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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 , a 2 > and b = < b1 ,b2 > , then
Dot product = a ⋅ b = a1b1 + a 2 b2
In 3D, if a = < a1 , a 2 , a3 > and b = < b1 , b2 , b3 > , then
Dot product = a ⋅ b = a1b1 + a 2 b2 + 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 θ , 0 ≤ θ ≤ π .
θ
Then
cos θ =
a⋅b
|a| |b|
Then we can write the dot product as
a ⋅ b = | a | | b | cos θ
Example 2: Find the angle θ 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 θ =
π
a ⋅ b = | a | | b | cos( ) =| a | | b | (0) = 0
2
π
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
θ
a
w = proja b
The vector in red w = proja 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 θ =
adjacent
| w | | ka | k | a |
=
=
=
hypotenuse | b |
|b|
|b|
Also, the definition of the dot product says that
cos θ =
a⋅b
|a| |b|
Hence, we can say that
k|a|
a⋅b
=
|b|
|a| |b|
Solving for k gives
k=
|b| a⋅b
a⋅b
=
| a | | a | | b | | a |2
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To get the vector projection, we compute the vector w. This gives the following result.
w = ka =
a⋅b
| a |2
a
Summarizing, we obtain the following results.
Scalar and Vector Projection
a⋅b
Scalar Projection of b onto a:
comp a b =
Vector Projection of b onto a:
proja b = (comp a b ) a =
| a |2
a⋅b
| a |2
a
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 |2
We see that
a ⋅ b = (0)(−2) + (2)(1) + (3)(1) = 0 + 2 + 3 = 5
and that
2
| a | 2 = ⎛⎜ (0) 2 + (2) 2 + (3) 2 ⎞⎟ =
⎝
⎠
( 13 )2 = 13 .
Thus the scalar projection is
Scalar projection of b onto a = comp a b =
a⋅b
| a |2
=
5
.
13
Hence, the vector projection is
Vector projection of b onto a = (comp a b) a =
5
10 15
< 0,2,3 >=< 0, , > .
13
13 13
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