The * cross product* of and is

Strictly speaking, you should take the first equality as the definition and the second equality as a helpful way to remember how to compute it. Why?

Determinants are defined for matrices all of whose entries are
numbers. In more advanced courses, you may see that
"numbers" can in general be elements in an algebraic
structure called a * commutative ring with
identity*. The problem with the determinant above is that the
elements of the first row are vectors, while the other elements are
numbers. It is not clear what *single* algebraic structure
contains the elements of the matrix, or if the properties of
determinants which hold for numerical matrices will hold for matrices
like the one above.

We will see this casual use of determinants in other places --- for
example, when we discuss the * curl* of a vector
field. Fortunately, with a little care in using these shortcuts
everything works as you'd expect.

* Example.* Compute .

Here are some algebraic properties of the cross product.

* Proposition.* Let
, , and be 3-dimensional vectors and let k be a
number.

(a) .

(b) .

(c) .

(d) .

(e) (Triple scalar product)

(f) is orthogonal to and to .

* Proof.* The idea in each case is to write the
vectors in terms of components, then compute.

For example, here's the proof of (a):

For (e), I have

Then using this formula with replaced with , replaced with , and replaced with , I have

The third and fourth equalities used the fact that swapping two rows multiplies a determinant by -1.

Now it's easy to prove (f). Since the determinant of a matrix with two equal rows is 0,

This proves that is perpendicular to . A similar argument proves the result for .

I'll show below that has a geometric interpretation: Its absolute value is the volume of the parallelepiped determined by , , and .

Property (c) shows that the cross product is not commutative. In fact, it is also not associative: In general, .

* Example.* Show that

Thus,

Thus,

Hence,

The next result gives part of the geometric interpretation of the cross product. It's routine --- just writing vectors out in terms of components and computing --- but pretty technical. You might want to skip the proof and try to understand the statement.

* Proposition.* Let and be 3-dimensional vectors. Then

is the angle from to satisfying .

* Proof.* First, note that

So

I'll use the last result in the middle of the following computation:

Taking the square root on both sides, and noting that implies that , I have

So far, I know that is a vector which is perpendicular to both and , and whose length is . This almost determines ; the only question is which of the two possible perpendicular vectors it could be:

In this picture, turns out to be the perpendicular vector pointing "upward"; the one pointing "downward" is actually .

* Definition.* An *ordered* set of vectors
in is * positively oriented* (or has a *
right-hand orientation*) if

(That is, make a matrix with the vectors *in the given order*
as its rows and take the determinant.)

If the determinant is negative, the ordered set of vectors is * negatively oriented* (or has a *
left-hand orientation*).

"Ordered set" means that if you keep the three vectors the
same but change the order in which they're listed, you have a
*different* ordered set.

* Example.* Show that the ordered set is positively oriented.

Hence, the set is positively oriented.

* Proposition.* If and are nonzero and nonparallel, then is a vector whose length is , and whose direction is
perpendicular to and , so that is positively oriented.

In other words, it is like the convention with the positive x, y, and
z-axes in : If you curl the fingers of your *right
hand* through the smaller angle from to , your thumb points in the
direction of .

* Proof.* Using the triple scalar product,

Now . Since and are nonzero, and are nonzero. Since Since and aren't parallel, .

Therefore, .

This shows that the set is positively oriented. The other assertions have been proven above.

* Example.* Find two unit vectors perpendicular
to both and .

Now

So the two unit vectors perpendicular to both and are .

Geometrically, the length of is the area of the parallelogram determined by and .

As the picture shows, is the length of the base of the parallelogram and is the altitude of the parallelogram. Consequently, their product is the area of the parallelogram, which is just

* Example.* The verices of a parallelogram,
listed counterclockwise, are , , , and . Find the area of the parallelogram.

and . Then

The area is

* Example.* Find the area of the parallelogram
whose vertices are , , , . What is the area of ?

The parallelogram is pictured below:

and are adjacent sides of the parallelogram. In order to take their cross product, regard them as 3-dimensional vectors with zero z-components: and .

Then

The area of the parallelogram is . The area of the triangle is half the area of the parallelogram: 2.

has the following geometric interpretation: Its absolute value gives the volume of the parallelepiped determined by , , and :

To see this, observe that if is the angle between and , then

is the area of the base, while is the altitude. Hence, their product is the volume of the parallelepiped (up to sign).

* Example.* Find the volume of the parallelepiped
determined by the vectors , , .

Copyright 2017 by Bruce Ikenaga