# Explanation of the cross product

1. Jul 11, 2011

### Forrest T

Can someone please explain to me the motivation for these definitions of the cross product?
Let A = (a1, a2, a3), B = (b1, b2, b3). Let A and B be the magnitudes of A and B, respectively, and let $\theta$ be the angle between the vectors.

A X B = AB sin($\theta$)

A X B = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

As I understand it, the cross product was developed for application to situations in physics, in which two vectors produce a result perpendicular to both of them. Therefore, the cross product would be orthogonal to both vectors: A $\cdot$ (A X B) = 0, and B $\cdot$ (A X B) = 0. However, I do not understand the other properties of the cross product which led to these definitions. Thank you for your help.

2. Jul 11, 2011

### Studiot

Good morning Forrest.

A vector (in this sense) is an ordered triple. That is it is three numbers bracketed together.

{p,q,r} where p, q and r are real numbers is a vector. You can regard p,q and r as its components.

That is exactly what you have in the second formula. The numbers given by the expression are the components of the vector which results from the cross product.

Your first formula needs further consideration, because as written it is incomplete.
ABsin($\theta$) is not a vector.

It is the magnitude of the resultant vector.

If you want to make this formula into a vector one you must add a unit vector pointing in a direction given by a right handed system.

Edit
A X B = AB sin(θ) n

As it happens this magnitude is also the area of the parallelogram which has A and B as sides. This fact is sometimes used when we want directed areas and we use the associated unit vector to represent the unit normal to the area.

There are many good representations to this in this forum ad the question has been asked many times.
Have you tried the forum search facility?

Last edited: Jul 11, 2011
3. Jul 11, 2011

### chogg

Hello Forrest,

Excellent question! I recently gave some thoughts on vector products (including the cross product) here:
I'd like to expand on my thoughts a bit.

First: the kinds of physical examples that motivate using a cross product are typically some kind of area, or moment. Perhaps the best example is angular momentum: remember that Kepler's law says the area swept out per unit time is constant. So we will need a way to calculate this area! Specifically, the area swept out per unit time, with respect to some origin, will be (proportional to) an "area product" $r \wedge v$ between the position vector $r$ and the velocity vector $v=\dot{r}$.

With that application in the back of our mind: how do we go about constructing an "area product" $a \wedge b$? Let's start with a few properties we'd want any product to have:
• $a\wedge(b+c) = a\wedge b + a\wedge c$ (distributive over addition: areas add just like vectors do)
• $(\alpha a)\wedge b = a\wedge (\alpha b) = \alpha(a \wedge b)$ (linear in each argument: scale up an argument, and you scale up the area by the same factor)
Now, for an area product in particular, we can figure out a few additional properties we'd want. The strategy will be to figure out the area for certain special cases that we know, and then see if we can generalize.
• $a\wedge a = 0$. Basically, you need two different vectors to span a nonzero area. Interestingly, this property implies that $a \wedge b = - b \wedge a$. In other words, if you make an area product, you automatically get a sense of orientation to go along with it. (The area from $a$ to $b$ is minus the area from $b$ to $a$).
• $a\wedge b = ab$ whenever $a\cdot b=0$. This is because whenever two vectors are perpendicular, our "area product" is just the familiar area of a rectangle.

Now we already have your first definition implied. To see why, let's choose a convenient basis to make the calculations easier. (We can choose any basis we want, of course, without affecting the final result.) Our basis should have the following properties:
• Orthonormal: $e_1 \cdot e_1 = 1$ and $e_1 \cdot e_2 = 0$, etc.
• $e_1$ is aligned along $a$, so we can write $a = |a| e_1$
• $e_2$ is in the plane formed by $a$ and $b$, so we can write $b = |b|(\cos\theta e_1 + \sin\theta e_2)$
Now we simply form our area product, remembering that $e_1 \wedge e_1 = 0$:
$$a \wedge b = |a||b|\sin\theta (e_1 \wedge e_2)$$

Here, the $(e_1 \wedge e_2)$ factor is shorthand for "the plane spanned by $e_1$ and $e_2$". If we decide to replace that factor with a normal to that plane, then we have the cross product.

As for your second definition: now we'll let the basis vectors be completely arbitrary. Write
\begin{align} a &= a_1e_1 + a_2e_2 + a_3e_3 \\ b &= b_1e_1 + b_2e_2 + b_3e_3 \end{align}
Simply do the brute force multiplication, then simplify by remembering that, for instance, $e_1 \wedge e_2 = -e_2 \wedge e_1$ and collecting like terms. You'll end up with three terms, one of which is
$$(a_1b_2 - a_2b_1)(e_1 \wedge e_2)$$
This is where your second definition (the one in terms of components) comes from. You can consider $(a_1b_2 - a_2b_1)$ to be the "$(e_1 \wedge e_2)$-component" of the area product. And of course, if you turned your area product into a cross product by taking surface normals, it would be the "$e_3$-component" of your cross product.

So to summarize: we wanted an "area product", so we could do computations with things like angular momentum (and torques, and electromagnetic fields, and so on). We wanted it to have a few friendly properties we'd want in pretty much any product (distributivity and linearity). And we insisted that it give the right answer in simple special cases (zero area for parallel edges, and "rectangle area" for perpendicular vectors). From these requirements, your definitions popped out. We got the "geometric" one by choosing a particular convenient basis, and we got the "component" one by checking the result for an arbitrary basis.

4. Jul 11, 2011

### EWH

Very good explanation, Chogg. Anything using the cross product is really describing an oriented area (with the orientation describing a direction of rotation - go east then north and you're going around the block counterclockwise as seen from above, go north then east and you're going clockwise, so a^b= -b^a). In 3D (and maybe 7D) we can represent that area with a vector normal to that plane of rotation, but it isn't a general solution. It does not work in 2D or 4D, for example. And the axial vectors we get are not the same sort of vectors as the ones we started with, leading to endless confusion.