component Form of the Dot Product (2/2) Substituti ng 2
2
2
2
u = u1 + u 2 + u3 , and
2
2
2
v = v1 + v2 + v3
2
v − u = (v1 − u1 ) 2 + (v2 − u2 ) 2 + (v3 − u3 ) 2
we obtain after Simplfying u ⋅ v = u1v1 + u2 v2 + u3v3 Similarly in 2 - space :
2
u ⋅ v = u1v1 + u 2v2
The formula is also valid if u=0 or v=0.
Finding the Angle Between Vectors
If u and v are nonzero vectors then
u ⋅ v = u v cosθ it also can be written as
u ⋅v cosθ = u v
(1)
Example 2 Dot Product Using [3]
Example 3 A Geometric Problem
Theorem 3.1
Example 4 Finding Dot products from Components
Orthogonal Vectors
Perpendicular vectors are also called orthogonal vectors. In light of Theorem 3.l.1b, two nonzero vectors are orthogonal if and only if their dot product is zero. To indicate that u and v are orthogonal vectors we write u v.
Example 5 A Vector Perpendicular to a Line
Show that in 2-space the nonzero vector n=(a,b) is perpendicular to the line ax+by+cz=0. Solution Let P1 ( x1 , y1 ) and P2 ( x2 , y 2 ) be distinct points on the line, so that ax1 + by1 + c = 0 ax2 + by 2 + c = 0
(6)
Since the vector P1 P2 = ( x2 − x1 , y2 − y1 ) runs along the line (Figure 3.3.5), we need only show that n and P1P2 are perpendicular. But on subtracting the equations in (6) we obtain a( x2 − x1 ) + b( y2 − y1 ) = 0 which can be expressed in the form ( a, b) ⋅ ( x2 − x1 , y2 − y1 ) = 0 or n ⋅ P1 P2 = 0 Thus, n and P1P2 are perpendicular.
Theorem 3.2 Properties of the Dot Product If u, v and w are vectors in 2- or 3-space and k is a scalar, then:
An Orthogonal Projection (1/2)
To "decompose" a vector u into a sum of two terms, one parallel to a specified nonzero vector a and the other perpendicular to a. Figure 3.3.6: Drop a perpendicular from the tip of u to the line through a, and construct the vector w1 from Q. Next form the difference: w2 = u − w1 then w1 + w2 = w1 + (u − w1 ) = u
An Orthogonal Projection (2/2)
The vector w1 is called the orthogonal projection of u on a or sometimes the vector component of u along a. It is denoted by proj u (7) a
The vector w1 is called the vector component of u orthogonal to a. Since we have w2 = u − w1 , this vector can be written in notation (7) as
w2 = u − proja u
Theorem 3.3
An Orthogonal Projection (cont) A formula for the length of the vector component of u along a can be obtained by writing
proja u =
u⋅a a
whitch yields
2
a =
u⋅a a
2
a =
u⋅a a
2
a
u⋅a proja u = a
If θ denotes the angle between u and a, then u ⋅ a = u a cos θ, so that (10) can also be written as proja u = u cosθ
Example 6 Vector Component of u Along a Let u = ( 2,−1,3) and a = ( 4,−1,2). Find the vector component of u along a and the vector component of u orthogonal to a. Solution :
u ⋅ a = ( 2)( 4) + ( −1)( −1) + (3)( 2) = 15 a
2
= 4 2 + ( −1) 2 + 2 2 = 21
Thus, the vector component of u along a is u ⋅a 15 20 5 10 proj a u = a = − = − ( 4 , 1 , 2 ) ( , , ) 21 7 7 7 2 a and the vector component of u orthogonal to a is u − proj a u = ( 2,−1,3) − ( 207 ,− 57 , 107 ) = ( − 67 ,− 72 , 117 ) Verify tha t the vector u − proj a u and a are perpendicu lar by showing that their dot product is zero.
Example 7 Distance Between a Point and a Line (1/2) Find a formula for the distance D between point P0 ( x0 , y0 ) and the line ax + by + c = 0. Solution :
Let Q( x1 , y1 ) be any point on the line and position the vector n = (a, b) so that its initial point is at Q. By virtue of Example5, the vector n is perpendicular to the line (Fig 3.3.8). As indicated in the figure, the distance D is equal to the length of the orthogonal projection of QP0 on n; thus, D = projn QP0 = But
QP0 ⋅ n n
QP0 = ( x0 − x1 , y0 − y1 ),
QP0 ⋅ n = a( x0 − x1 ) + b( y0 − y1 ),
n = a 2 + b2
Example 7 Distance Between a Point and a Line (2/2) Solution (count)
so that D=
a ( x0 − x1 ) + b( y0 − y1 ) 2
2
(12)
a +b Since the point Q( x1 , y1 ) lies on the line, its coordinates satisfy the equation of the line, so ax1 + by1 + c = 0 or c = −ax1 − by1 Substituti ng this expression in (12) yields the formula D=
ax0 + by0 + c 2
a +b
2
(13)
Example 8 Using the Distance Formula It follows from Formula (13) that the distance D from the point (1,-2) to the line 3x + 4y - 6 = 0 is
D=
(3)(1) + 4( −2) − 6 2
3 +4
2
=
− 11
11 = 25 5
4 Cross Product
Cross Product of Vectors
Recall from Section 3 that the dot product of two vectors in 2-space or 3space produces a scalar. We will now define a type of vector multiplication that produces a vector as the product, but which is applicable only in 3-space.
Definition
Example 1 Calculating a Cross Product
Theorem 4.1 Relationships Involving Cross Product and Dot Product
Example 2 u v Is Perpendicular to u and to v
Theorem 4.2 Properties of Cross Product
Example 3 Standard Unit Vectors
Example 3 Standard Unit Vectors (count) Obtaining the following results: i i j
i=0 j=k i = -k
j j=0 j k=i k j = -i
k k=0 k i=j i k = -j
Figure 3.4.2 is helpful for remembering these results. Referring to this diagram, the cross product of two consecutive vectors going clockwise is the next vector around, and the cross product of two consecutive vectors going counterclockwise is the negative of the next vector around.
Determinant Form of Cross Product (1/2)
A cross product can be represented symbolically in the form of 3 3 determinant:
For example :
if u = (1,2,-2) and v = (3,0,1), then i j k u × v = 1 2 − 2 = 2i − 7 j − 6k 3 0
1
Determinant Form of Cross Product (2/2)
Warning: It is not true in general that u (v w) = (u v) w
For example i
and so that
(j j)= i 0= 0 (i j) j = k j= -i i (j j) (i j) j
We known from Theorem 3.4.1 that u v is orthogonal to both u and v. If u and v are nonzero vectors, it can be shown that the direction of u v can be determined using the following right-hand rule.
Geometric Interpretation of Cross Product
Lf u and v are vectors in 3-space, then the norm of u x v has a useful geometric interpretation. Lagrange's identity, given in Theorem 3.4.1, states that 2
u×v = u = u = u
2 2 2
2
v − (u ⋅ v ) 2 2
v −u 2
2
2
v cos 2 θ 2
v (1 − cos θ) = u
2
2
v sin 2 θ
since 0 ≤ θ ≤ π, it follows that sinθ ≥ 0, so u × v = u v sin θ
Theorem 4.3 Area of a Parallelogram If u and v are vectors in 3-space, then u × v is equal to the area of the parallelogram determined by u and v.
Example 4 Area of a Triangle
Definition
(7)
Example 5 Calculating a Scalar Triple Product (1/2)
Example 5 Calculating a Scalar Triple Product (2/2) Note : The symbol (u v) w make no sense since we cannot form the cross product of a scale and a vector. It follows from (7) that u(v
w)= w · ( u
v)= v · ( w
u)
This relationship can be remembered by moving the vector u, v, and w clockwise around the vertices of the triangle in Figure 3.4.6.
