vector products

8/12/2019 Vector Products

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The Dot Product

&The Cross Product


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Mathematicians are often a little weird.


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They are frequently obsessive-compulsive.


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Give them a new toy, and they immediately

want to do arithmetic with it.


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At one point, their new toy was vectors.


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Weve seen how to add and subtract vectors,


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But how do you multiply vectors?


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In 1773, Joseph Lagrange was working on a

problem involving tetrahedrons, and he came up

with two ways to multiply vectors.


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Today, we call these two methods the dot product

and the cross product.


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And as with many ideas in mathematics, over time

they underwent both evolution and mutation.


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And as with many ideas in mathematics, over time

they underwent both evolution and mutation.

Ideas are also subject to the doctrine of survival ofthe fittest.


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The end result was two ways of multiplying vectors

that are:


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The end result, though, was two ways of multiplying

vectors that are:

a.Extremely useful, and


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The end result, though, was two ways of multiplying

vectors that are:

a.Extremely useful, and

b.Rather bizarre in appearance.


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Well define the dot product as follows:

1 2 3 1 2 3

,

cos , .

If a a i a j a k and b b i b j b k

then a b a b where isthe anglebetweenthevectors

= + + = + +

=

a

b


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What does this definition mean?

cosa b a b =

b

a

cosb


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What does this definition mean?

cosa b a b =

We can think of the dot product as the length of

vector a times the length of the component ofvector b that is parallel to vector a.

b

a

cosb


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cosa b a b =

Alternatively, we could think of it as the length ofvector b times the length of the component of

vector a that is parallel to vector b.

b

a

cosb


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cosa b a b =

The dot product gives us a way of multiplying twovectors such that the result is a number (or scalar),

not another vector.

b

a

cosb


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cosa b a b =

The problem, though, is that we need to know theangle between the vectors in order to accomplish

this.

b

a

cosb


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cosa b a b =

The problem, though, is that we need to know theAngle between the vectors in order to accomplish

This.

And this is something we often dont know.

b

a

cosb


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cosa b a b =

Fortunately, however, we have a theorem to help us.

b

a

cosb


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[ ]1

1 1 2 2 3 3 1 2 3 2

3

: cos

b

Theorem a b a b a b a b a b a a a b

b

= = + + =

cosa b a b =

b

a

cosb


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[ ]1

1 1 2 2 3 3 1 2 3 2

3

: cos

b

Theorem a b a b a b a b a b a a a b

b

= = + + =

b

a

a b

2 222 cos .a b a b a b = +

Proof: By the law of cosines,


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b

a

a b

2 2 21 1 2 2 3 3( ) ( ) ( )a b a b a b + +

But this implies that,

2 2 2 2 2 21 2 3 1 2 3 2 cos .a a a b b b a b = + + + + +


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b

a

a b

2 2 2 2 2 21 1 1 1 2 2 2 2 3 3 3 32 2 2a a b b a a b b a a b b + + + + +

Expanding the left side of the equation yields,

2 2 2 2 2 21 2 3 1 2 3 2 cos .a a a b b b a b = + + + + +


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b

a

a b

1 1 2 2 3 3

2 2 2 2 cos .a b a b a b a b =

Next, subtracting like terms from each side gives,


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b

a

a b

EXAMPLE:

2 3 4a i j k = + +

2 2b i j k = +

(2)( 1) (3)(2) (4)( 2) 4a b = + + =

14 4 4

cos cos 104.34 9 16 1 4 4 3 29 3 29

a b

a b

= = = = + + + +


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A fairly immediate and important consequence of whatweve seen is the following:

cos 0 90 0 0.a b a b if and only if or a or b = = = = =


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A fairly immediate and important consequence of what

weve seen is the following:

cos 0 90 0 0.a b a b if and only if or a or b = = = = =

In other words, we can determine if two vectors areperpendicular or not simply by looking to see if the dot

product is zero. Furthermore, the zero vector is considered

perpendicular to all vectors.As a final note, the dot product is also known as the

scalar product.


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The Cross Product


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The cross product of two vectors is defined as

follows:

1 2 3 1 2 3

2 3 1 3 1 21 2 3

2 3 1 3 1 21 2 3

2 3 3 2 1 3 3 1 1 2 2 1

,

( . ) ( . ) ( . ) .

If a a i a j a k and b b i b j b k

i j ka a a a a a

then a b a a a i j k b b b b b b

b b b

a b a b i a b a b j a b a b k

= + + = + +

= = +

= +


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The cross product of two vectors is another vector.


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Also, the cross product is perpendicular to both vector a

and vector b.


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( ) [ ]

2 3 1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1

2 3 11 2 3 1 3 2 1 3 2 3 12 3 21

,

( . ) ( . ) ( . )

0

If a a i a j a k and b b

a

i b j b k then

a a b a

a b a a b

a b a b a a b a b a a b a

a a ba a b

b

aa ab a b

= + + = + +

= +

+ +

+

= =


and vector b.


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( ) [ ]

2 3 1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1

2 3 11 2 3 1 3 2 1 3 2 3 12 3 21

,

( . ) ( . ) ( . )

0

If a a i a j a k and b b

a

i b j b k then

a a b a

a b a a b

a b a b a a b a b a a b a

a a ba a b

b

aa ab a b

= + + = + +

= +

+ +

+

= =


and vector b.

The proof that the cross product is perpendicular to vector b

is similar.


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The cross product, , obeys a right-hand rule.a b

a

b

a b


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The cross product is also known as the

vector product. Furthermore, just like the

dot product, so does the cross product have

interesting properties.


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sin , ,a b a b where is the angle between the two vectors =

One of the more important ones is below:






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sin , ,a b a b where is the angle between the two vectors =

One of the more important ones is below:





The proof is amazingly simple!

P f


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Proof:

2 2 2 2 2 2 2 22 3 2 3 2

2 2 2 22 3 3 2 1 3 3 1 1 2 2 1

2 2 2 2 2 2 2 2 2 2 2 21 2 1 3 2 1 2 3 3 1 3 2

1 2 1 2 1 3 1 3 2 3 2 3

3 3 2 1 3 1 3 1 3 3 1

2 2 2 2

1 2 1 2 1

1

2 2 1

2 2

1

2

( . ) (

2

2

. ) ( . )

2 2 2

a b a a b b a b a b a a b b a b

a b

a b a b a b a b a b a b a b

a b a b a b a b a b a b

a a b b a a b b a a b

a b

a a b b b

b

a

= + +

= + + + + +

=

= + + +

+ +

+

( ) ( )

2 2 2 2

2 2 3 3

2 2 2 2 2 2 21 2 3 1 2 3 1 1 2 2

2 2 2 2 2 2 2 2 2 2 2 2

1 2 1 3 2 1 2 3 3 1 3 2

1 2 1 2 1 3 1 3 2 3 2 3

2 22

222

2 2 2 2 2 21 1 2 2 3 3

3

2

2

2

3

2

2

co

( )( ) (

2 2 2

co

s

s

)

a b a b a

a

a b a b a b a b a b a ba a b

a b a b

a a a b b b a b a b a b

b a a

b a b a b a b

bb b a a b b

a b a b

+

= + + + + +

=

+ + + + + +

=

+

=

( )2 22 22 21 cos sin .a b a b = =

Th


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Thus,

2 22 2sin .a b a b =

And since,

sin 0 0 ,for

We have that,

sin .a b a b =


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One application of this result is a formula for the area

of a parallelogram.

sinArea a b a b= =

a

b

sinheight b =


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EXAMPLE (Cross Product):

2 3 4a i j k = + +

2 2b i j k = +


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EXAMPLE (Cross Product):

2 3 4a i j k = + +

2 2b i j k = +

2 3 4 14 0 7 14 71 2 2

i j k

a b i j k i k = = + + = +


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EXAMPLE (Area of aParallelogram):

2 3 4a i j k = + +

2 2b i j k = +


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