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
In 3D, if a =
 a1b1  a2b2
 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
Example 1: Given a  2i  j  2k and b  i  3 j  2k
find
a. a  b
Solution:
b. b  a
c.
aa
d. | a | 2
e. (a  b)b
f. a  (2v )
Angle Between Two Vectors
Given two vectors a and b separated by an angle
 , 0  .

Then
ab
cos  
|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:
Parallel Vectors
Two vectors a and b are parallel if 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
0
a angle) of 90
ab  0
b
Orthogonal Vectors
a
Note: If two vectors a and b are orthogonal, they

intersect at the angle   2 and

a  b  | a | | b | cos( ) | a | | b | (0)  0
2
Example 3: Determine whether the two vectors a
and b are orthogonal, parallel, or neither.
3 1
a. a  2, 18  , b  , 
2 6
Solution:
b. a  4i  5 j  6k , b  8i  10 j  12k
c. a  4i  5 j  6k , b  5 j  6k
Projections
Suppose we are given the vectors a and b in the
following diagram
b

w  proj a b
a
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,
adjacent
|w|
k
cos  


hypotenuse
|b| |b|
Also, the definition of the dot product says that
ab
cos  
|a| |b|
Hence, we can say that
k
ab

|b| |a| |b|
Solving for k gives
|b| ab
ab
k

1 |a| |b| |a|
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
.
|a|
This gives the following result.
a
ab a
ab
wk


a
2
|a| |a| |a| |a|
Summarizing, we obtain the following results.
Scalar and Vector Projection
Scalar Projection of b onto a:
ab
comp a b 
|a|
Vector Projection of b onto a:
proja b 
ab
| a |2
a
Example 4: Find the scalar and vector projections of
b onto a if a  0,2,3  and b =  2,1,1 
Solution: (In typewritten notes)