• question_answer Unit vectors a, b and c are coplanar. A unit vector d is perpendicular to them. If $(\mathbf{a}\times \mathbf{b})\times (\mathbf{c}\times \mathbf{d})=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}$ and the angle between a and b is ${{30}^{o}}$, then c is    [Roorkee Qualifying 1998] A) $\frac{(\mathbf{i}-2\mathbf{j}+2\mathbf{k})}{3}$ B) $\frac{(2\mathbf{i}+\mathbf{j}-\mathbf{k})}{3}$ C) $\frac{(-\mathbf{i}+2\mathbf{j}-2\mathbf{k})}{3}$ D) $\frac{(-\mathbf{i}+2\mathbf{j}+\mathbf{k})}{3}$

Solution :

• Since $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ are coplanar, hence $[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]=0$
• Given $(\mathbf{a}\times \mathbf{b})\times (\mathbf{c}\times \mathbf{d})=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}$
• $\Rightarrow [(\mathbf{a}\times \mathbf{b})\,.\,\mathbf{d}]\,\mathbf{c}-[(\mathbf{a}\times \mathbf{b})\,.\,\mathbf{c}]\mathbf{d}=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}$
• $\Rightarrow [(|\mathbf{a}||\mathbf{b}|\sin 30{}^\circ )\,\mathbf{\hat{n}}\,.\,\mathbf{d}]\,\mathbf{c}-0=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}$
• $\Rightarrow \left[ (1)(1)\left( \frac{1}{2} \right) \right][|\mathbf{\hat{n}}||\mathbf{d}|\cos \theta ]\,\mathbf{c}=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}$
• $\Rightarrow [(\mathbf{a}\times \mathbf{b})\,.\,\mathbf{d}]\,\mathbf{c}-[(\mathbf{a}\times \mathbf{b})\,.\,\mathbf{c}]\mathbf{d}=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}$,
• Where $\mathbf{\hat{n}}$ and $\mathbf{d}$ are unit perpendicular vector and angle between $\mathbf{\hat{n}}$ and $\mathbf{d}$ may be 0 or $\pi$.
• When $\theta =0{}^\circ ,$ $\mathbf{c}=\frac{1}{3}[\mathbf{i}-2\mathbf{j}+2\mathbf{k}]$
• When $\theta =\pi ,$ $\mathbf{c}=\frac{1}{3}[-\mathbf{i}+2\mathbf{j}-2\mathbf{k}]$.

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