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

{\bf Question}

Let the vertices of a triangle $ABC$ have the following position
vectors relative to some origin $O$:

\begin{eqnarray*} {\bf{OA}}={\bf{i}}+2{\bf{j}}+{\bf{k}},\\
{\bf{OB}}=2{\bf{i}}+{\bf{j}},\\
{\bf{OC}}={\bf{i}}-{\bf{j}}+2{\bf{k}}. \end{eqnarray*}

\begin{description}
\item[(i)]
Show these vectors and the triangle on a rough sketch.

\item[(ii)]
Find the angle between ${\bf{AB}}$ and ${\bf{AC}}$. Repeat the
calculation for ${\bf{BA}}$ and ${\bf{BC}}$. Hence deduce the
three angles within the triangle.

\item[(iii)]
Calculate the area of the triangle $ABC$ using an appropriate
vector product.

\end{description}
\medskip

{\bf Answer} 
\begin{description}
\item[(i)]

${}$

\setlength{\unitlength}{.5in}
\begin{picture}(7,4)

\put(0,1){\vector(1,0){5}}

\put(0,1){\vector(3,2){3}}

\put(0,1){\vector(3,1){6}}

\put(5,1){\line(1,2){1}}

\put(3,3){\line(1,0){3}}

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

\put(3,3){\makebox(0,0)[b]{$A$}}

\put(5,1.1){\makebox(0,0)[l]{$B$}}

\put(6,3){\makebox(0,0)[l]{$C$}}

\put(2,.5){\makebox(0,0)[l]{${\bf{OB}}=2{\bf{i}}+{\bf{j}}+0{\bf{k}}$}}

\put(3,3){\makebox(0,0)[r]{${\bf{OA}}={\bf{i}}+2{\bf{j}}+{\bf{k}}\
$}}

\put(3,2){\makebox(0,0)[l]{${\bf{OC}}={\bf{i}}-{\bf{j}}+2{\bf{k}}$}}
\end{picture}

(or topologically equivalent...)

\item[(ii)]
\begin{eqnarray*} {\bf{AB}}={\bf{AO}}+{\bf{OB}} & = &
-{\bf{OA}}+{\bf{OB}}\\ & = &
-{\bf{i}}-2{\bf{j}}-{\bf{k}}+2{\bf{i}}+{\bf{j}}\\ & = &
{\bf{i}}-{\bf{j}}-{\bf{k}} \end{eqnarray*}

\begin{eqnarray*} {\bf{AC}}={\bf{OC}}-{\bf{OA}} & = &
{\bf{i}}-{\bf{j}}+2{\bf{k}}-{\bf{i}}-2{\bf{j}}-{\bf{k}}\\ & = &
-3{\bf{j}}+{\bf{k}} \end{eqnarray*}

\begin{eqnarray*} {\bf{BC}}={\bf{OC}}-{\bf{OB}} & = &
{\bf{i}}-{\bf{j}}+2{\bf{k}}-2{\bf{i}}-{\bf{j}}\\ & = &
-{\bf{i}}-2{\bf{j}}+2{\bf{k}} \end{eqnarray*}

$\angle CAB$ given by

$${\bf{AC}} \cdot {\bf{AB}}=|{\bf{AC}}||{\bf{AB}}| \cos(\angle
CAB)$$

\begin{eqnarray*} {\bf{AC}} \cdot {\bf{AB}} & = & (0,-3,1) \cdot
(+1,-1,-1)\\ & = & +3-1\\ & = & +2 \end{eqnarray*}

$|{\bf{AC}}|=\sqrt{0^2+9+1}=\sqrt{10}$

$|{\bf{AB}}|=\sqrt{1+1+1}=\sqrt{3}$

Therefore $\cos(\angle
CAB)=\ds\frac{+2}{\sqrt{10}\sqrt{3}}=\ds\frac{+2}{\sqrt{30}}=0.51639$

$\Rightarrow \angle
CAB=\arccos\left(\ds\frac{+2}{\sqrt{30}}\right)=68.583^{\circ}$

${\bf{BA}}=-{\bf{AB}}=-{\bf{i}}+{\bf{j}}+{\bf{k}};\
{\bf{BC}}=-{\bf{i}}-2{\bf{j}}+2{\bf{k}}$

Therefore ${\bf{BA}} \cdot {\bf{BC}}=(-1,1,1)(-1,-2,2)=1-2+2=1$

$|{\bf{AB}}|=\sqrt{3}$

$|{\bf{AB}}|=\sqrt{1+4+4}=3$

Therefore $\angle
CBA=\arccos\left(\ds\frac{1}{3\sqrt{3}}\right)=\arccos(0.19245)=78.904^{\circ}$

$\angle BCA=180-\arccos\left(\ds\frac{2}{\sqrt{30}}\right)-
\arccos\left(\ds\frac{1}{3\sqrt{3}}\right)=32.51$

\item[(iii)]
\begin{eqnarray*} \rm{Area\ of}\ \bigtriangledown & = &
\ds\frac{1}{2}|{\bf{AB}} \times {\bf{AC}}|\\ & = &
\ds\frac{1}{2}\left|\left|\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ 1 & -1 & -1\\ 0 & -3 & 1 \end{array} \right| \right|\\
& = & \ds\frac{1}{2}\left|{\bf{i}}\left|\begin{array}{cc}-1 &
-1\\-3 & 1 \end{array}\right|-{\bf{j}}\left|\begin{array}{cc}1 &
-1\\0 & 1 \end{array}\right|+{\bf{k}}\left|\begin{array}{cc}1 &
-1\\0 & -3 \end{array}\right|\right|\\ & = &
\ds\frac{1}{2}|-4{\bf{i}}-{\bf{j}}-3{\bf{k}}|\\ & = &
\ds\frac{1}{2}\sqrt{16+1+9}\\ & = &
\ds\frac{\sqrt{26}}{2}=\sqrt{\ds\frac{13}{2}}=2.549
\end{eqnarray*}

\end{description}

\end{document}