Chap12_Sec3 - Dot Product.ppt

VECTORS AND
THE GEOMETRY OF SPACE
12

VECTORS AND THE GEOMETRY OF SPACE
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 will discuss
in this section.

Another is the cross product,
which we will discuss in Section
12.4

12.3
The Dot Product
In this section, we will learn about:
Various concepts related to the dot product
and its applications.

If a = ‹a1, a2, a3› and b = ‹b1, b2, b3›, then
the dot product of a and b is the number a • b
given by:
a • b = a1b1 + a2b2 + a3b3
THE DOT PRODUCT Definition 1

Thus, to find the dot product of a and b,
we multiply corresponding components
and add.
DOT PRODUCT

The result is not a vector.
It is a real number, that is, a scalar.
 For this reason, the dot product is sometimes
called the scalar product (or inner product).
SCALAR PRODUCT

Though Definition 1 is given for three-
dimensional (3-D) vectors, the dot product
of two-dimensional vectors is defined in
a similar fashion:
‹a1, a2› ∙ ‹b1, b2› = a1b1 + a2b2
DOT PRODUCT

‹2, 4› ∙ ‹3, – 1› = 2(3) + 4(–1) = 2
‹–1, 7, 4› ∙ ‹6, 2, –½› = (–1)(6) + 7(2) + 4(–½)
= 6
(i + 2j – 3k) ∙ (2j – k) = 1(0) + 2(2) + (–3)(–1)
= 7
DOT PRODUCT Example 1

The dot product obeys many of the laws
that hold for ordinary products of real
numbers.
 These are stated in the following theorem.
DOT PRODUCT

If a, b, and c are vectors in V3 and c is
a scalar, then
PROPERTIES OF DOT PRODUCT
2
1. =| |
2.
3. ( )
4. ( ) ( ) ( )
5. 0 0
c c c

  
     
    
 
a a a
a b b a
a b c a b a c
a b a b a b
a
Theorem 2

These properties are easily proved
using Definition 1.
 For instance, the proofs of Properties 1 and 3
are as follows.
DOT PRODUCT PROPERTIES

a ∙ a
= a1
2 + a2
2 + a3
2
= |a|2
DOT PRODUCT PROPERTY 1 Proof

a • (b + c)
= ‹a1, a2, a3› ∙ ‹b1 + c1, b2 + c2, b3 + c3›
= a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3)
= a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3
= (a1b1 + a2b2 + a3b3) + (a1c1 + a2c2 + a3c3)
= a ∙ b + a ∙ c
DOT PRODUCT PROPERTY 3 Proof

The proofs of the remaining
properties are left as exercises.
DOT PRODUCT PROPERTIES

The dot product a • b can be given
a geometric interpretation in terms of
the angle θ between a and b.
 This is defined to be the angle between
the representations of a and b that start
at the origin, where 0 ≤ θ ≤ π.
GEOMETRIC INTERPRETATION

In other words, θ is the angle between
the line segments and here.
 Note that if a and b
are parallel vectors,
then θ = 0 or θ = π.
GEOMETRIC INTERPRETATION
OA OB

The formula in the following theorem
is used by physicists as the definition
of the dot product.
DOT PRODUCT

If θ is the angle between the vectors
a and b, then
a ∙ b = |a||b| cos θ
DOT PRODUCT—DEFINITION Theorem 3

If we apply the Law of Cosines to triangle OAB
here, we get:
|AB|2 = |OA|2 + |OB|2 – 2|OA||OB| cos θ
 Observe that
the Law of Cosines
still applies in
the limiting cases
when θ = 0 or π, or
a = 0 or b = 0
DOT PRODUCT—DEFINITION Proof—Equation 4

However,
|OA| = |a|
|OB| = |b|
|AB| = |a – b|
DOT PRODUCT—DEFINITION Proof

So, Equation 4 becomes:
|a – b|2 = |a|2 + |b|2 – 2|a||b| cos θ
DOT PRODUCT—DEFINITION Proof—Equation 5

Using Properties 1, 2, and 3 of the dot
product, we can rewrite the left side of
the equation as follows:
|a – b|2 = (a – b) ∙ (a – b)
= a ∙ a – a ∙ b – b ∙ a + b ∙ b
= |a|2 – 2a ∙ b + |b|2

Therefore, Equation 5 gives:
|a|2 – 2a ∙ b + |b|2 = |a|2 + |b|2 – 2|a||b| cos θ
 Thus,
–2a ∙ b = –2|a||b| cos θ
or
a ∙ b = |a||b| cos θ

If the vectors a and b have lengths 4
and 6, and the angle between them is π/3,
find a ∙ b.
 Using Theorem 3, we have:
a ∙ b = |a||b| cos(π/3)
= 4 ∙ 6 ∙ ½
= 12
DOT PRODUCT Example 2

The formula in Theorem 3
also enables us to find the angle
between two vectors.
DOT PRODUCT

If θ is the angle between the nonzero
vectors a and b, then
cos
| || |



a b
a b
NONZERO VECTORS Corollary 6

Find the angle between the vectors
a = ‹2, 2, –1› and b = ‹5, –3, 2›
NONZERO VECTORS Example 3

Also,
a ∙ b = 2(5) + 2(–3) +(–1)(2) = 2
NONZERO VECTORS Example 3
2 2 2
2 2 2
| | 2 2 ( 1) 3
and
| | 5 ( 3) 2 38
    
    
a
b

Thus, from Corollary 6, we have:
 So, the angle between a and b is:
NONZERO VECTORS
2
cos
| || | 3 38


 
a b
a b
Example 3
1 2
cos 1.46 (or 84 )
3 38
   
 
 
 

Two nonzero vectors a and b are called
perpendicular or orthogonal if the angle
between them is θ = π/2.
ORTHOGONAL VECTORS

Then, Theorem 3 gives:
a ∙ b = |a||b| cos(π/2) = 0
 Conversely, if a ∙ b = 0, then cos θ = 0;
so, θ = π/2.
ORTHOGONAL VECTORS

The zero vector 0 is considered to be
perpendicular to all vectors.
 Therefore, we have the following method for
determining whether two vectors are orthogonal.
ZERO VECTORS

Two vectors a and b are orthogonal
if and only if
a ∙ b = 0
ORTHOGONAL VECTORS Theorem 7

Show that 2i + 2j – k is perpendicular
to 5i – 4j + 2k.
 (2i + 2j – k) ∙ (5i – 4j + 2k)
= 2(5) + 2(–4) + (–1)(2)
= 0
 So, these vectors are perpendicular
by Theorem 7.
ORTHOGONAL VECTORS Example 4

As cos θ > 0 if 0 ≤ θ < π/2 and cos θ < 0
if π/2 < θ ≤ π, we see that a ∙ b is positive
for θ < π/2 and negative for θ > π/2.
 We can think of a ∙ b as measuring the extent
to which a and b point in the same direction.
DOT PRODUCT

The dot product a ∙ b is:
 Positive, if a and b point in the same general direction
 Zero, if they are
perpendicular
 Negative, if they point
in generally opposite
directions
DOT PRODUCT

In the extreme case where a and b
point in exactly the same direction,
we have θ = 0.
 So, cos θ = 1 and
a ∙ b = |a||b|
DOT PRODUCT

If a and b point in exactly opposite
directions, then θ = π.
 So, cos θ = –1 and
a ∙ b = –|a| |b|
DOT PRODUCT

The direction angles of a nonzero vector a
are the angles α, β, and γ (in the interval
[0, π]) that a makes with the positive x-, y-,
and z-axes.
DIRECTION ANGLES

The cosines of these direction angles—cos α,
cos β, and cos γ—are called the direction
cosines of the vector a.
DIRECTION COSINES

Using Corollary 6 with b replaced by i,
we obtain:
DIRECTION ANGLES & COSINES Equation 8
1
cos
| || | | |
a


 
a i
a i a

This can also be seen directly from
the figure.
DIRECTION ANGLES & COSINES

Similarly, we also have:
3
2
cos cos
| | | |
a
a
 
 
a a
Equation 9

By squaring the expressions
in Equations 8 and 9 and adding,
we see that:
DIRECTION ANGLES & COSINES Equation 10
2 2 2
cos cos cos 1
  
  

We can also use Equations 8 and 9
to write:
a = ‹a1, a2, a3›
= ‹|a| cos α, |a| cos β, |a| cos γ›
= |a|‹cos α, cos β, cos γ›

Therefore,
 This states that the direction cosines of a
are the components of the unit vector in
the direction of a.
1
cos ,cos ,cos
| |
  

a
a
Equation 11

Find the direction angles of the vector
a = ‹1, 2, 3›

 So, Equations 8 and 9 give:
DIRECTION ANGLES & COSINES Example 5
2 2 2
| | 1 2 3 14
   
a
1 2 3
cos cos cos
14 14 14
   

 Therefore,
1
1
1
1
cos 74
14
2
cos 58
14
3
cos 37
14






 
  
 
 
 
  
 
 
 
  
 
 
Example 5

The figure shows representations and
of two vectors a and b with the same initial
point P.
PROJECTIONS
PQ PR

Let S be the foot of the perpendicular
from R to the line containing .
PROJECTIONS
PQ

Then, the vector with representation is
called the vector projection of b onto a and is
denoted by proja b.
 You can think of it as a shadow of b.
VECTOR PROJECTION
PS

The scalar projection of b onto a
(also called the component of b along a)
is defined to be the signed magnitude
of the vector projection.
SCALAR PROJECTION

This is the number |b| cos θ, where θ
is the angle between a and b.
 This is denoted
by compa b.
 Observe that
it is negative
if π/2 < θ ≤ π.
PROJECTIONS

The equation
a ∙ b = |a||b| cos θ = |a|(|b| cos θ)
shows that:
 The dot product of a and b can be interpreted
as the length of a times the scalar projection of b
onto a.
PROJECTIONS

Since
the component of b along a can be
computed by taking the dot product of b
with the unit vector in the direction of a.
PROJECTIONS
| | cos
| | | |


  
a b a
b b
a a

We summarize these ideas
as follows.
PROJECTIONS

Scalar projection of b onto a:
Vector projection of b onto a:
 Notice that the vector projection
is the scalar projection times
the unit vector in the direction of a.
PROJECTIONS
a
comp
| |


a b
b
a
a 2
proj
| | | | | |
 
 
 
 
 
a b a a b
b a
a a a

Find the scalar and vector projections of:
b = ‹1, 1, 2› onto a = ‹–2 , 3, 1›
PROJECTIONS Example 6

Since
the scalar projection of b onto a is:
PROJECTIONS Example 6
2 2 2
| | ( 2) 3 1 14
    
a
a
( 2)(1) 3(1) 1(2)
comp
| | 14
3
14
   
 

a b
b
a

The vector projection is that scalar projection
times the unit vector in the direction of a:
PROJECTIONS
a
3 3
proj
| | 14
14
3 9 3
, ,
7 14 14
 
 
a
b a
a
Example 6

One use of projections occurs
in physics in calculating work.
APPLICATIONS OF PROJECTIONS

In Section 6.4, we defined the work done
by a constant force F in moving an object
through a distance d as:
W = Fd
 This, however, applies only when the force is
directed along the line of motion of the object.
CALCULATING WORK

However, suppose that the constant force
is a vector pointing in some other
direction, as shown.
CALCULATING WORK
PR

F

If the force moves the object from
P to Q, then the displacement vector
is .
CALCULATING WORK
PQ

D

The work done by this force is defined to be
the product of the component of the force
along D and the distance moved:
W = (|F| cos θ)|D|
CALCULATING WORK

However, from Theorem 3,
we have:
W = |F||D| cos θ
= F ∙ D
CALCULATING WORK Equation 12

Therefore, the work done by a constant
force F is:
 The dot product F ∙ D, where D is
the displacement vector.
CALCULATING WORK

A wagon is pulled a distance of 100 m along
a horizontal path by a constant force of 70 N.
The handle of the wagon is held at an angle
of 35° above the horizontal.
 Find the work
done by the force.
CALCULATING WORK Example 7

Suppose F and D are the force and
displacement vectors, as shown.

Then, the work done is:
W = F ∙ D = |F||D| cos 35°
= (70)(100) cos 35°
≈ 5734 N∙m
= 5734 J

A force is given by a vector F = 3i + 4j + 5k
and moves a particle from the point P(2, 1, 0)
to the point Q(4, 6, 2).
 Find the work done.

The displacement vector is
So, by Equation 12, the work done is:
W = F ∙ D
= ‹3, 4, 5› ∙ ‹2, 5, 2›
= 6 + 20 + 10 = 36
 If the unit of length is meters and the magnitude
of the force is measured in newtons, then the work
done is 36 joules.
CALCULATING WORK
2,5,2
PQ
 
D
Example 8

Chap12_Sec3 - Dot Product.ppt

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Chap12_Sec3 - Dot Product.ppt