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Section 9.3: The Dot Product
Practice HW from Stewart Textbook (not to hand in)
p. 655 # 3-8, 11, 13-15, 17, 23-26
Dot Product of Two Vectors
The dot product of two vectors gives a scalar that is computed in the following manner.
In 2D, if a =  a1 , a2  and b =  b1 ,b2  , then
Dot product = a  b  a1b1  a2b2
In 3D, if a =  a1 , a2 , a3  and b =  b1 , b2 , b3  , then
Dot product = a  b  a1b1  a2b2  a3b3
Properties of the Dot Product p, 654
Let a, b, and c be vectors, k be a scalar.
1. a  b  b  a
2. a  (b  c )  a  b  a  c
3. 0  a  0
4. k (a  b)  (k a )  b  a  ( k b)
5. a  a  | a | 2
2
Example 1: Given a  2i  j  2k and b  i  3 j  2k , find
a. a  b
b. b  a
c. a  a
d. | a | 2
e. (a  b)b
f. a  (2v )
Solution:
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Angle Between Two Vectors
Given two vectors a and b separated by an angle  , 0     .

Then
cos  
ab
|a| |b|
Then we can write the dot product as
a  b  | a | | b | cos 
Example 2: Find the angle  between the given vectors a = < 3, 1 > and b = < 2, -1 >.
Solution:
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Parallel Vectors
Two vectors a and b are parallel of there is a scalar k where a = k b .
y
Parallel Vectors
2a
a
-a
x
Orthogonal Vectors
Two vectors a and b are orthogonal (intersect at a 90 0 angle) of
ab  0
b
Orthogonal Vectors
a
Note: If two vectors a and b are orthogonal, they intersect at the angle  

a  b  | a | | b | cos( ) | a | | b | (0)  0
2

2
and
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Example 3: Determine whether the two vectors a and b are orthogonal, parallel, or
neither.
3 1
, 
2 6
b. a  4i  5 j  6k , b  8i  10 j  12k
c. a  4i  5 j  6k , b  5 j  6k
a. a  2, 18  , b 
Solution:
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Projections
Suppose we are given the vectors a and b in the following diagram
b

a
w  proj a b
The vector in red w  proj a b is called the vector projection of the vector b onto the
vector a. Since w is a smaller vector in length the vector a, it is “parallel” to a and hence
is a scalar multiple of a. Thus, we can write w = k a, The scalar k is know as the scalar
projection of vector b onto the vector a (also known as the component of b along a). We
assign the scalar k the notation
k = comp a b
Our goal first is to find k. From the definition of a right triangle,
cos  
adjacent
|w|
k


hypotenuse
|b| |b|
Also, the definition of the dot product says that
cos  
ab
|a| |b|
Hence, we can say that
k
ab

|b| |a| |b|
Solving for k gives
k
|b| ab
ab

1 |a| |b| |a|
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To get the vector projection, we compute the vector w. To get a vector of length k that
represents the length of w, we multiply k by the unit vector in the direction of a given by
a
. This gives the following result.
|a|
a
ab a
ab
wk


a
| a | | a | | a | | a |2
Summarizing, we obtain the following results.
Scalar and Vector Projection
Scalar Projection of b onto a:
comp a b 
Vector Projection of b onto a:
proja b 
ab
|a|
ab
a
| a |2
Example 4: Find the scalar and vector projections of b onto a if a  0,2,3  and
 2,1,1 
Solution: The scalar projection of of b onto a is given by the formula
comp a b 
ab
|a|
We see that
a  b  (0)( 2)  (2)(1)  (3)(1)  0  2  3  5
and that
| a | (0) 2  (2) 2  (3) 2  13 .
Thus the scalar projection is
Scalar projection of b onto a = comp a b 
ab
5

.
|a|
13
Hence, the vector projection is
Vector projection of b onto a = proja b 
ab
|a|
2

5
10 15
 0,2,3  0, ,  .
13
13 13
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