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If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$, then
$\frac{[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]}$
Options:
$\frac{[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]}$
Solution:
1567 Upvotes
Verified Answer
The correct answer is:
1
Since, $(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c}$
$$
\begin{aligned}
& (\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a} \\
& (\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b} \\
& \text { and }[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}]=2[\mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c}] \\
& \text { and }[\mathbf{b} \times \mathbf{c ~ c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]=[\mathbf{a} \mathbf{b} \mathbf{c}]^2 \\
& {[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})} \\
& \text { So, } \frac{(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]} \\
& =\frac{[[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c}[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a}[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b}]}{2[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \mathbf{b}]^2} \\
& =\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]^3[\mathbf{a} \mathbf{b} \mathbf{c}]}{2[\mathbf{a} \mathbf{b} \mathbf{c}]^3}=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{2}=1 \\
& \left\lceil\because[\mathbf{a} \mathbf{b} \mathbf{c}]=\left|\begin{array}{lll}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right|=2\right\rfloor \\
&
\end{aligned}
$$
$$
\begin{aligned}
& (\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a} \\
& (\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b} \\
& \text { and }[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}]=2[\mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c}] \\
& \text { and }[\mathbf{b} \times \mathbf{c ~ c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]=[\mathbf{a} \mathbf{b} \mathbf{c}]^2 \\
& {[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})} \\
& \text { So, } \frac{(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]} \\
& =\frac{[[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c}[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a}[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b}]}{2[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \mathbf{b}]^2} \\
& =\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]^3[\mathbf{a} \mathbf{b} \mathbf{c}]}{2[\mathbf{a} \mathbf{b} \mathbf{c}]^3}=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{2}=1 \\
& \left\lceil\because[\mathbf{a} \mathbf{b} \mathbf{c}]=\left|\begin{array}{lll}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right|=2\right\rfloor \\
&
\end{aligned}
$$
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