11.2 Vectors and the in Three Dimensions

In Section 1.1 we defined the idea of a two dimensional vector. In this section, we extend this concept to three dimensions.

Geometrically, a three-dimensional vector can be considered as an arrow with both a length and a direction. An arrow is a directed line segment with a starting point and an ending point. Algebraically, a three-dimensional vector is an ordered triple of numbers.

Definition: A three-dimensional vector is an ordered triple a a1,, a 2 a 3  of real numbers. The numbers a1,, a 2 a 3 are called the components of a .

First figue is a vector starting at the origin (0,0,0), and the ending point is (1,3,2). A vector does not have to start at origin.

Given the points A( x1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) , the vector with starting point A and ending point B with representation AB ,

a AB  x2  x 1,, y 2  y 1 z 2  z 1 

For example, as shown in the second figure above,

AB axxa1,,,, 2  yya 3  zz aaa 1 2 3 

while OP a1,, a 2 a 3  starts at origin, we call it position vector of the point P a1,, a 2 a 3  .

The magnitude or length of the vector a a1,, a 2 a 3  , denoted by aa, is

2 2 2 a a1  a 2  a 3

Vector Operations: If ,b b1,, b 2 b 3  , and c is a (number), then

a. a b  a1  b 1,, a 2  b 2 a 3  b 3

b. ca ca1,, ca 2 ca 3  Example: Given a 2,  3,5  and b  6,2,  2  . Find ab 2

ab2  2,  3,5  2 6,2,  2 2,  3,5  12,4,  4 14,  7,9 

Vector addition, multiplication of a vector by a scalar are illustrated geometrically in the following figures.

Unit Vector: A is a vector whose length is 1.

In general, if a  0 , then the unit vector that has the same direction as a is 1 a ua aa

We would like to introduce three special unit vectors: i1,0,0  , j  0,1,0  , k  0,0,1 

If a a1,, a 2 a 3  , then we can write

a a1,, a 2 a 3  a 1 ,0,0  0,,0 a 2  0,0, a 3 

=a1 1,0,0   a 2  0,1,0   a 3  0,0,1 

=a1 i a 2 j a 3 k

For instance,  2,3,4   2i  3 j  4 k

Example: Find the unit vector in the direction of the vector 2i  3 j  4 k

Solution: The given vector has length:

2i  3 j  4 k  (  2)2  3 2  4 2  29 So the unit vector with the same direction is 1 2 3 4 2i  3 j  4 k   i  j  k 29 29 29 29

Definition: The dot product of two nonzero vectors a and b is the number

a b a b cos where  is the between the vectors ab and , 0  . If one of them is zero vector, naturally, we define ab 0 .

ab Remark: From the above definition, we have cos  ab

The component formula for the dot product is given as below.

The dot product of a a1,, a 2 a 3  and b b1,, b 2 b 3  is

a b a1 b 1  a 2 b 2  a 3 b 3

From the above formula and definition, we have the following important theorem:

a b0  a  b

Example: Find the angle between the vectors a 2,2,  1  and b 5,  3,2  .

2 Solution: Since a 222  2   1  3, b 52  (  3) 2  2 2  38 ,

ab2(5)  2(  3)  (  1)(2)  2

ab 2 cos  ab 3 38

So, the angle between the given two vectors is

102  cos 1.46 (or 84 ) 3 38

Example: Prove that 3,2,4  2,3,  3  .

Solution: 3,2,4  2,3,  3 3(2)  2(3)  4( 3) 0 3,2,4  2,3,  3 The direction of a nonzero vector a a1,, a 2 a 3  are the angles ,  and  in the interval [0, ] that a makes with the positive x,, y  z  axes (see the following figure).

aaa We have cos12 ,cos   ,cos   3 a a a

Example: Find the direction angles of the vector a 1,2,3 

Solution: a 12  2 2  3 2  14 , so

aa1 2a 3 cos12  ,cos    ,cos  3  a14 a 14 a 14

11  0  1  2  0  1  3  0 cos   74 ,   cos    58 ,   cos    37 14   14   14 

Projections: is easily illustrated by the following figures. Vector projection of ba onto is denoted by proja b . The scalar projection of (also called the component of ba along ) is defined to be the number b cos , where  is the angle between ba and , denoted by

compa b .

ab Scalar projection of : comp b  a a

a a b a a b Vector projection of : projaa b comp b   a a a a a 2 Example: Find the scalar projection and vector projection of ba1,1,2  onto   2,3,1  .

Solution:

Since a (  2)2  3 2  1 2  14 , the scalar projection of ba onto is

ab ( 2)(1)  3(1)  1(2) 3 comp b    a a 14 14

The vector projection is this scalar projection times the unit vector in the direction of a : aa3 3 3 3 9 3 proj b comp b   a 2,3,1 , ,  aaaa14 14 14 7 14 14

One of the applications of dot product is in physics in calculating work. The work done by a constant force F in moving an object through a distance d is W Fd , but this applied only when the force is directed along the line of motion of the object. Suppose, however, the constant force is a vector F PR pointing in some other direction as in the following figure. If the force moves the object from P to Q , the displacement vector is D PQ , so the work done by this force is defined to be the product of the component of the force along D , and the distance moved:

WFD ( cos )

By the definition of dot product, we have

WFDFD( cos )

Example. A force is given by a vector F i 24 j  k and moves a particle from the point P(1,1,0) to Q(3,5,4) . Find the work done.

Solution: The displacement D PQ 3  1,5  1,4  0  2,4,4  . The work done is

WFD 1,2,4   2,4,4  2  8  16  26