\documentclass[12pt]{article} % BB Jan 11 2013.
\textheight 9.0in
\textwidth  6.0in
\topmargin -1.0in  
\headsep    0.4in  
\oddsidemargin  0.25in  
\evensidemargin 0.25in  

\def\sqr#1#2{{\vcenter{\hrule height.#2pt
   \hbox{\vrule width.#2pt height#1pt \kern#1pt
      \vrule width.#2pt}
   \hrule height.#2pt}}}
\def\square{{\mathchoice\sqr64\sqr64\sqr{4.2}3\sqr{3.0}3}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}
% \pagestyle{empty}

\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{\bf Levi-Cevita Tensor Homework 1} \\ (January 23, 2013) \\
\end{center}
\setcounter{equation}{0}
\bigskip

\begin{enumerate}

\item Use the $3D$ identity
\begin{eqnarray} \label{epsdel}
  \sum_{i=1}^3 \epsilon_{ijk}\epsilon_{ilm} =
  \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}
\end{eqnarray} 
to calculate
\begin{eqnarray} 
  (\hat{a}\times\hat{b})\cdot(\hat{a}\times\hat{b})
\end{eqnarray} 
where $\hat{a}$ and $\hat{b}$ are unit vectors. Eliminate all 
$\hat{a}\cdot\hat{b}$ in favor of $\cos\theta$. 

Solution: Let
\begin{eqnarray} \label{hataxhatb}
  \hat{a}=\left(\matrix{\tilde{a}_1\cr\tilde{a}_2\cr\tilde{a}_3}\right)
  ~~{\rm and}~~ \hat{b}=
  \left(\matrix{\tilde{b}_1\cr\tilde{b}_2\cr\tilde{b}_3}\right)\,.
\end{eqnarray} 
Then
\begin{eqnarray} 
  &~& (\hat{a}\times\hat{b})\cdot(\hat{a}\times\hat{b})\ = \\ \nonumber
  &~& \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \sum_{l=1}^3
  \sum_{m=1}^3 \epsilon_{ijk}\epsilon_{ilm}
  \tilde{a}_j\tilde{b}_k\tilde{a}_l\tilde{b}_m\ = 
\end{eqnarray} 
\begin{eqnarray} \nonumber
  \sum_{j=1}^3 \sum_{k=1}^3 \sum_{l=1}^3 \sum_{m=1}^3 \left(
  \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}\right)\,
  \tilde{a}_j\tilde{b}_k\tilde{a}_l\tilde{b}_m\ = 
\end{eqnarray} 
\begin{eqnarray} \nonumber
  \sum_{l=1}^3 \sum_{m=1}^3 \left(
  \tilde{a}_l\tilde{b}_m\tilde{a}_l\tilde{b}_m -
  \tilde{a}_m\tilde{b}_l\tilde{a}_l\tilde{b}_m\right)&=&\\ \nonumber
  \left(\hat{a}\cdot\hat{a}\right)\,\left(\hat{b}\cdot\hat{b}\right)-
  \left(\hat{a}\cdot\hat{b}\right)\,\left(\hat{b}\cdot\hat{a}\right)\
  &=&\\ \nonumber 1 - \cos^2(\theta)\ =\ \sin^2(\theta)\,.&~&
\end{eqnarray} 

\item Show $\vec{a}\times\vec{b}=a\,b\,\sin\theta$ (book p.21).

From (\ref{hataxhatb}) we have $\hat{a}\times\hat{b}=\sin\theta$.
Hence,
\begin{eqnarray} 
  \vec{a}\times\vec{b}\ =\ a\,b\,\hat{a}\times\hat{b}\ =\
  a\,b\,\sin\theta\,.
\end{eqnarray} 

\end{enumerate}
\end{document}
