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

{\bf Question}

Find the scalar product ${\bf{a}} \cdot {\bf{b}}$ and hence find
the angle between the vectors ${\bf{a}}$ and ${\bf{b}}$ given
that:

\begin{description}
\item[(i)]
${\bf{a}}=7{\bf{i}}-3{\bf{j}}+{\bf{k}}\
{\bf{b}}=-{\bf{i}}+2{\bf{j}}+2{\bf{k}}$

\item[(ii)]
${\bf{a}}=2{\bf{i}}-2{\bf{j}}+{\bf{k}}\
{\bf{b}}=-3{\bf{i}}-3{\bf{j}}+4{\bf{k}}$

\item[(iii)]
${\bf{a}}={\bf{i}}+2{\bf{j}}+3{\bf{k}}\
{\bf{b}}=-2{\bf{i}}-4{\bf{j}}-6{\bf{k}}$

\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
\begin{eqnarray*} {\bf{a}} \cdot {\bf{b}} & = & (7,-3,1) \cdot
(-1,2,2)\\ & = & (7 \times -1)+(-3 \times 2)+(1 \times 2)\\ & = &
-7-6+2 = \un{-11} \end{eqnarray*}

Now ${\bf{a}} \cdot {\bf{b}}=|{\bf{a}}||{\bf{b}}| \cos \theta$

$\Rightarrow \cos \theta=\ds\frac{({\bf{a}} \cdot
{\bf{b}})}{|{\bf{a}}||{\bf{b}}|}$

So need

$|{\bf{a}}|=\sqrt{7^2+(-3)^2+1^2}=\sqrt{59}$

$|{\bf{b}}|=\sqrt{(-1)^2+2^2+2^2}=\sqrt{9}=3$

Therefore $\cos \theta=-\ds\frac{11}{3\sqrt{59}} \Rightarrow
\theta=$

\item[(ii)]
\begin{eqnarray*} {\bf{a}} \cdot {\bf{b}} & = &
(-2{\bf{i}}-2{\bf{j}}+{\bf{k}}) \cdot
(-3{\bf{i}}-3{\bf{j}}+4{\bf{k}})\\ & = & (2,-2,1) \cdot
(-3,-3,4)\\ & = & (2 \times -3)+(-2 \times -3)+(1 \times 4)\\ & =
& -6+6+4 = \un{4} \end{eqnarray*}

$|{\bf{a}}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3$

$|{\bf{b}}|=\sqrt{(-3)^2+(-3)^2+4^2}=\sqrt{34}$

$\Rightarrow \cos \theta=\ds\frac{({\bf{a}} \cdot
{\bf{b}})}{|{\bf{a}}||{\bf{b}}|}=\ds\frac{4}{3\sqrt{34}}
\Rightarrow \theta =$

\item[(iii)]
\begin{eqnarray*} {\bf{a}} \cdot {\bf{b}} & = &
({\bf{i}}+2{\bf{j}}+3{\bf{k}}) \cdot
(-2{\bf{i}}-4{\bf{j}}-6{\bf{k}})\\ & = & (1 \times -2)+(2 \times
-4)+(3 \times -6)\\ & = & -2-8-18 = -28 \end{eqnarray*}

$|{\bf{a}}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

$|{\bf{b}}|=\sqrt{(2^2+4^2+6^2}=\sqrt{56}$

$\Rightarrow \cos \theta=\ds\frac{({\bf{a}} \cdot
{\bf{b}})}{|{\bf{a}}||{\bf{b}}|}=\ds\frac{-28}{\sqrt{14}}{\sqrt{56}}=-1
\Rightarrow \theta=\arccos(-1)=\un{\pi}$

Could have spotted this from anti-parallel vectors.
\end{description}

\end{document}