Transcript Document
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
ab
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