\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand\ds{\displaystyle}
\begin{document}

\parindent=0pt

QUESTION

The point $A(2,3,1),\ B(0,1,2)$ and $C(2,-1,-1)$ lie on a plane.

\begin{description}

\item[(i)]
Write down the vectors $\vec{AB}$ and $\vec{AC}$.

\item[(ii)]
Obtain a unit vector perpendicular to the plane.

\item[(iii)]
Derive the vector equation of the plane.

\item[(iv)]
Find the perpendicular distance from the point $D(5,2,2)$ to the
plane.

\item[(v)]
Obtain the coordinates of the point at which the perpendicular
from $D$ to the plane intersects the plane.

\end{description}

\bigskip

ANSWER

$A(2,3,1),\hspace{1cm}B(0,1,2),\hspace{1cm}C(2,-1,-1)$

\begin{description}

\item[(i)]
$\vec{AB}=(0-2,1-3,2-1)=(-2,-2,1)\\
\vec{AC}=(2-2,-1-3,-1-1)=(0,-4,-2)$

\item[(ii)]
$\vec{AB}\times\vec{AC}=(-2,-2,1)\times(0,-4,-2)=(4-(-4),0-4,8-0)=(8,-4,8)$.
To find the unit vector divide by the magnitude.\\ Now
$\ds|\vec{AB}\times\vec{AC}|=(8^2+(-4)^2+8^2)^\frac{1}{2}=(64+16+64)^\frac{1}{2}=12,$\\
therefore the unit vector is
$\ds\frac{1}{12}(8,-4,8)=\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)$

\item[(iii)]
The vector equation of the plane is
$\textbf{r.n}=c=\textbf{a.n}$\\ Therefore
$\ds\textbf{r.n}=(2,3,1).\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)
=\frac{4}{3}-1+\frac{2}{3}=1$\\ i.e.
$\ds\textbf{r}.\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)=1$
or $\textbf{r}.(2,-1,2)=3$

\item[(iv)]
$ $

\begin{center}
\epsfig{file=17x-2000-2.eps, width=40mm}
\end{center}


Distance from $D$ to plane
=$\ds\left|\vec{DA}.\hat{\mathbf{n}}\right|=
\left|(-3,1,-1).\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)\right|=
\left|-2-\frac{1}{3}-\frac{2}{3}\right| =|-3|=3$

\item[(v)]
Coordinates of $N$ are

$\ds(5,2,2)-3\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)=(5,2,2)-(2,-1,2)=(3,3,0)$

(Other methods could have been used).
\end{description}




\end{document}