Transcript Document

Section 9.3
The Dot Product

Goals
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Introduce the dot product of two vectors and
explain its significance to work.
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Discuss the dot product and perpendicularity.
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Give properties of the dot product.
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Introduce projections.
Introduction
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So far we have added two vectors and
multiplied a vector by a scalar.
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The question arises: Is it possible to
multiply two vectors so that their product
is a useful quantity?
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One such product is the dot product,
which we consider in this section.
Work
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Recall that the work done by a constant
force F in moving an object through a
distance d is W = Fd.
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However this applies only when the force
is directed along the line of motion.
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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)
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If the force moves the object from P to Q,
then the displacement vector is D  PQ.
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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)
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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
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This product is called the dot product
because of the dot in the notation a ∙ b.
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a ∙ b is a scalar, not a vector.
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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
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If the vectors a and b have lengths 4 and 6,
and the angle between them is π/3, find
a ∙ b.
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Solution According to the definition,
a ∙ b = |a||b| cos(π/3) = 4∙ 6 ∙ ½ = 12
Perpendicular Vectors
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Two nonzero vectors a and b are called
perpendicular or orthogonal if the angle
between them is θ = π/2.
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For such vectors we have
a ∙ b = |a||b| cos(π/2) = 0
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Conversely, if a ∙ b = 0, then cos θ = 0, so
θ = π/2.
Perpendicular Vectors (cont’d)
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Since the zero vector 0 is considered to be
perpendicular to all vectors, we have
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Further, by properties of the cosine, a ∙ b is
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positive for θ < π/2, and
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negative for θ > π/2,
as the next slide illustrates:
Perpendicular Vectors (cont’d)
Perpendicular Vectors (cont’d)
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We can think of a ∙ b as measuring the
extent to which a and b point in the same
general direction.
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The dot product a ∙ b is…
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positive if a and b point in the same general
direction,
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0 if they are perpendicular, and
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negative if they point in generally opposite
directions.
Perpendicular Vectors (cont’d)
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In the extreme cases where…
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a and b point in exactly the same direction,
we have θ = 0, so cos θ = 1 and
a ∙ b = |a||b|
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a and b point in exactly opposite directions,
then θ = π, so cos θ = –1 and
a ∙ b = –|a||b|
Component Form
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Suppose we are given two vectors in
component form:
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We want to find a convenient expression
for a ∙ b in terms of these components.
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An application of the Law of Cosines
gives the result on the next slide:
Component Form (cont’d)
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Here are some examples:
Example
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Show that 2i + 2j – k is perpendicular to
5i – 4j + 2k.
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Solution Since
(2i + 2j – k)∙ (5i – 4j + 2k) =
2(5) + 2(– 4) + (– 1)(2) = 0,
these vectors are perpendicular.
Example
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Find the angle between a  2,2, 1 and
b  5, 3,2 .
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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
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So the angle between a and b is
  cos
1
 
2
 1.46 (or 84 )
3 38
Example
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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.
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Solution The displacement vector is
D  PQ  2,5,2 , so the work done is
W  F  D  36
Properties
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The dot product obeys many of the laws
that hold for ordinary products of real
numbers:
Projections
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The next slide shows representations
PQ and PR of two vectors a and b with
the same initial point P.
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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)
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The scalar projection of
b onto a is the length
|b|cos θ of the vector
projection.
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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)
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To summarize:
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For example we find the scalar projection
and vector projection of b  1,1,2 onto
a  2,3,1 :
Solution
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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
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Definition of dot product in terms of work
done
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Orthogonality and the dot product
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The dot product in component form
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Properties of the dot product
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Projections