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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Evaluate the triple scalar products ${\bf{a}} \cdot {\bf{b}}
\times {\bf{c}}$ and ${\bf{b}} \times {\bf{a}} \cdot {\bf{c}}$
given that:

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

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

\end{description}

\medskip

{\bf Answer}
\begin{eqnarray*} {\bf{a}} \cdot {\bf{b}} \times {\bf{c}} &amp; = &amp;
(a_1{\bf{a}}+a_2{\bf{a}}+a_3{\bf{a}}) \cdot
\left|\begin{array}{ccc} {\bf{i}} &amp; {\bf{j}} &amp; {\bf{k}}\\ b_1 &amp;
b_2 &amp; b_3\\ c_1 &amp; c_2 &amp; c_3\end{array}\right|\\ &amp; = &amp;
\left|\begin{array}{ccc} a_1 &amp; a_2 &amp; a_3\\ b_1 &amp; b_2 &amp; b_3\\ c_1 &amp;
c_2 &amp; c_3\end{array}\right| \end{eqnarray*} (think about it!)

\begin{description}
\item[(i)]
\begin{eqnarray*} {\bf{a}} \cdot {\bf{b}} \times {\bf{c}} &amp; = &amp;
\left|\begin{array}{ccc} 2 &amp; -1 &amp; -3\\ 0 &amp; 0 &amp; 3\\ 1 &amp; 2 &amp;
2\end{array}\right|\\ &amp; = &amp; (2 \times 0 \times 2)-(2 \times 2
\times 3)-(0 \times -1 \times 2)\\ &amp; &amp; +(0 \times 2 \times -3) +
(1 \times -1 \times 3)-(1 \times 0 \times -3)\\ &amp; = &amp;
0-12-0+0-3+0\\ &amp; = &amp; \un{-15} \end{eqnarray*}

Now \begin{eqnarray*} ({\bf{b}} \times{\bf{a}}) \cdot {\bf{c}} &amp; =
&amp; -({\bf{a}} \times{\bf{b}}) \cdot {\bf{c}}\\ &amp; = &amp;  -{\bf{a}}
\cdot ({\bf{b}} \times{\bf{c}})\\ &amp; = &amp; \un{+15} \end{eqnarray*}

by relevance to the above determinant.

\newpage
\item[(ii)]
\begin{eqnarray*}  &amp; &amp; {\bf{a}} \cdot {\bf{b}} \times {\bf{c}}\\ &amp; = &amp;
\left|\begin{array}{ccc} 1 &amp; 2 &amp; 1\\ 2 &amp; 1 &amp; 1\\ 4 &amp; 2 &amp;
2\end{array}\right|\\ &amp; = &amp; (1 \times 1 \times 2)-(1 \times 2
\times 1)-(2 \times 2 \times 2)\\ &amp; &amp; +(2 \times 2 \times 1) + (4
\times 2 \times 1)-(4 \times 1 \times 1)\\ &amp; = &amp; 2-2-8+4+8-4\\ &amp; =
&amp; \un{0}
\end{eqnarray*}

Hence $({\bf{b}} \times{\bf{a}}) \cdot {\bf{c}} = -({\bf{a}}
\times{\bf{b}}) \cdot {\bf{c}}  = -0=\un{0}$

\end{description}
\end{document}
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