\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}}=2{\bf{i}}+3{\bf{j}}-{\bf{k}},\\
{\bf{OB}}={\bf{i}}+{\bf{j}}+{\bf{k}},\\
{\bf{OC}}={\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]{${\bf{OA}}=2{\bf{i}}+3{\bf{j}}-{\bf{k}}$}}

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

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

\item[(ii)]
\begin{eqnarray*} {\bf{AB}}={\bf{AO}}+{\bf{OB}} & = &
{\bf{OB}}-{\bf{OA}}\\ & = &
{\bf{i}}+{\bf{j}}+{\bf{k}}-2{\bf{i}}-3{\bf{j}}+{\bf{k}}\\ & = &
-{\bf{i}}-2{\bf{j}}+2{\bf{k}} \end{eqnarray*}

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

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

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

$|{\bf{AC}}|=\sqrt{4+4+9}=\sqrt{17}$

$\Rightarrow$ \begin{eqnarray*} \cos(\theta) & = &
\ds\frac{(-1,-2,2) \cdot (-2,-2,3)}{3\sqrt{17}}\\ & = &
\ds\frac{2+4+6}{3\sqrt{17}}=\ds\frac{4}{\sqrt{17}} \end{eqnarray*}

$\theta=\arccos(0.970143)=14.04^{\circ}$

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

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

${\bf{BA}} \cdot {\bf{BC}}=|{\bf{BA}}||{\bf{BC}}|\cos \theta$

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

$|{\bf{BC}}|=\sqrt{1+1}=\sqrt{2}$

Therefore

\begin{eqnarray*} \cos\theta & = & \ds\frac{(1,2,-2) \cdot
(-1,0,1)}{3\sqrt{2}}\\ & = & \ds\frac{-1-2}{3\sqrt{2}}\\ & = &
-\ds\frac{1}{\sqrt{2}}\\ \theta & = & \un{135^{\circ}}
\end{eqnarray*}

Other angle = $\angle ACB=180-135-14.04=30.86^{\circ}$

\newpage
\item[(iii)]
\begin{eqnarray*} \rm{Area\ of}\ \bigtriangledown ABC & = &
\ds\frac{1}{2}\left|\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ -1 & -2 & 2\\ -2 & -2 & 3
\end{array} \right|\\ & = &
\ds\frac{1}{2}|{\bf{i}}(-2\times 3-(-2)\times2)\\ & &
-{\bf{j}}((-1) \times 3-(-2)\times 2)\\ & & +{\bf{k}}((-1) \times
(-2) - (-2) \times (-2)|\\ & = &
\ds\frac{1}{2}|-2{\bf{i}}-{\bf{j}}-2{\bf{k}}|\\ & = &
\ds\frac{1}{2}\sqrt{4+1+4}\\ & = & \un{\ds\frac{3}{2}}
\end{eqnarray*}

\end{description}


\end{document}
