Updated: February 4, 2016
Calculus III
Section 12.3
Math 232
Calculus III
Brian Veitch •
Fall 2015 •
Northern Illinois University
12.3 The Dot Product
Definition 1: The Dot Product
Let ~a =< a1 , a2 , a3 > and ~b =< b1 , b2 , b3 >. Then the Dot Product is
~a · ~b = a1 b1 + a2 b2 + a3 b + 3
Note: The dot product is a scalar (NOT ANOTHER VECTOR)
Properties of the Dot Product
1. ~a · ~b = |~a|2
2. ~a · ~b = ~b · ~a
3. ~a · (~b + ~v ) = ~a · ~b + ~a · ~v
4. 0 · ~a = 0 (the zero vector)
~
~
5. (c~a) · b = c ~a · b
Theorem 1
Let θ be the angle between vectors ~a and
~b. Then
~a · ~b = |~a||~b| cos(θ) or cos(θ) =
1
~a · ~b
|~a||~b|
Updated: February 4, 2016
Calculus III
Section 12.3
Example 1
√
√
Find the angle between the vectors ~a =< 1, 3 > and ~b =< 3, 1 >.
cos(θ) =
~a · ~b
|~a||~b|
1. Find the dot product ~a · ~b
√
√
√
√
√
~a · ~b =< 1, 3 > · < 3, 1 >= 1 3 + 31 = 2 3
2. Find |~a| and |~b|
|~a| =
|~b| =
3. Use the formula cos(θ) =
q
√
√
(1)2 + ( 3)2 = 4 = 2
q√
√
( 3)2 + (1)2 = 4 = 2
~a · ~b
|~a||~b|
√
√
2 3
3
cos(θ) =
=
(2)(2)
2
√ !
3
θ = cos−1
= π/6 (30 degrees)
2
Example 2
Find the angle between the vectors ~a =< 2, 2, −1 > and ~b =< 5, −3, 2 >.
2
Updated: February 4, 2016
Calculus III
Section 12.3
~a · ~b
10 − 6 − 2
2
= √
= √
938
3 38
|~a||~b|
2
−1
√
θ = cos
≈ 84 degrees
3 38
cos(θ) =
Definition 2: Orthogonal
Vectors ~a and ~b are Orthogonal or Perpendicular if ~a · ~b = 0
If ~a · ~b > 0, the angle is acute.
If ~a · ~b = 0, the angle is right.
If ~a · ~b < 0, the angle is obtuse.
Example 3
Let ~a =< 2, 2 − 1 > and ~b =< 5, −4, 2 >. Show ~a and ~b are orthogonal.
~a · ~b = 2(5) + 2(−4) − 1(2) = 0 orthogonal
3
Updated: February 4, 2016
Calculus III
Section 12.3
Direction Cosines and Direction Angles
Definition 3: Direction Cosines and Direction Angles
Consider the graph below:
The Direction Angle of a vector ~a are
the angles α, β, and γ on [0, π] that ~a
makes with the positive x, y, z axes.
The Direction Cosines are cos(α),
cos(β), and cos(γ).
a1
⇒ a1 = ~a cos(α)
|~a|
a2
cos(β) =
⇒ a2 = ~a cos(β)
|~a|
a3
cos(γ) =
⇒ a3 = ~a cos(γ)
|~a|
cos(α) =
~a =< a1 , a2 , a3 >=< |~a| cos(α), |~a|cos(β), |~a|cos(γ) >
Also note that
~a
=< cos(α), cos(β), cos(γ) > is a unit vector in the direction of
|~a|
vector ~a.
Example 4
Find the direction angles of ~a =< 1, 2, 3 >
1. First thing to do is find |~a|
|~a| =
√
12 + 22 + 32 =
√
14
2. To find the direction angles we need to solve for α, β, and γ using
2
3
1
cos(α) = √ , cos(β) = √ , cos(γ) = √
14
14
14
4
Updated: February 4, 2016
Calculus III
Section 12.3
3. Solving for the angles we get
−1
α = cos
−1
β = cos
−1
γ = cos
5
1
√
14
2
√
14
3
√
14
≈ 74 degrees
≈ 58 degrees
≈ 37 degrees
Updated: February 4, 2016
Calculus III
Section 12.3
Projections
Definition 4: Vector Projection of ~b onto ~a
It’s much easier to visualize a vector projection in 2D than 3D. Let ~a = P~Q, ~b = P~R,
and ~c = P~S. Vector ~c is called the vector projection of ~b onto ~a. Think of vector ~c as
the shadow of ~b on ~a if you shined a light straight down over ~b.
Vector Projection of ~b onto ~a:
~c = proj~a~b =
!
~a · ~b
~a
|~a|2
Scalar Projection of ~b onto ~a
comp~a~b =
~a · ~b
|~a|
You can think of comp~a~b as the length of ~c with a ± to determine direction.
If the angle between vectors ~a and ~b is greater than 90 degrees, the picture would look
like this:
In the above graph comp~a~b is negative.
Example 5
Find the scalar and vector projection of ~b =< 1, 1, 2 > onto ~a =< −2, 3, 1 >
Let me show you the vector ~c we are trying to find.
6
Updated: February 4, 2016
Calculus III
Section 12.3
< −2, 3, 1 > · < 1, 1, 2 >
3
|~c| = comp ~a~b = p
=√
14
(−2)2 + (3)2 + (1)2
The signed length of ~c is the scalar projection of ~b onto ~a.
< −2, 3, 1 > · < 1, 1, 2 >
~
√
~c = proj~a b =
< 1, 1, 2 >
( 14)2
3
< 1, 1, 2 >
14
6 9 3
=< − , ,
>
14 14 14
=
7