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\begin{document}


{\bf Question}

If ${\bf a}=a_1{\bf i}+a_2{\bf j}+a_3{\bf k}$, ${\bf b}=b_1{\bf
i}+b_2{\bf j}+b_3{\bf k}$ and ${\bf c}=c_1{\bf i}+c_2{\bf
j}+c_3{\bf k}$, show that:
\begin{description}
\item[(i)]
$a\cdot(a\times b)=0$\ and\ $b\cdot(a\times b)=0$;
\item[(ii)]
$a\cdot(b\times c)=(a\times b)\cdot c$;
\item[(iii)]
$a\times(b\times c)=(a\cdot c)b-(a\cdot b)c$.
\end{description}
Illustrate these results when  ${\bf a}={\bf i}+{\bf j}+{\bf k}$,
${\bf b}={\bf i}-2{\bf j}+3{\bf k}$ and ${\bf c}={\bf i}-4{\bf
k}$.



{\bf Answer}

\begin{description}
\item[(i)]
${\bf a} \times {\bf b} = \left|
\begin{array}{ccc} {\bf{i}} &amp; {\bf{j}} &amp;
{\bf{k}}\\ a_1 &amp; a_2 &amp; a_3 \\ b_1 &amp; b_2 &amp; b_3 \end{array} \right|
= \left( \begin{array}{c} {a_2b_3 - a_3b_2} \\ {a_3b_1 - a_1b_3}
\\ {a_1b_2 - a_2b_1} \end{array} \right).$
\begin{eqnarray*}
{\bf a} \cdot ({\bf a} \times {\bf b}) &amp; = &amp; \left(
\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array} \right) \cdot
\left(\begin{array}{c} a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\
a_1b_2-a_2b_1 \end{array} \right) \\ &amp; = &amp; a_1a_2b_3 - a_1a_3b_2 +
a_2a_3b_1 - a_1a_2b_3 + a_1a_3b_2 - a_2a_3b_1 = 0 \end{eqnarray*}
\begin{eqnarray*}
{\bf b} \cdot ({\bf a} \times {\bf b}) &amp; = &amp;
\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array} \right) \cdot
\left(\begin{array}{c} a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\
a_1b_2-a_2b_1 \end{array} \right)\\ &amp; = &amp; a_2b_1b_3 - a_3b_1b_3 +
a_3b_1b_2 - a_1b_2b_3 + a_1b_2b_3 - a_2b_1b_3 = 0 \end{eqnarray*}
\item[(ii)]
${\bf b} \times {\bf c} = \left|
\begin{array}{ccc} {\bf{i}} &amp; {\bf{j}} &amp;
{\bf{k}}\\ b_1 &amp; b_2 &amp; b_3 \\ c_1 &amp; c_2 &amp; c_3 \end{array} \right|
= \left( \begin{array}{c} {b_2c_3 - b_3c_2} \\ {b_3c_1 - b_1c_3}
\\ {b_1c_2 - b_2c_1} \end{array} \right).$

Show that ${\bf a}\cdot({\bf b}\times {\bf c})=({\bf a}\times {\bf
b})\cdot {\bf c}$

\begin{eqnarray*}{\rm LHS:\ } {\bf a} \cdot ({\bf b} \times {\bf c}) &amp; = &amp;
\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array} \right) \cdot
\left(\begin{array}{c}  {b_2c_3 - b_3c_2} \\ {b_3c_1 - b_1c_3} \\
{b_1c_2 - b_2c_1}\end{array}\right) \\ &amp; = &amp; a_1b_2c_3 - a_1b_3c_2
+ a_2b_3c_1 - a_2b_1c_3 + a_3b_1c_2 - a_3b_2c_1
\end{eqnarray*}
\begin{eqnarray*}{\rm RHS:\ } ({\bf a} \times {\bf b}) \cdot {\bf c} &amp; = &amp;
\left(\begin{array}{c}  {a_2b_3 - a_3b_2} \\ {a_3b_1 - a_1b_3}
{a_1b_2 - a_2b_1}\end{array}\right)\cdot \left(\begin{array}{c}
c_1\\ c_2 \\ c_3 \end{array} \right) \\ &amp; = &amp; a_2b_3c_1 -
a_3b_2c_1 + a_3b_1c_2 - a_1b_3c_2 + a_1b_2c_3 - a_2b_1c_3
\end{eqnarray*}
Note that $$LHS = RHS$$
\item[(iii)]
${\bf a}\times({\bf b}\times {\bf c})=({\bf a}\cdot {\bf c}){\bf
b}-({\bf a}\cdot {\bf b}){\bf c}$

LHS:
\begin{eqnarray*}{\bf a}\times({\bf b}\times {\bf c}) &amp; = &amp;
\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array} \right)
\times\left(\begin{array}{c} {b_2c_3 - b_3c_2} \\ {b_3c_1 -
b_1c_3} \\ {b_1c_2 - b_2c_1}\end{array}\right) \\ &amp; = &amp; \left|
\begin{array}{ccc} {\bf{i}} &amp; {\bf{j}} &amp;
{\bf{k}}\\ a_1 &amp; a_2 &amp; a_3 \\ {b_2c_3 - b_3c_2} &amp; {b_3c_1 -
b_1c_3} &amp; {b_1c_2 - b_2c_1}\end{array} \right| \\ &amp; = &amp; \left(
\begin{array}{c} {a_2b_1c_2 - a_2b_2c_1 - a_3b_3c_1 + a_3b_1c_3} \\
{a_3b_2c_3 - a_3b_3c_2 - a_1b_1c_2 + a_1b_2c_1} \\ {a_1b_3c_1 -
a_1b_1c_3 - a_2b_2c_3 + a_2b_3c_2} \end{array} \right)
\end{eqnarray*}
RHS:
\begin{eqnarray*}({\bf a}\cdot {\bf c}){\bf
b}-({\bf a}\cdot {\bf b}){\bf c}  &amp; = &amp; \left[ \left(
\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array} \right) \cdot
\left(\begin{array}{c} c_1 \\ c_2 \\ c_3 \end{array}
\right)\right] \left( \begin{array}{c} b_1 \\ b_2 \\ b_3
\end{array} \right)  - \left[ \left(
\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array} \right) \cdot
\left(\begin{array}{c} b_1 \\ b_2 \\ b_3 \end{array}
\right)\right] \left( \begin{array}{c} c_1 \\ c_2 \\ c_3
\end{array} \right) \\ &amp; = &amp; (a_1c_1 + a_2c_2 + a_3c_3) \left(
\begin{array}{c} b_1 \\ b_2 \\ b_3 \end{array} \right) -
(a_1b_1+a_2b_2+a_3b_3)\left(\begin{array}{c}c_1\\c_2\\c_3\end{array}\right)\\
&amp; = &amp; \left( \begin{array}{c} {a_1b_1c_1 + a_2b_1c_2 + a_3b_1c_3 -
a_1b_1c_1 - a_2b_2c_1 - a_3b_3c_1}
\\ {a_1b_2c_1 + a_2b_2c_2 + a_3b_2c_3 - a_1b_1c_2 - a_2b_2c_2 - a_3b_3c_2}
\\ {a_1b_3c_1b + a_2b_3c_2 + a_3b_3c_3 - a_1b_1c_3 - a_2b_2c_3 - a_3b_3c_3}
\end{array} \right)
\end{eqnarray*}
Note that $$LHS = RHS$$
\end{description}
If ${\bf a} = \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array}
\right) $ , ${\bf b} = \left( \begin{array}{c} 1 \\ -2 \\ 3
\end{array} \right) $, ${\bf c} = \left( \begin{array}{c} 1 \\ 0 \\ -4 \end{array}
\right) $

then ${\bf a} \times {\bf b} =\left| \begin{array}{ccc} {\bf{i}} &amp;
{\bf{j}} &amp; {\bf{k}}\\ 1 &amp; 1 &amp; 1 \\ 1 &amp; -2 &amp; 3 \end{array} \right|
= \left( \begin{array}{c} 5 \\ -2 \\ -3 \end{array} \right) $

and ${\bf b} \times {\bf c} =\left| \begin{array}{ccc} {\bf{i}} &amp;
{\bf{j}} &amp; {\bf{k}}\\ 1 &amp; -2 &amp; 3 \\ 1 &amp; 0 &amp; {-4} \end{array}
\right| = \left( \begin{array}{c} 8 \\ 7 \\ 2 \end{array} \right)
$
\begin{description}
\item[(i)]
${\bf a} \cdot ({\bf a} \times {\bf b}) = \left(
\begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right)
\cdot \left( \begin{array}{c} 5 \\ -2
\\ -3 \end{array} \right) = 5 - 2 - 3 = 0.$

${\bf b} \cdot ({\bf a} \times {\bf b}) = \left(
\begin{array}{c} 1 \\ -2 \\ 3 \end{array} \right)
\cdot \left( \begin{array}{c} 5 \\ -2
\\ -3 \end{array} \right) = 5 + 4 - 9 = 0.$

\item[(ii)]
${\bf a} \cdot ({\bf b} \times {\bf c}) = \left(
\begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right)
\cdot \left( \begin{array}{c} 8 \\ 7
\\ 2 \end{array} \right) = 8 + 7 + 2 = 17.$

$({\bf a} \times {\bf b}) \cdot {\bf c} = \left(
\begin{array}{c} 5 \\ -2 \\ -3 \end{array} \right)
\cdot \left( \begin{array}{c} 1 \\ 0
\\ 4 \end{array} \right) = 5 + 0 + 12 = 17.$

So ${\bf a}\cdot({\bf b}\times {\bf c})=({\bf a}\times {\bf
b})\cdot {\bf c}$

\item[(iii)]
${\bf a}\times({\bf b}\times {\bf c}) = \left(
\begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right)
\times \left( \begin{array}{c} 8 \\ 7
\\ 2 \end{array} \right) =  \left| \begin{array}{ccc} {\bf{i}} &amp;
{\bf{j}} &amp; {\bf{k}}\\ 1 &amp; 1 &amp; 1 \\ 8 &amp; 7 &amp; 2 \end{array} \right| =
\left( \begin{array}{c} -5 \\ 6 \\ -1 \end{array} \right)$

\begin{eqnarray*}({\bf a}\cdot {\bf c}){\bf
b}-({\bf a}\cdot {\bf b}){\bf c}  &amp; = &amp; \left[ \left(
\begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) \cdot
\left(\begin{array}{c} 1 \\ 0 \\ -4 \end{array} \right)\right]
\left( \begin{array}{c}1 \\ -2 \\ 3 \end{array} \right)  - \left[
\left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) \cdot
\left(\begin{array}{c} 1 \\ -2 \\ 3 \end{array} \right)\right]
\left( \begin{array}{c} 1 \\ 0 \\ -4 \end{array} \right) \\ &amp; = &amp;
(1 + 0 - 4) \left( \begin{array}{c} 1 \\ -2 \\ 3 \end{array}
\right) - (1 - 2 + 3) \left( \begin{array}{c} 1 \\ 0 \\ -4
\end{array} \right)
\\ &amp; = &amp; -3\left( \begin{array}{c} 1 \\ -2 \\ 3 \end{array} \right)
- 2 \left( \begin{array}{c} 1 \\ 0 \\ -4 \end{array} \right)
\\ &amp; = &amp; \left( \begin{array}{c} -5 \\ 6 \\ -1 \end{array} \right)
\end{eqnarray*}
So $${\bf a}\times({\bf b}\times {\bf c})=({\bf a}\cdot {\bf
c}){\bf b}-({\bf a}\cdot {\bf b}){\bf c}$$
\end{description}


\end{document}
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