\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

\begin{description}
\item[(a)] Are the two lines $x = 3y + 7, \, \, \, y = z-1$ and
$x-1 = 3y - 5 = 3z+1$ parallel? (Give a reason.)
\item[(b)] At what point does the line ${\bf r} = \frac{1}{2}{\bf
i} + u{\bf j} + {\bf k}$ $u\epsilon{\bf R}$ cut the ellipse ${\bf
r} = \sin t {\bf i} + \cos t {\bf j} + 2\sin t{\bf k}, \, \,
t\epsilon{\bf R}$
\item[(c)] What surface does the equation ${\bf r} = 4(\cos u{\bf i}
+ \sin u {\bf j}) + v{\bf k}, \, \, \, u,v\epsilon{\bf R}$?
\item[(d)] Given a geometrical interpretation of the equation
$|{\bf r} - {\bf a}|^2 = f^2$ when ${\bf a}$ is a given vector and
$f$ is a constant.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] Put the lines is standard form $$\frac{x}{1} = \frac{y +
\frac{7}{3}}{\frac{1}{3}} = \frac{z +2}{\frac{1}{3}} \, \, : \, \,
\frac{x-1}{1} = \frac{y-\frac{5}{3}}{\frac{1}{3}} =
\frac{z+\frac{1}{3}}{\frac{1}{3}} $$

They are parallel.  Both have direction vector $\ds\left(1,
\frac{1}{3}, \frac{1}{3} \right)$

${}$
\item[(b)] ${\bf r} = \frac{1}{2}{\bf i} + u{\bf j} + {\bf k}$
meets

${\bf r} = \sin t{\bf i} + \cos t{\bf j} + 2\sin t{\bf k}$

Where $\sin t = \frac{1}{2}$ and $\cos t = u$

So $\cos t = \pm \frac{{\sqrt 3}}{2}$

So the line cuts the ellipse at two points $\ds\left( \frac{1}{2},
\pm \frac{\sqrt 3}{2}, 1\right)$

${}$
\item[(c)] ${\bf r} = 4(\cos u{\bf i} + \sin u{\bf j}) + v{\bf k}$
represents a cylinder centred on the z-axis with radius 4.

${}$
\item[(d)] $|{\bf r-a}|^2 = f^2$ represents a sphere, centre ${\bf
a}$ radius $|f|$
\end{description}

\end{document}
