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{\bf Question}

Given that ${\bf a}$ is an arbitary vector and ${\bf e}$ is a
fixed unit vector, determine the magnitude and direction of ${\bf
e \times(a \times e)}$ and deduce that $${\bf a = (a \cdot e) e +
e \times(a \times e)}$$ explaining this result.

${ }$

Consider any three non-collinear vectors $a,b,c.$  By resolving
${\bf a}$ along ${\bf b, c}$ and ${\bf b \times c}$ use the above
result to prove that $${\bf a \times (b \times c) = a \cdot c
\cdot b - a \cdot b \cdot c}$$

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{\bf Answer}

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${\bf a = a \cdot e \, e + a \cdot f \, f}$

So ${\bf a \times e} = {\bf (a \cdot f )f \times e = -(a \cdot f)}
{\bf g}$

Thus ${\bf e \times (a \times e) = -(a \cdot f) e\times g - (a
\cdot f)f}$

So ${\bf a = \beta b + \gamma c + \delta(b \times c)}$

\begin{eqnarray*} a \times (b \times c) & = & \beta b \times (b
\times c) + \gamma(c \times (b \times c)) \\ & = &  \beta b^2
[(\hat{b} \cdot c) \hat{b} - {\bf c}] + \gamma c^2[{\bf b} -
(\hat{c} \cdot b) \hat{c}] \\ & = & \beta(b \cdot c){\bf b} +
\gamma(c \cdot c){\bf b} - (\beta(b \cdot b){\bf c} + \gamma(c
\cdot b){\bf c})
\end{eqnarray*}

Now \begin{eqnarray*} a \cdot c & = & \beta(b \cdot c) + \gamma(c
\cdot c) {\rm \ \ \ since\ } (b \times c) \cdot c = 0 \\ a \cdot b
& = & \beta(b \cdot b) + \gamma(c \cdot b)\\ {\rm So \ \ } a
\times (b \times c) & = & (a \cdot c){\bf b} - (a \cdot b) {\bf c}
\end{eqnarray*}



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