If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}},…

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 $\frac{[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]}$
  1. 0
  2. 1
  3. $\sqrt{3}$
  4. $\sqrt{2}$

Solution

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} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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