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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • 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) \[\frac{(-\mathbf{i}+2\mathbf{j}+\mathbf{k})}{3}\]

    Correct Answer: A

    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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