The Dot Product & The Cross Product

Mathematicians are often a little weird.

They are frequently obsessive-compulsive.

Give them a new toy, and they immediately want 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: G G If a = a1iˆ + a2 ˆj + a3 kˆ and b = b1iˆ + b2 ˆj + b3 kˆ, G G G G then a ⋅ b = a b cos θ , whereθ is the anglebetweenthe vectors.

G b θ

G a

What does this definition mean?

G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

What does this definition mean? We can think of the dot product as the length of vector a times the length of the component of vector b that is parallel to vector a. G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

Alternatively, we could think of it as the length of vector b times the length of the component of vector a that is parallel to vector b.

G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

Alternatively, we could think of it as the length of vector b times the length of the component of vector a that is parallel to vector b. The result is the same either way. G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

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

G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

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

G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

The problem, though, is that we need to know the Angle between the vectors in order to accomplish This. And this is something we often don’t know. G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

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

G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

G G G G Theorem : a ⋅ b = a b cos θ = a1b1 + a2b2 + a3b3 = [ a1

G G G G a ⋅ b = a b cos θ

G b θ

G b cosθ

G a

a2

⎡b1 ⎤ a3 ] ⎢⎢b2 ⎥⎥ ⎢⎣b3 ⎥⎦

G G G G Theorem : a ⋅ b = a b cos θ = a1b1 + a2b2 + a3b3 = [ a1

Proof: By the law of cosines, G G a −b

2

G2 G2 G G = a + b − 2 a b cos θ .

G G a −b

G b θ

G a

a2

⎡b1 ⎤ a3 ] ⎢⎢b2 ⎥⎥ ⎢⎣b3 ⎥⎦

But this implies that, (a1 − b1 ) 2 + (a2 − b2 ) 2 + (a3 − b3 ) 2 =

a12

+ a2 + 2

a3 + b12 2

G G + b2 + b3 − 2 a b cos θ . 2

G G a −b

G b θ

2

G a

Expanding the left side of the equation yields,

a12 − 2a1b1 + b12 + a2 2 − 2a2b2 + b2 2 + a32 − 2a3b3 + b32 =

a12

+ a2 + 2

a3 + b12 2

G G + b2 + b3 − 2 a b cos θ . 2

G G a −b

G b θ

2

G a

Next, subtracting like terms from each side gives, G G −2a1b1 − 2a2b2 − 2a3b3 = −2 a b cos θ .

G G a −b

G b θ

G a

And finally, we divide each side by -2.

G G G G a1b1 + a2b2 + a3b3 = a b cos θ = a ⋅ b .

G G a −b

G b θ

G a

As a corollary, we get the following formula for finding the angle between two vectors. G G a1b1 + a2b2 + a3b3 a ⋅b = cos θ . G G = G G a b a b 0 ≤θ ≤π

G G a −b

G b θ

G a

EXAMPLE:

G a = 2iˆ + 3 ˆj + 4kˆ G b = −iˆ + 2 ˆj − 2kˆ G G a ⋅ b = (2)(−1) + (3)(2) + (4)(−2) = −4 G G −4 −4 a ⋅b −1 ⎛ −4 ⎞ = = = ⇒ = cos cos θ θ G ⎜ ⎟ ≈ 104.3° G 4 + 9 + 16 1 + 4 + 4 3 29 a b ⎝ 3 29 ⎠

G G a −b

G b θ

G a

A fairly immediate and important consequence of what we’ve seen is the following: G G G G G G G G a ⋅ b = a b cos θ = 0 if and only if θ = 90° or a = 0 or b = 0.

A fairly immediate and important consequence of what we’ve seen is the following: G G G G G G G G a ⋅ b = a b cos θ = 0 if and only if θ = 90° or a = 0 or b = 0.

In other words, we can determine if two vectors are perpendicular or not simply by looking to see if the dot product is zero. Furthermore, the zero vector is considered perpendicular to all vectors.

A fairly immediate and important consequence of what we’ve seen is the following: G G G G G G G G a ⋅ b = a b cos θ = 0 if and only if θ = 90° or a = 0 or b = 0.

In other words, we can determine if two vectors are perpendicular 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.”

The Cross Product

The cross product of two vectors is defined as follows: G G ˆ If a = a1iˆ + a2 ˆj + a3 k and b = b1iˆ + b2 ˆj + b3 kˆ, ˆj iˆ G G then a × b = a1 a2 b1 b2

kˆ a2 a3 = b2 b3

a3 a ˆi − 1 b3 b1

a3 a ˆj + 1 b3 b1

= (a2b3 − a3 .b2 )iˆ − (a1b3 − a3 .b1 ) ˆj + (a1b2 − a2 .b1 )kˆ.

a2 ˆ k b2

The cross product of two vectors is defined as follows: G G ˆ If a = a1iˆ + a2 ˆj + a3 k and b = b1iˆ + b2 ˆj + b3 kˆ, ˆj iˆ G G then a × b = a1 a2 b1 b2

kˆ a2 a3 = b2 b3

a3 a ˆi − 1 b3 b1

a3 a ˆj + 1 b3 b1

= (a2b3 − a3 .b2 )iˆ − (a1b3 − a3 .b1 ) ˆj + (a1b2 − a2 .b1 )kˆ.

Isn’t that cool!

a2 ˆ k b2

The cross product of two vectors is another vector.

The cross product of two vectors is another vector. Also, the cross product is perpendicular to both vector a and vector b.

The cross product of two vectors is another vector. Also, the cross product is perpendicular to both vector a and vector b.

G G If a = a iˆ + a2 ˆj + a3 kˆ and b = b1iˆ + b2 ˆj + b3 kˆ, then G G G a ⋅ a × b = a1 (a2b3 − a3 .b2 ) + a2 [ −(a1b3 − a3 .b1 ) ] + a3 (a1b2 − a2 .b1 )

(

)

= a1a2b3 − a1a3b2 − a1a2b3 + a2 a3b1 + a1a3b2 − a2 a3b1 = 0

The cross product of two vectors is another vector. Also, the cross product is perpendicular to both vector a and vector b.

G G If a = a iˆ + a2 ˆj + a3 kˆ and b = b1iˆ + b2 ˆj + b3 kˆ, then G G G a ⋅ a × b = a1 (a2b3 − a3 .b2 ) + a2 [ −(a1b3 − a3 .b1 ) ] + a3 (a1b2 − a2 .b1 )

(

)

= a1a2b3 − a1a3b2 − a1a2b3 + a2 a3b1 + a1a3b2 − a2 a3b1 = 0 The proof that the cross product is perpendicular to vector b is similar.

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

G a

G 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.

The cross product is also known as the “vector product.” Furthermore, just like the dot product, so does the cross product have interesting properties. One of the more important ones is below:

G G G G a × b = a b sin θ , where θ is the anglebetweenthetwo vectors,

The cross product is also known as the “vector product.” Furthermore, just like the dot product, so does the cross product have interesting properties. One of the more important ones is below:

G G G G a × b = a b sin θ , where θ is the anglebetweenthetwo vectors,

The proof is amazingly simple!

Proof: G G a ×b

2

= (a2b3 − a3 .b2 ) 2 + (a1b3 − a3 .b1 ) 2 + (a1b2 − a2 .b1 ) 2

= a2 2b32 − 2a2 a3b2b3 + a32b2 2 + a12b32 − 2a1a3b1b3 + a32b12 + a12b2 2 − 2a1a2b1b2 + a2 2b12 = a12b2 2 + a12b32 + a2 2b12 + a2 2b32 + a32b12 + a32b2 2 − 2a1a2b1b2 − 2a1a3b1b3 − 2a2 a3b2b3 = a12b12 + a2 2b2 2 + a32b32 + a12b2 2 + a12b32 + a2 2b12 + a2 2b32 + a32b12 + a32b2 2 − 2a1a2b1b2 − 2a1a3b1b3 − 2a2 a3b2b3 − a12b12 − a2 2b2 2 − a32b32 = (a12 + a2 2 + a32 )(b12 + b2 2 + b32 ) − (a1b1 + a2b2 + a3b3 ) 2 2 G2 G2 G G 2 G2 G2 G G = a b − a ⋅ b = a b − a b cos θ

(

G = a

2

G2 G b − a

)

2

G2 G 2 b cos θ = a

(

2

G b

)

2

(

)

G 1 − cos θ = a 2

2

G2 2 b sin θ .

Thus, G G a ×b

2

G = a

2

G2 2 b sin θ .

And since, sin θ ≥ 0 for 0 ≤ θ ≤ π ,

We have that,

G G G G a × b = a b sin θ .

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

G G G G Area = a b sin θ = a × b

G b

G height = b sin θ

G a

EXAMPLE (Cross Product):

G a = 2iˆ + 3 ˆj + 4kˆ G b = −iˆ + 2 ˆj − 2kˆ

EXAMPLE (Cross Product):

G a = 2iˆ + 3 ˆj + 4kˆ G b = −iˆ + 2 ˆj − 2kˆ iˆ ˆj kˆ G G a × b = 2 3 4 = −14iˆ + 0 ˆj + 7 kˆ = −14iˆ + 7 kˆ −1 2 −2

EXAMPLE (Area of aParallelogram):

G a = 2iˆ + 3 ˆj + 4kˆ G b = −iˆ + 2 ˆj − 2kˆ

EXAMPLE (Area of aParallelogram):

G a = 2iˆ + 3 ˆj + 4kˆ G b = −iˆ + 2 ˆj − 2kˆ

G G Area = a × b = −14iˆ + 7 kˆ = 245 = 7 5 ≈ 15.65