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### Cross Products; Rotating in Three Dimensions

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Date: 10/26/2001 at 00:00:24
From: RJ
Subject: Matrices (Cross-Product)

Hi,

We are currently studying matrices and vectors, and we just came to
the concept of the cross product. Our entire class understands how
it works, but we don't understand why or the proof behind it. Could
you give us any insight as to why it works?  Thanks.
```

```
Date: 10/26/2001 at 03:37:33
From: Doctor Mitteldorf
Subject: Re: Matrices (Cross-Product)

Dear RJ,

One important thing to know about cross-products is that they're a
special accident of 3 dimensions. The cross-product of two vectors is
about rotation in the plane defined by those two vectors. Components
of the product are not really vector components - they're really
directions of a plane surface. The three components of a cross product
are really not (x,y,z) but rather (yz,zx,xy).

If we lived in a 4-dimensional world, there would be a cross-product
object, but it wouldn't be a vector. It would be a six-component
thing, called a "tensor," with all possible pairs of the 4 coordinates
for components:  wx,wy,wz,xy, and xz.

(w1)     (w2)    (   0       w2x1-x2w1  w2y1-y2w1  w2z1-z2w1)
(x1)  X  (x2) =  (w1x2-x1w2     0       x2y1-y2x1  x2z1-z2x1)
(y1)     (y2)    (w1y2-y1w2  x1y2-x2y1      0      y2z1-z2y1)
(z1)     (z2)    (w1z2-z1w2  x1z2-z1x2  y1z2-z1y2      0    )

In other words, How many ways are there to rotate in 3 dimensions?
There are 3, but it's only an accident that this number is the same as
the number of dimensions. It's really the number of pairs of different
dimensions that is the answer, and in 3 dimensions this happens to be
the same number. In 4 dimensions, there aren't 4 pairs of dimensions,
but 6.

(How many such pairs are there in 5 dimensions?)

So it's a shorthand and a convenience that we write the cross-product
as a vector, rather than as an "anti-symmetric tensor."  Can you
induce what the word "anti-symmetric" means here?

Think of the cross product of vector A and vector B as a rotation in
the plane defined by A and B.

...But maybe this isn't what you were looking for. Please write back
and tell me what you meant by "how it works" and whether there's a
specific proof you don't understand.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra
High School Linear Algebra

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