the dot product the cross product - doc benton dot product & the cross product. ... vector b...

The Dot Product&

The Cross Product

Mathematicians are often a little weird.

They are frequently obsessive-compulsive.

Give them a new toy, and they immediatelywant to do arithmetic with it.

At one point, their new toy was “vectors.”

We’ve seen how to add and subtract vectors, …

But how do you multiply vectors?

In 1773, Joseph Lagrange was working on a problem involving tetrahedrons, and he came up with two ways to multiply vectors.

Today, we call these two methods the “dot product”and the “cross product.”

And as with many ideas in mathematics, over time they underwent both evolution and mutation.

And as with many ideas in mathematics, over time they underwent both evolution and mutation.

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

The end result was two ways of multiplying vectors that are:

The end result, though, was two ways of multiplying vectors that are:

a.Extremely useful, and

The end result, though, was two ways of multiplying vectors that are:

a.Extremely useful, and

b.Rather bizarre in appearance.

The Dot Product

We’ll define the dot product as follows:

1 2 3 1 2 3ˆ ˆˆ ˆ ˆ ˆ ,

cos , .

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

thena b a b where is theanglebetweenthevectorsθ θ

= + + = + +

⋅ =

θ a

b

What does this definition mean?

cosa b a b θ⋅ =

θ

b

acosb θ

What does this definition mean?

cosa b a b θ⋅ =

θ

We can think of the dot product as the length ofvector a times the length of the component of vector b that is parallel to vector a.

b

acosb θ

cosa b a b θ⋅ =

θ

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

acosb θ

cosa b a b θ⋅ =

θ

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.

The result is the same either way.

b

acosb θ

cosa b a b θ⋅ =

θ

The dot product gives us a way of multiplying twovectors such that the result is a number (or scalar),not another vector.

b

acosb θ

cosa b a b θ⋅ =

θ

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

b

acosb θ

cosa b a b θ⋅ =

θ

The problem, though, is that we need to know theAngle between the vectors in order to accomplishThis.And this is something we often don’t know.

b

acosb θ

cosa b a b θ⋅ =

θ

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

b

acosb θ

[ ]1

1 1 2 2 3 3 1 2 3 2

3

: cosb

Theorem a b a b a b a b a b a a a bb

θ⎡ ⎤⎢ ⎥⋅ = = + + = ⎢ ⎥⎢ ⎥⎣ ⎦

θ

cosa b a b θ⋅ =

b

acosb θ

[ ]1

1 1 2 2 3 3 1 2 3 2

3

: cosb

Theorem a b a b a b a b a b a a a bb

θ⎡ ⎤⎢ ⎥⋅ = = + + = ⎢ ⎥⎢ ⎥⎣ ⎦

θ

b

a

a b−

2 22 2 cos .a b a b a b θ− = + −

Proof: By the law of cosines,

θ

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 θ= + + + + + −

θ

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 θ= + + + + + −

θ

b

a

a b−

1 1 2 2 3 32 2 2 2 cos .a b a b a b a b θ− − − = −

Next, subtracting like terms from each side gives,

θ

b

a

a b−

1 1 2 2 3 3 cos .a b a b a b a b a bθ+ + = = ⋅

And finally, we divide each side by -2.

θ

b

a

a b−

1 1 2 2 3 3 cos .a b a b a ba ba b a b

θ+ +⋅

= =

As a corollary, we get the following formula for findingthe angle between two vectors.

0 θ π≤ ≤

θ

b

a

a b−

EXAMPLE:

ˆˆ ˆ2 3 4a i j k= + +ˆˆ ˆ2 2b i j k= − + −

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

14 4 4cos cos 104.34 9 16 1 4 4 3 29 3 29

a ba b

θ θ − ⎛ ⎞⋅ − − −= = = ⇒ = ≈ °⎜ ⎟+ + + + ⎝ ⎠

A fairly immediate and important consequence of what we’ve seen is the following:

cos 0 90 0 0.a b a b if and only if or a or bθ θ⋅ = = = ° = =



In other words, we can determine if two vectors areperpendicular or not simply by looking to see if the dotproduct is zero. Furthermore, the zero vector is consideredperpendicular to all vectors.



In other words, we can determine if two vectors areperpendicular or not simply by looking to see if the dotproduct is zero. Furthermore, the zero vector is consideredperpendicular to all vectors.

As a final note, the dot product is also known as the“scalar product.”

The Cross Product

The cross product of two vectors is defined asfollows:

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

ˆ ˆˆ ˆ ˆ ˆ ,

ˆˆ ˆˆˆ ˆ

ˆˆ ˆ( . ) ( . ) ( . ) .


i j ka a a a a a

then a b a a a i j kb b b b b b

b b b

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

= + + = + +

× = = − +

= − − − + −

The cross product of two vectors is defined asfollows:

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

ˆ ˆˆ ˆ ˆ ˆ ,

ˆˆ ˆˆˆ ˆ

ˆˆ ˆ( . ) ( . ) ( . ) .


i j ka a a a a a

then a b a a a i j kb b b b b b

b b b

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

= + + = + +

× = = − +

= − − − + −

Isn’t that cool!

The cross product of two vectors is another vector.


Also, the cross product is perpendicular to both vector aand vector b.


( ) [ ]

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

ˆ ˆˆ ˆ ˆ ˆ ,

( . ) ( . ) ( . )

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

= + + = + +

⋅ × = − + −

+ −− +

− + −

= − =



( ) [ ]

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

ˆ ˆˆ ˆ ˆ ˆ ,

( . ) ( . ) ( . )

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

= + + = + +

⋅ × = − + −

+ −− +

− + −

= − =


The proof that the cross product is perpendicular to vector bis similar.

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

a b

a b×

The cross product is also known as the “vector product.” Furthermore, just like the dot product, so does the cross product have interesting properties.

sin , ,a b a b where is theanglebetweenthetwovectorsθ θ× =

One of the more important ones is below:


sin , ,a b a b where is theanglebetweenthetwovectorsθ θ× =

One of the more important ones is below:


The proof is amazingly simple!

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 12 2 2 2

1 2 1 2 1

1

2 2 1

2 21

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 ba a b b a a b b a a b

a b

a a b b b

b

a

× = − + − + −

= + + + + +

−

=

= − + + − +

− −

+ +

+

−

( ) ( )

2 2 2 22 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 21 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 b

a 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

θ

+

= + + + + − +

= −

+ + + + + +

− − −

= −

⋅

−

+

−

= −

−

( )2 22 22 21 cos sin .a b a bθ θ θ= − =

Thus,2 22 2sin .a b a b θ× =

And since,

sin 0 0 ,forθ θ π≥ ≤ ≤

We have that,

sin .a b a b θ× =

One application of this result is a formula for the areaof a parallelogram.

sinArea a b a bθ= = ×

a

b sinheight b θ=

EXAMPLE (Cross Product):

ˆˆ ˆ2 3 4a i j k= + +ˆˆ ˆ2 2b i j k= − + −

EXAMPLE (Cross Product):

ˆˆ ˆ2 3 4a i j k= + +ˆˆ ˆ2 2b i j k= − + −

ˆˆ ˆˆ ˆˆ ˆ ˆ2 3 4 14 0 7 14 7

1 2 2

i j ka b i j k i k× = = − + + = − +

− −

EXAMPLE (Area of aParallelogram):

ˆˆ ˆ2 3 4a i j k= + +ˆˆ ˆ2 2b i j k= − + −

EXAMPLE (Area of aParallelogram):

ˆˆ ˆ2 3 4a i j k= + +ˆˆ ˆ2 2b i j k= − + −

ˆˆ14 7 245 7 5 15.65Area a b i k= × = − + = = ≈

the dot product the cross product - doc benton dot product & the cross product. ... vector b...

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