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


{\bf Question}

\begin{description}
\item[(a)] Find the cartesian equation of the plane containing the
point $(1,0,-1)$ parallel to both the lines $$\frac{x-1}{2} = y =
\frac{z+2}{3}$$ $$\frac{x+4}{3} = \frac{y-1}{2} = z$$ Find the
shortest distance between each line and the plane.
\item[(b)] The triangle $ABC$ has each side extended by the same
factor $\lambda$, as in the diagram.  (So $|AB'| = \lambda|AB|$
etc.) Find the ratio of the areas of triangles $A'B'C'$ and $ABC.$

\setlength{\unitlength}{.5in}
\begin{picture}(4,6)
\put(5,1){\line(0,1){4.65}}

\put(1,3){\line(3,2){4}}

\put(1,3){\line(2,-1){4}}

\put(1,3){\line(3,-1){3.85}}

\put(5,1){\line(-1,6){.67}}

\put(5,5.65){\line(-1,-1){3}}

\put(0.7,3){$A'$}

\put(5,0.7){$C'$}

\put(5.1,5.5){$B'$}

\put(1.8,2.75){$A$}

\put(4.4,4.7){$B$}

\put(4.4,1.5){$C$}

\end{picture}


\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] A normal to the plane is $$(2,1,3) \times (3,2,1) =
(-5,7,1)$$ So the plane has equation $$-5x + 7y +z = k$$It
contains $(1,0,-1)$ so $k = -6$

Therefore the equation is $$-5x + 7y +z = -6$$

Distance of ${\bf p}$ from ${\bf a \cdot n }=k$ is $\ds
\frac{|{\bf a \cdot p} - k|}{|{\bf a}|}$

So the distance of $l_1$ from $\pi$ is the distance of $(1,0,-2)$
from  $\pi$

$\ds = \frac{|-7 + 6|}{\sqrt{75}} = \frac{1}{\sqrt{75}} =
\frac{1}{5 \sqrt 3}$

So the distance of $l_2$ from $\pi$ is the distance of $(-4,1,0)$
from  $\pi$

$\ds = \frac{|27 + 6|}{\sqrt{75}} = \frac{33}{\sqrt{75}} =
\frac{11 \sqrt 3}{5}$


${}$
\item[(b)] Let $A$ be the origin $\vec{AB} = {\bf b} \, \,
\vec{AC} = {\bf c}$

\begin{eqnarray*} \vec{A'B'} = \vec{A'A} + \vec{AB'} & = &
(\lambda -1) {\bf c} + \lambda{\bf b} \\ \vec{A'C'} = \vec{A'C} +
\vec{CC'} & = & \lambda c + (\lambda -1) (\bf{c-b}) \\ & = &
(2\lambda -1){\bf c} - (\lambda -1){\bf b}\end{eqnarray*}

The area of $A'B'C'$ is

\begin{eqnarray*} \frac{1}{2}|(\vec{A'B'} \times \vec{A'C'})| & = &
\frac{1}{2} \left|((\lambda -1) {\bf c} + \lambda {\bf b}) \times
((2\lambda -1) {\bf c}(\lambda -1) {\bf b})\right| \\ & = &
\frac{1}{2} |+(\lambda -1)^2 + \lambda(2\lambda -1)||{\bf B \times
C}| \\ & = & \frac{1}{2} |3 \lambda^2 - 3\lambda +1|\, |{\bf b
\times c}|
\end{eqnarray*} Therefore the required ration is $|3\lambda^2 -
3\lambda +1|$
\end{description}


\end{document}