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

{\bf Question}

Determine unit vectors that are normal to both vectors ${\bf{a}}$
and ${\bf{b}}$ when:

\begin{description}
\item[(i)]
${\bf{a}}=3{\bf{i}}+5{\bf{j}}-2{\bf{k}}\
{\bf{b}}={\bf{i}}+{\bf{j}}+{\bf{k}}$

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

\end{description}

Are the results unique?
\medskip

{\bf Answer}

First a vector normal to both ${\bf{a}}$ and ${\bf{b}}$ is given
by ${\bf{a}} \times {\bf{b}}$. The corresponding unit vector is
$\ds\frac{{\bf{a}} \times {\bf{b}}}{|{\bf{a}} \times {\bf{b}}|}$

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

\begin{eqnarray*} {\bf{a}}\times{\bf{b}} & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ 3 & 5 & -2\\ 1 & 1 & 1\end{array}\right|\\ &
= & {\bf{i}}(5 \times 1)-{\bf{i}}(1 \times -2)-{\bf{j}}(3 \times
1)\\ & & +{\bf{k}}(3 \times 1)+{\bf{j}}(1 \times -2)-{\bf{k}}(1
\times 5)\\ & = &
5{\bf{i}}+2{\bf{i}}-3{\bf{j}}+3{\bf{k}}-2{\bf{j}}-5{\bf{k}}\\& = &
7{\bf{i}}-5{\bf{j}}-2{\bf{k}} \end{eqnarray*}

Now $|{\bf{a}} \times
{\bf{b}}|=\sqrt{7^2+(-5)^2+(-2)^2}+\sqrt{49+25+4}=\sqrt{78}$

Thus \un{$\ds\frac{{\bf{a}} \times {\bf{b}}}{|{\bf{a}} \times
{\bf{b}}|}=\ds\frac{7}{\sqrt{78}}{\bf{i}}-\ds\frac{5}{\sqrt{78}}{\bf{j}}
-\ds\frac{2}{\sqrt{78}}{\bf{k}}$}

\item[(ii)]

\begin{eqnarray*} & & {\bf{a}}\times{\bf{b}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ -4 & 0 & 2\\ 0 & 1 & -3\end{array}\right|\\
& = & {\bf{i}}(0 \times -3)-{\bf{i}}(1 \times 2)-{\bf{j}}(-3
\times -4)\\ & & +{\bf{k}}(-4 \times 1)+{\bf{j}}(0 \times
2)-{\bf{k}}(0 \times 0)\\ & = & -2{\bf{i}}-12{\bf{j}}-4{\bf{k}}
\end{eqnarray*}

Now $|{\bf{a}} \times
{\bf{b}}|=\sqrt{2^2+12^2+4^2}+\sqrt{4+144+16}=\sqrt{164}$

Thus \un{$\ds\frac{{\bf{a}} \times {\bf{b}}}{|{\bf{a}} \times
{\bf{b}}|}=-\ds\frac{2}{\sqrt{164}}{\bf{i}}-\ds\frac{12}{\sqrt{164}}{\bf{j}}
-\ds\frac{4}{\sqrt{164}}{\bf{k}}$}

\end{description}
These results are \un{not} unique, you could have the negative
sign of them. Also, there are \un{two} vectors perpendicular to
any pair (these vectors being ${\bf{a}} \times {\bf{b}}$ and
${\bf{b}} \times {\bf{a}}$ [$=-{\bf{a}} \times {\bf{b}}$]).

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

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

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

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

\put(0,-.2){\line(1,0){.3}}

\put(.3,-.2){\line(0,1){.4}}

\put(0,.2){\line(1,0){.2}}

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\put(0,2){\makebox(0,0)[b]{${\bf{a}} \times {\bf{b}}$}}

\put(0,-2){\makebox(0,0)[t]{${\bf{b}} \times {\bf{a}}$}}

\put(2,1){\makebox(0,0)[l]{${\bf{a}}$}}

\put(1,2){\makebox(0,0)[l]{${\bf{b}}$}}

\end{picture} \vspace{1in}

\end{document}