3. Simple Example of Dot Product
x
y
From the given Vectors:
A
A = 6i + 8j
B
B = 8i
Find A • B
4. Solution to Simple Dot Product
Use the given equation: A · B = |A| |B| cos Θ
x
y
We must then find the angle between the vectors (Θ)
A
B
Θ
A · B = (10)(8) cos 53.1 =
8
6
Θ
Θ = tan-1
(8/6)= 53.1°
48.03
|A| = sqrt (62
+ 82
) = 10
-Find |A|
-Find |B| |B| = sqrt ( 82
) = 8
5. Laws of Operations
• Commutative
A · B = B · A
• Associative with respect to scalar multiplication
a (A · B) = (a A) · B = A · (a B)
• Distributive with respect to vector addition
A · (B + D) = (A · B) + (A · D)
6. Since the cosine of 90° = 0, and the cosine of 0° = 1,
from A · B = cos Θ, we can derive:
i · i = 1
j · j = 1
k · k = 1
Since the angle between two i direction vectors would be
0, the equation would be: i · i = |i| |i| cos 0° = (1)(1)(1) = 1
Dot Product Identities
Since the angle between two different
direction vectors would be 90°, the equation
would be: i · j = |i| |j| cos 90° = (1)(1)(0) = 0
i · j = 0
i · k = 0
k · j = 0
j · i = 0
k · i = 0
j · k = 0
7. Components of Dot Product
Consider the dot product of two 3 dimensional vectors expressed in Cartesian
vector form:
A · B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk)
= AxBx( i · i ) + AxBy( i · j ) + AxBz( i · k )
+ AyBx( j · i ) + AyBy( j · j ) + AyBz( j · k )
+ AxBx( k · i ) + AzBy( k · j ) + AzBz( k · k )
Using the laws from the previous page, we see that this equation
can be reduced to:
A · B = AxBx + AyBy + AzBz
Therefore, to determine the Dot Product of two vectors, multiply their
corresponding x, y, z components and sum their products.
8. Example of Finding Dot Product
Using Cartesian Components
Given two vectors:
x
y
z
A
A = 2i + 3j – 4k
B
B = -3i + 2k
Find A • B
9. Solution to Component Dot Product
Use the formula A · B = AxBx + AyBy + AzBz
A · B = (2)(-3) + (3)(0) + (-4)(2) =
A = 2i + 3j – 4k
B = -3i + 2k
Ax = 2; Ay = 3; Az = -4
Bx = -3; By = 0; Bz = 2
-14
10. Applications
The Dot product has useful applications in mechanics
1. The angle formed between two vectors or intersecting lines
Θ = cos -1
(A • B)
|A| |B|
The angle between the tails of vectors A and B can be determined by
solving A · B = |A| |B| cos Θ, for Θ.
11. B
Θ
B = -3i + 2k
Example Finding Angle Between Vectors
Find the angle between the
two vectors in the previous
example
x
y
z
A = 2i + 3j – 4k
A
12. Solution for Finding Angle
Use the equation: Θ = cos -1
(A • B)
|A| |B|
136.13°
A · B = AxBx + AyBy + AzBz = (2)(-3) + (3)(0) + (-4)(2) = -14
We will then find the magnitude of A and B:
|A| = sqrt (22
+ 32
+ (-4)2
) = 5.385
|B| = sqrt ( (-3)2
+ 02
+ 22
) = 3.606
Θ = cos -1
-14
(5.385)(3.606)
=
First, we will find (A • B) using the component form:
B = -3i + 2k
A = 2i + 3j – 4k
13. The cross product can be viewed as the "projection" of the
magnitude of A on a line perpendicular to vector B (Asinθ),
multiplied by the magnitude of B.
The magnitude of this product can be positive or zero depending on
angle θ. θ ranges from 0 to 180
0
.
The vector cross product is a minimum when the two vectors are in
line, θ = 0
0
. It is a maximum when θ = 90
0
.
Vector Cross Product
The vector cross product yields a vector quantity and its
magnitude is defined as:
17. and
19 Solve for the angle between vectors
A 98
o
B 278
o
C 57
o
D 85
o
E 124
o
