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

{\bf Question}

Express the following vectors as the product of a scalar and a
unit vector.

\begin{description}
\item[(i)]
${\bf{a}}=2{\bf{i}}-{\bf{j}}+3{\bf{k}}$

\item[(ii)]
${\bf{a}}=3{\bf{i}}-3{\bf{j}}+{\bf{k}}$

\item[(iii)]
${\bf{a}}=\ds\frac{-\sqrt{71}}{9}{\bf{i}}-\ds\frac{1}{3}{\bf{j}}-\ds\frac{1}{9}{\bf{k}}$

\end{description}

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$|{\bf{a}}|=\sqrt{2^2+(-1)^2+3^2}=\sqrt{4+1+9}=\sqrt{14}$

Therefore
${\bf{a}}=\hat={\bf{a}}|{\bf{a}}|=\sqrt{14}\left(\underbrace{\ds\frac{2{\bf{i}}}{\sqrt{14}}
-\ds\frac{{\bf{j}}}{\sqrt{14}}+3\ds\frac{{\bf{k}}}{\sqrt{14}}}\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ | \hat{{\bf{a}}}|=1$

\item[(ii)]
$|{\bf{a}}|=\sqrt{3^2+(-3)^2+1^2}=\sqrt{19}$

Therefore
${\bf{a}}=\hat={\bf{a}}|{\bf{a}}|=\sqrt{19}\left(\underbrace{\ds\frac{3{\bf{i}}}{\sqrt{19}}
-3\ds\frac{{\bf{j}}}{\sqrt{19}}+\ds\frac{{\bf{k}}}{\sqrt{19}}}\right)$

\item[(iii)]
$|{\bf{a}}|=\sqrt{\ds\frac{71}{81}+\ds\frac{1}{9}+\ds\frac{1}{81}}=\ds\frac{\sqrt{81}}{9}=1$

Thus ${\bf{a}}$ is already a unit vector so
${\bf{a}}=\hat={\bf{a}}=-\ds\frac{\sqrt{71}}{9}{\bf{i}}-\ds\frac{1}{3}{\bf{j}}-
\ds\frac{1}{9}{\bf{k}}$

\end{description}
\end{document}