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\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{\bf Vectors 2} (Classwork January 9, 2013) \\
\end{center}
\setcounter{equation}{0}

\rightline{Group \#~~~~~~~~}

\leftline{Participating students (print):} \vskip 60pt

\centerline{------------------------------------------------------}

\bigskip

\begin{enumerate}
\item Write down the {\bf commutative} law of vector addition
\begin{equation}
  \vec{a} + \vec{b}\ =\ \vec{b} + \vec{a}\,.
\end{equation}
\item Write down the {\bf associative} law of vector addition
\begin{equation}
  \left(\,\vec{a}+\vec{b}\,\right)+\vec{c}\ =\
  \vec{a}+\left(\,\vec{b}+\vec{c}\,\right)
\end{equation}
\item How is the positively chosen angle $\theta$ between two $nD$ 
vectors $\vec{a}$, $\vec{b}$ defined?
\begin{equation}
  \cos(\theta)\ =\ \hat{a}\cdot\hat{b}\,,~~~0\le\theta\le\pi\,.
\end{equation} 

\eject

\item Write down the velocity for a $nD$ position vector
\begin{equation}
  \vec{r}\ =\ \left(\matrix{x_1(t)\cr .\cr .\cr .\cr x_n(t)}\right)
  \,, ~~~~\vec{v}\ =\ \left(\matrix{\frac{dx_1}{d\,t}\cr .\cr .\cr .
                       \cr \frac{dx_n}{d\,t}}\right)
\end{equation} \smallskip
\item Draw (millimeter paper provided) 
\begin{eqnarray} 
  \vec{r}(t)=\vec{r}_0+\vec{v}\,t
\end{eqnarray} 
with (in arbitrary units)
\begin{equation} \label{sab}
  \vec{r}_0=\left(\matrix{2\cr 4}\right)\,,~~
  \vec{v}_0=\left(\matrix{4\cr 2}\right)\,,~~0\le t\le 2\,.
\end{equation} \bigskip
\item Calculate the work (in SI units $[J]$) for 
\begin{equation}
  \vec{F}= \left(\matrix{2\cr 1}\right)\,[N]\,,~~
  \triangle\vec{r}= \left(\matrix{1\cr 3}\right)\,[m]\,.
\end{equation} 
\begin{equation}
  W\ =\ (2+3)\,[N\,m]\ =\ 5\,[J]\,.
\end{equation} 

\eject

\item Describe the the surface swept out by $\vec{r}$ for \vskip -60pt
\begin{eqnarray} \label{a}
  \left(\vec{r}-\vec{a}\right)\cdot\vec{a} &=& 0\,,\\
  \left(\vec{r}-\vec{a}\right)\cdot\vec{r} &=& 0\,,\label{b}
\end{eqnarray} 
where $\vec{a}$ is a constant non-zero $nD$ vector (compare
exercise 1.2.2 of the book).

The trick is to write $\vec{r}$ as
\begin{eqnarray}
  \vec{r}\ =\ r_{\perp}\,\hat{a}_{\perp} + r_{\|}\,\hat{a}~~{\rm with}
  ~~\hat{a} = \frac{\vec{a}}{a}~~~
\end{eqnarray} 
where $a=\left|\vec{a}\right|>0$ and $\hat{a}_{\perp}$ is any unit 
vector perpendicular to $\vec{a}$, i.e., $\hat{a}_{\perp}\cdot\hat{a}
=0$.

Calculation for (\ref{a}):
\begin{eqnarray} \nonumber
  0 &=& \left(\vec{r}-\vec{a}\right)\cdot\vec{a}\ =\ 
  \left(r_{\|}-a\right)\,a \\ &\Rightarrow& r_{\|}\ =\ a
\end{eqnarray} 
and the solution is
\begin{eqnarray}
  \vec{r}\ =\ r_{\perp}\,\hat{a}_{\perp} + a\,\hat{a}
\end{eqnarray} 
with arbitrary values for $r_{\perp}$. The is the $(n-1)\,D$ plane
perpendicular to $\vec{a}$ and located at the tip of $\vec{a}$.

Calculation for (\ref{b}):
\begin{eqnarray} \nonumber
  0 &=& \left(\vec{r}-\vec{a}\right)\cdot\vec{r}\ =\ 
  r_{\perp}^2+\left(r_{\|}-a\right)\,r_{\|} \\ &\Rightarrow& 
  r_{\perp}^2\ =\ \left(a-r_{\|}\right)\,r_{\|} \nonumber
\end{eqnarray} 
with the solution 
\begin{eqnarray} 
  r_{\perp}\ =\ \pm\sqrt{\left(a-r_{\|}\right)\,r_{\|}}
\end{eqnarray} 
for $0\le r_{\|}\le a$. The part $+\sqrt{\left(a-r_{\|}\right)\,r_{\|}}$
is drawn below (the $-\sqrt{}$ root is included by $\hat{a}_{\perp}\to -
\hat{a}_{\perp}$).

\end{enumerate}

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=1.0]{perp}
% \caption{} \label{fig:2b}
\end{center}
\end{figure}

\end{document}
