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


{\bf Question}

Two lines are given in space by the following equations.
$$\frac{x}{2} = \frac{y-1}{2} = z; \hspace{.2in} x+1 = y-2 =
\frac{z+4}{2}$$  Find the equations of the following planes
\begin{description}
\item[(i)] The plane containing the first line and parallel to the
second line
\item[(ii)] The plane containing the second line and parallel to
the first line
\item[(iii)] The plane containing the first line and passing
through the origin
\item[(iv)] The plane containing the first line and the common
perpendicular
\item[(v)] The plane containing the second line and the common
perpendicular
\item[(vi)] The angle between the planes (iv) and (v).
\end{description}

\vspace{.25in}

{\bf Answer}

\setlength{\unitlength}{.5in}
\begin{picture}(4,2)
\put(0,0){\vector(2,1){3}}

\put(0.5,1.5){\line(2,-1){3}}

\put(2.5,0.5){\line(0,1){.75}}

\put(0,0.2){$l$}

\put(.2,1.5){$m$}

\put(3.2,1.5){$a$}

\put(3.6,0){$b$}

\end{picture}

\begin{description}
\item[(i)(ii)] The plane containing $l$ and that is parallel to
$m$ and the plane containing $m$ and that is parallel to $l$ both
have ${\bf a \times b}$ as a normal vector.

$(2,3,1) \times (1,1,2) = (3,-3,0)$

So the equation of plane (i) is $x - y = -19$ (contains (0,1,0))

So the equation of plane (ii) is $x - y = -3$ (contains (-1,2,-4))

${}$

\item[(iii)] Let the equation be $ax + by + cz=0$

It contains the first line and so contains $(0,1,0)$.  Thus $b=0$.

Its normal is perpendicular to ${\bf a}$ so $(a,0,c) \cdot (2,3,1)
= 0 \Rightarrow 2a + c = 0$  choose $a=1 \,c=-2$

Thus the equation is $x - 2z=0$

${}$

\item[(iv)] Let the equation be $ax+by+cz=k$.

The normal $(a,b,c)$ is perpendicular to ${\bf a}$ and to
$(3,-3,0)$.

So $2a + 3b + c = 0$ and $ 3a-3b = 0$

Thus $a = b\, c = -4b$ so choose $a=b=1$ and $c=-4$

The equation therefore is $x+y-4z = 1$ (contains (0,1,0))

${}$

\item[(v)]Let the equation be $ax+by+cz=k$ (as in (iv))

$\left.\begin{array}{rcl} a+b+2c & = & 0 \\ 3a -3b & = & 0
\end{array}\right\}a = b =-c$ choose $ a = b = -c = 1$

So the plane is $x+y-z = 5$ as it contains (-1,2,-4)

${}$

\item[(vi)] \begin{eqnarray*} \cos \theta & = & \frac{(9,1,-4)
\cdot (1,1,-1)|}{|(1,1,-4)||(1,1,-1)|} \\ & = & \frac{6}{\sqrt
{18} \sqrt{3}} = \frac{2}{\sqrt6} \\ \theta & = & 35^\circ \\ {\rm
or\ \ } \theta & = & 0.6155 {\rm \ \ radians} \end{eqnarray*}


\end{description}
\end{document}