11.2 Vectors and the Dot Product 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 scalar (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
ab2 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 unit vector 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: i1,0,0 , 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 angle 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 b0 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 ,
ab2(5) 2( 3) ( 1)(2) 2
ab 2 cos ab 3 38
So, the angle between the given two vectors is
102 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 angles 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 cos12 ,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 cos12 ,cos ,cos 3 a14 a 14 a 14
11 0 1 2 0 1 3 0 cos 74 , cos 58 , cos 37 14 14 14
Projections: Vector projection 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 ba1,1,2 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