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\begin{document}

{\bf Question}

Find the vector equation of the line that passes through the point
$A$ with position vector ${\bf{a}}=-2{\bf{i}}-{\bf{j}}+3{\bf{k}}$
and is normal to both the vectors
${\bf{b}}={\bf{i}}+2{\bf{j}}+3{\bf{k}}$ and
${\bf{c}}=-{\bf{i}}+{\bf{j}}-{\bf{k}}$.

\medskip

{\bf Answer}

The direction vector of the line must be normal to

${\bf{b}}={\bf{i}}+2{\bf{j}}+3{\bf{k}}$ and
${\bf{c}}=-{\bf{i}}+{\bf{j}}-{\bf{k}}$

Let the direction vector be ${\bf{d}}$. Thus ${\bf{d}}$ must be
parallel to either ${\bf{b}} \times {\bf{c}}$ or ${\bf{c}} \times
{\bf{b}}$. Either one will do as we can just trivially alter the
scalar parameter in the final vector equation of the line, if
needs be.

So choose

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

Thus if ${\bf{r}}={\bf{a}}+\lambda{\bf{d}}$ where ${\bf{a}}$ is a
point on the line, we have

$$\un{{\bf{r}}=-2{\bf{i}}-{\bf{j}}+3{\bf{k}}+\lambda(-5{\bf{i}}-2{\bf{j}}+3{\bf{k}})}$$


\end{document}