If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors, then match the items of List-I with…

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors, then match the items of List-I with those of List-II.
The correct answer is A $\quad$ B $\quad$ B $\quad$ D
  1. III IV V II
  2. IV V II III
  3. I IV V III
  4. I IV II III

Solution

$\begin{aligned} & \text { (A) }[\mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}] \\ & =(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a}) \cdot(\mathbf{a} \times \mathbf{b}) \\ & =\{\mathbf{d} \times(\mathbf{c} \times \mathbf{a})\} \cdot(\mathbf{a} \times \mathbf{b}) \\ & =\{(\mathbf{d} \cdot \mathbf{a}) \mathbf{c}-(\mathbf{d} \cdot \mathbf{c}) \mathbf{a}\} \cdot(\mathbf{a} \times \mathbf{b}) \\ & =\{[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c}-[\mathbf{b} \mathbf{c} \mathbf{c} \mathbf{\mathbf { a }}\} \cdot(\mathbf{a} \times \mathbf{b}) \\ & =\{[\mathbf{a} \mathbf{b} \mathbf{\mathbf { c }} \times \mathbf{c}-0\} \cdot(\mathbf{a} \times \mathbf{b}) \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]\{\mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})\} \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{c} \mathbf{a} \mathbf{b}] \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]^2\end{aligned}$ (B) $ \begin{aligned} & {[\mathbf{a} \times \mathbf{b} \mathbf{a} \times \mathbf{c} \cdot \mathbf{b}]} \\ & =\{(\mathbf{a} \times \mathbf{b}) \times(\mathbf{a} \times \mathbf{c})\} \cdot \mathbf{b} \\ & =\{\mathbf{d} \times(\mathbf{a} \times \mathbf{c})\} \cdot \mathbf{b} \quad[\text { let } \mathbf{d}=\mathbf{a} \times \mathbf{b}] \\ & =\{(\mathbf{d} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{d} \cdot \mathbf{a}) \mathbf{c}\} \cdot \mathbf{b} \\ & =\{(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\} \mathbf{a} \cdot \mathbf{b}-\{(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a}\} \mathbf{c} \cdot \mathbf{b} \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \cdot \mathbf{b}]-[\mathbf{a} \mathbf{b} \mathbf{a}][\mathbf{c} \cdot \mathbf{b}] \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \cdot \mathbf{b}]-0 \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \cdot \mathbf{b}] \end{aligned} $ (C) $ \begin{aligned} & {[\mathbf{a}+\mathbf{b} \mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a}]} \\ & =(\mathbf{a}+\mathbf{b}) \cdot\{(\mathbf{b}+\mathbf{c}) \times(\mathbf{c}+\mathbf{a})\} \\ & =(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{b} \times \mathbf{c}+\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}) \\ & =(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{b} \times \mathbf{c}+\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a})[\mathbf{c} \times \mathbf{c}=0] \\ & =\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})+\mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{c} \times \mathbf{a}) \\ & +\mathbf{b} \cdot(\mathbf{b} \times \mathbf{c})+\mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a}) \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]+0+0+0+0+[\mathbf{b} \mathbf{c}] \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]+[\mathbf{a} \mathbf{b} \mathbf{c}] \\ & =2[\mathbf{a} \mathbf{b} \mathbf{c}] \end{aligned} $ (D) $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three mutually perpendicular units vector $ \begin{aligned} & \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=0 \\ & =[(\mathbf{a}+\mathbf{b}+\mathbf{c}) \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}] \\ & =(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot\{(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a}) \\ & =(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot(\mathbf{a} \times \mathbf{b}) \quad[\because \mathbf{b} \times \mathbf{c}=\mathbf{a} \text { and } \mathbf{c} \times \mathbf{a}=\mathbf{b}] \\ & =\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b})+\mathbf{b} \cdot(\mathbf{a} \times \mathbf{b})+\mathbf{c} \cdot(\mathbf{a} \times \mathbf{b}) \\ & =0+0+[\mathbf{c} \mathbf{a} \mathbf{b}] \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}] \end{aligned} $ Hence, $\mathrm{A} \rightarrow \mathrm{I}, \mathrm{B} \rightarrow \mathrm{IV}, \mathrm{C} \rightarrow \mathrm{II}, \mathrm{D} \rightarrow$ III

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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