Section 9.3
The Dot Product

Goals

Introduce the dot product of two vectors and
explain its significance to work.

Discuss the dot product and perpendicularity.

Give properties of the dot product.

Introduce projections.
Introduction

So far we have added two vectors and
multiplied a vector by a scalar.

The question arises: Is it possible to
multiply two vectors so that their product
is a useful quantity?

One such product is the dot product,
which we consider in this section.
Work

Recall that the work done by a constant
force F in moving an object through a
distance d is W = Fd.

However this applies only when the force
is directed along the line of motion.

Suppose, however, that the constant force
is a vector F  PR pointing in some other
direction, as shown on the next slide:
Work (cont’d)
Work (cont’d)

If the force moves the object from P to Q,
then the displacement vector is D  PQ.

The work W done by F is defined as the
magnitude of the displacement, |D| ,
multiplied by the magnitude of the
applied force in the direction of the
motion, namely PS  F cos .
Work (cont’d)


Thus W  D  F cos   F D cos
We use this expression to define the dot
product of two vectors even when they
don’t represent force or displacement:
Remarks

This product is called the dot product
because of the dot in the notation a ∙ b.

a ∙ b is a scalar, not a vector.


Sometimes the dot product is called the scalar
product.
In the example of finding work done, it
makes no sense for θ to be greater than
π/2, but in our general definition we allow
θ to be any angle from 0 to π.
Example

If the vectors a and b have lengths 4 and 6,
and the angle between them is π/3, find
a ∙ b.

Solution According to the definition,
a ∙ b = |a||b| cos(π/3) = 4∙ 6 ∙ ½ = 12
Perpendicular Vectors

Two nonzero vectors a and b are called
perpendicular or orthogonal if the angle
between them is θ = π/2.

For such vectors we have
a ∙ b = |a||b| cos(π/2) = 0

Conversely, if a ∙ b = 0, then cos θ = 0, so
θ = π/2.
Perpendicular Vectors (cont’d)

Since the zero vector 0 is considered to be
perpendicular to all vectors, we have

Further, by properties of the cosine, a ∙ b is

positive for θ < π/2, and

negative for θ > π/2,
as the next slide illustrates:
Perpendicular Vectors (cont’d)
Perpendicular Vectors (cont’d)

We can think of a ∙ b as measuring the
extent to which a and b point in the same
general direction.

The dot product a ∙ b is…

positive if a and b point in the same general
direction,

0 if they are perpendicular, and

negative if they point in generally opposite
directions.
Perpendicular Vectors (cont’d)

In the extreme cases where…

a and b point in exactly the same direction,
we have θ = 0, so cos θ = 1 and
a ∙ b = |a||b|

a and b point in exactly opposite directions,
then θ = π, so cos θ = –1 and
a ∙ b = –|a||b|
Component Form

Suppose we are given two vectors in
component form:

We want to find a convenient expression
for a ∙ b in terms of these components.

An application of the Law of Cosines
gives the result on the next slide:
Component Form (cont’d)

Here are some examples:
Example

Show that 2i + 2j – k is perpendicular to
5i – 4j + 2k.

Solution Since
(2i + 2j – k)∙ (5i – 4j + 2k) =
2(5) + 2(– 4) + (– 1)(2) = 0,
these vectors are perpendicular.
Example

Find the angle between a  2,2, 1 and
b  5, 3,2 .

Solution Let θ be the required angle.
Since
a  2  2   1  3 and
2
2
2
b  5   3   2  38
2
2
2
Solution (cont’d)
and since
a ∙ b = 2(5) + 2(– 3) + (– 1)(2) = 0,
the definition of dot product gives
ab
2
cos 

a b 3 38

So the angle between a and b is
  cos
1
 
2
 1.46 (or 84 )
3 38
Example

A force is given by a vector F = 3i + 4j + 5k
and moves a particle from P(2, 1, 0) to
Q(4, 6, 2). Find the work done.

Solution The displacement vector is
D  PQ  2,5,2 , so the work done is
W  F  D  36
Properties

The dot product obeys many of the laws
that hold for ordinary products of real
numbers:
Projections

The next slide shows representations
PQ and PR of two vectors a and b with
the same initial point P.

If S is the foot of the perpendicular from R
to the line containing PQ , then the vector
with representation PS is called the vector
projection of b onto a and is denoted by
proja b.
Projections (cont’d)
Projections (cont’d)

The scalar projection of
b onto a is the length
|b|cos θ of the vector
projection.

This is denoted by
compa b, and can also
be computed by taking the dot product of
b with the unit vector in the direction of a.
Projections (cont’d)

To summarize:

For example we find the scalar projection
and vector projection of b  1,1,2 onto
a  2,3,1 :
Solution


Since a   2   3  1  14, the scalar
projection of b onto a is
2
2
2
The vector projection is this scalar
projection times the unit vector in the
direction of a:
Solution (cont’d)
Review

Definition of dot product in terms of work
done

Orthogonality and the dot product

The dot product in component form

Properties of the dot product

Projections