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If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are non-coplanar, then the value of $\mathbf{a} \cdot\left\{\frac{\mathbf{b} \times \mathbf{c}}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}\right\}-\mathbf{b} \cdot\left\{\frac{\mathbf{c} \times \mathbf{a}}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})}\right\}$ is
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Verified Answer
The correct answer is:
$-\frac{1}{6}$
Given, [a b c] $\neq 0$, i.e., non-coplanar.
$$
\begin{aligned}
&=\mathbf{a} \cdot\left\{\frac{\mathbf{a} \times \mathbf{c}}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}\right\}-\mathbf{b} \cdot\left\{\frac{\mathbf{c} \times \mathbf{a}}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{c})}\right\} \\
&=\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}-\frac{\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})} \\
&=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{3[\mathbf{b} \mathbf{c} \mathbf{a}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{c} \mathbf{a} \mathbf{b}]}(\because \mathbf{a} \cdot(\mathbf{b} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \\
&=\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]=[\mathbf{a} \mathbf{b} \mathbf{c}],[\mathbf{c} \mathbf{a} \mathbf{b}]=[\mathbf{b} \mathbf{c} \mathbf{a}]}{3[\mathbf{a} \mathbf{b ~ c}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{b} \mathbf{c} \mathbf{a}]} \\
&=1 / 3-1 / 2=-1 / 6
\end{aligned}
$$
$$
\begin{aligned}
&=\mathbf{a} \cdot\left\{\frac{\mathbf{a} \times \mathbf{c}}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}\right\}-\mathbf{b} \cdot\left\{\frac{\mathbf{c} \times \mathbf{a}}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{c})}\right\} \\
&=\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}-\frac{\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})} \\
&=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{3[\mathbf{b} \mathbf{c} \mathbf{a}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{c} \mathbf{a} \mathbf{b}]}(\because \mathbf{a} \cdot(\mathbf{b} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \\
&=\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]=[\mathbf{a} \mathbf{b} \mathbf{c}],[\mathbf{c} \mathbf{a} \mathbf{b}]=[\mathbf{b} \mathbf{c} \mathbf{a}]}{3[\mathbf{a} \mathbf{b ~ c}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{b} \mathbf{c} \mathbf{a}]} \\
&=1 / 3-1 / 2=-1 / 6
\end{aligned}
$$
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