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{\bf Question}

Using the cross product, find a unit vector orthogonal to both
${\bf u}$ and ${\bf v}$ where:
\begin{description}
\item[(i)]
${{\bf u} = {\bf i} + {\bf j}}$ and ${{\bf v} = {\bf j} + 2{\bf
k}}$;
\item[(ii)]
${{\bf u}=3{\bf i}+2{\bf j}+{\bf k}}$ and ${{\bf v}=4{\bf i}+{\bf
j}+3{\bf k}}$;
\item[(iii)]
${{\bf u}={\bf j}+3{\bf k}}$ and ${{\bf v}=5{\bf i}+8{\bf j}+4{\bf
k}}$
\end{description}


{\bf Answer}

The cross product, ${\bf u} \times {\bf v}$ gives a vector
orthogonal to both ${\bf u}$ and ${\bf v}$:
\begin{description}
\item[(a)]
Let $\bf{n} = \bf{u} \times \bf{v} = \left|
\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ 1 & 1 & 0 \\ 0 & 1 & 2 \end{array} \right| = \left(
\begin{array}{c} 2 \\ -2 \\ 1 \end{array} \right).$
The length of {\bf n} is $|{\bf n}| = \sqrt{2^2 + (-2)^2 + 1^2} =
3$

A unit vector orthogonal to both ${\bf u}$ and ${\bf v}$ is $${\bf
\hat n} = \frac{{\bf n}}{|{\bf n}|} = \frac{1}{3}\left(
\begin{array}{c} 2 \\ -2 \\ 1 \end{array} \right) = \left(
\begin{array}{c} {\frac{2}{3}} \\ {\frac{-2}{3}} \\ {\frac{1}{3}} \end{array}
\right).$$
\item[(b)]
Let $\bf{n} = \bf{u} \times \bf{v} = \left|
\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ 3 & 2 & 1 \\ 4 & 1 & 3 \end{array} \right| = \left(
\begin{array}{c} 5 \\ -5 \\ -5 \end{array} \right).$
The length of {\bf n} is $|{\bf n}| = 5\sqrt 3$

A unit vector orthogonal to both ${\bf u}$ and ${\bf v}$ is $${\bf
\hat n} = \frac{{\bf n}}{|{\bf n}|} = \frac{1}{5\sqrt 3}\left(
\begin{array}{c} 5 \\ -5 \\ -5 \end{array} \right) = \left(
\begin{array}{c} {\frac{1}{\sqrt 3}} \\ {\frac{-1}{\sqrt 3}} \\ {\frac{-1}{\sqrt 3}} \end{array}
\right).$$

\item[(c)]
Let $\bf{n} = \bf{u} \times \bf{v} = \left|
\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ 0 & 1 & 3 \\ 5 & 8 & 4 \end{array} \right| = \left(
\begin{array}{c} -20 \\ 15 \\ -5 \end{array} \right).$
The length of {\bf n} is $|{\bf n}| = \sqrt {650} = 5\sqrt{26}$

A unit vector orthogonal to both ${\bf u}$ and ${\bf v}$ is $${\bf
\hat n} = \frac{{\bf n}}{|{\bf n}|} = \frac{1}{5\sqrt {26}}\left(
\begin{array}{c} -20 \\ 15 \\ -5 \end{array} \right) = \left(
\begin{array}{c} {\frac{-4}{\sqrt {26}}} \\ {\frac{3}{\sqrt {26}}} \\ {\frac{-1}{\sqrt {26}}} \end{array}
\right).$$
\end{description}


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