\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\begin{document}


{\bf Question}

\begin{description}
\item[(a)] Show that the plane through the point ${\bf a, b, c}$
has the equation  $${\bf r \cdot b \times c + r \cdot c \times a +
r \cdot a \times b = a \cdot b \times c}$$
\item[(b)] Prove that the shortest distance from the point ${\bf
a}$ to the line joining the points ${\bf b}$ and ${\bf c}$ is
given by $$\frac{|{\bf a \times b + b \times c + c \times
a}|}{|{\bf b - c}|}$$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}

\item[(a)] if the plane passes through the points ${\bf \, \, b,
\, c}$ it has a normal vector \begin{eqnarray*} {(\bf b - a)
\times (c - a)} & = & b \times c - a \times c - b \times a \\ & =
& b \times c + c \times a + a \times b \end{eqnarray*} Thus the
equation of the plane is $${\bf r} (b \times c + c \times a + a
\times b) = k$$ Now ${\bf r} = {\bf a}$ lies in the plane and so
$$k = {\bf a} (b \times c + c \times a + a \times b) = a \cdot b
\times c$$ therefore the equation is $${\bf r} \cdot(b \times c +
c \times a + a\times b) = a\cdot b\times c$$

\item[(b)] If a line has  equation $r = {\bf s} + t{\bf u}$ the
shortest distance of ${\bf p}$ form the line is $\ds \frac{|{\bf
(s - p) \times u|}}{|u|}$.  The equation of the line here is ${\bf
r = b} + t({\bf c - b})$

So the shortest distance of ${\bf a}$ from the line is
$$\frac{{\bf |(b-a) \times (c - b)|}}{{\bf | c-b|}} = \frac{|b
\times c + c \times a + a \times b|}{|{\bf c - b}|}$$

\end{description}

\end{document}