The document discusses the dot product of vectors, which is defined as the sum of the products of the corresponding components of the vectors. The dot product can be used to determine the angle between vectors and project one vector onto another. It is also related to the work done by a constant force acting along the direction of movement between two points. Examples are provided to demonstrate finding the dot product, angle between vectors, and projection of one vector onto another.

1. 6.2
Dot Product
of Vectors
Copyright © 2011 Pearson, Inc.

2. What you’ll learn about
 The Dot Product
 Angle Between Vectors
 Projecting One Vector onto Another
 Work
… and why
Vectors are used extensively in mathematics and
science applications such as determining the net effect
of several forces acting on an object and computing the
work done by a force acting on an object.
Copyright © 2011 Pearson, Inc. Slide 6.2 - 2

3. Dot Product
The dot product or inner product of u  u1,u2
and v  v1,v2 is
u v  u1v1  u2v2 .
Copyright © 2011 Pearson, Inc. Slide 6.2 - 3

4. Properties of the Dot Product
Let u, v, and w be vectors and let c be a scalar.
1. u·v = v·u
2. u·u = |u|2
3. 0·u = 0
4. u·(v + w) = u·v + u·w
(u + v) ·w = u·w + v·w
5. (cu)·v = u·(cv) = c(u·v)
Copyright © 2011 Pearson, Inc. Slide 6.2 - 4

5. Example Finding the Dot Product
Find the dot product.
4,3  1, 2
Copyright © 2011 Pearson, Inc. Slide 6.2 - 5

6. Example Finding the Dot Product
Find the dot product.
4,3  1, 2
4,3  1,2  (4)(1)  (3)(2)  10
Copyright © 2011 Pearson, Inc. Slide 6.2 - 6

7. Angle Between Two Vectors
If  is the angle between the nonzero vectors u and v,
then
cos 
u v
u v

and   cos-1 u v
u v
 

 
Copyright © 2011 Pearson, Inc. Slide 6.2 - 7

8. Example Finding the Angle Between
Vectors
Let u  2,3 and v  4, 1 .
Find the angle between the vectors u and v.
Copyright © 2011 Pearson, Inc. Slide 6.2 - 8

9. Example Finding the Angle Between
Vectors
Let u  2,3 and v  4, 1 .
Find the angle between the vectors u and v.
cos 
u v
u v
=
2,3  4, 1
2,3 4, 1









5
13 17
So,

  cos1 3



 131




 70¼
Copyright © 2011 Pearson, Inc. Slide 6.2 - 9

10. Orthogonal Vectors
The vectors u and v are orthogonal if and
only if u·v = 0.
Copyright © 2011 Pearson, Inc. Slide 6.2 - 10

11. Projection of u and v
If u and v are nonzero vectors, the projection of u
onto v is
projvu 
u v
2
v








v.
Copyright © 2011 Pearson, Inc. Slide 6.2 - 11

12. Work
If F is a constant force whose direction is the same as
the direction of AB, then the work W done by F in
moving an object from A to B is W | F || AB |
Copyright © 2011 Pearson, Inc. Slide 6.2 - 12

13. Quick Review
Find |u|.
1. u  1,2
2. u  4i  3 j
The points A and B lie on the circle x2  y2  4.
Find the component form of the vector AB.
3. A  (1, 3), B  (2,0)
4. A  (1, 3), B  (2,0)
5. Find a vector u with the given magnitude in the
direction of v. |u|  3, v  2,4
Copyright © 2011 Pearson, Inc. Slide 6.2 - 13

14. Quick Review Solutions
Find |u|.
1. u  1,2 5
2. u  4i  3 j 5
The points A and B lie on the circle x2  y2  4.
Find the component form of the vector AB.
3. A  (1, 3), B  (2,0) 1,  3
4. A  (1, 3), B  (2,0) 3,  3
5. Find a vector u with the given magnitude in the
direction of v. |u|  3, v  2,4 1.34,2.68
Copyright © 2011 Pearson, Inc. Slide 6.2 - 14