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\author{Neil Edelman}
\title{``Generalised'' Cross-Product}
\date{2007-05-03}

\begin{document}

\thispagestyle{empty}

\maketitle

For all those in MATH-133, here's a great trick that will save you
lots and lots of time. I came up with it when I was in MATH-133
several years ago.

Let's start easy; so you've got a vector and you need to get a
perpendicular vector.

\fig{axes}{This is the situation.}

Well you do it like this:

\begin{align}
\times (v) &=
\left| \begin{array}{cc}
\hat{x} &     \hat{y} \\
v_0 &         v_1
\end{array} \right| \\
&= v_1 \hat{x} - v_0 \hat{y}
\end{align}

But wait, it generalises:

\begin{align}
\times (x_0, \ldots, x_n) &=
\left| \begin{array}{ccc}
\hat{x_0} &     \cdots &     \hat{x_n} \\
x_{00} &        \cdots &     x_{0n} \\
\vdots &        &            \vdots \\
x_{n0} &        \cdots &     x_{nn}
\end{array} \right|
\end{align}

Which the cross-product is a sub-operation of dimension three.

{\it eg}

\begin{align}
\times (a, b, c) &=
\left| \begin{array}{cccc}
\hat{x} & \hat{y} & \hat{z} & \hat{t} \\
a_0 &     a_1 &     a_2 &     a_3 \\
b_0 &     b_1 &     b_2 &     b_3 \\
c_0 &     c_1 &     c_2 &     c_3
\end{array} \right|
\end{align}

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