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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    Let \[\mathbf{a}=\mathbf{i}-\mathbf{j},\,\,\mathbf{b}=\mathbf{j}-\mathbf{k},\,\,\mathbf{c}=\mathbf{k}-\mathbf{i}.\] If \[\mathbf{\hat{d}}\] is a unit vector such that \[\mathbf{a}\,.\,\mathbf{\hat{d}}=0=[\mathbf{b}\,\,\mathbf{c}\,\,\mathbf{\hat{d}}],\] then \[\mathbf{\hat{d}}\] is equal to [IIT 1995]

    A) \[\pm \frac{\mathbf{i}+\mathbf{j}-\mathbf{k}}{\sqrt{3}}\]

    B) \[\pm \frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}\]

    C) \[\pm \frac{\mathbf{i}+\mathbf{j}-2\mathbf{k}}{\sqrt{6}}\]

    D) \[\pm \,\,\mathbf{k}\]

    Correct Answer: C

    Solution :

    • Let \[\mathbf{a}=\mathbf{i}-\mathbf{j},\] \[\mathbf{b}=\mathbf{j}-\mathbf{k}\] and \[\mathbf{c}=\mathbf{k}-\mathbf{i}\]                   
    • Let \[\mathbf{\hat{d}}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k},\] \[|\mathbf{\hat{d}}|\,=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}=1\]     \[\Rightarrow a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1\]                                ......(i)                   
    • \[\mathbf{a}\,.\mathbf{\hat{d}}=0\Rightarrow {{a}_{1}}-{{a}_{2}}=0\]    ..(ii)                   
    • \[[\mathbf{b}\,\mathbf{c}\,\mathbf{\hat{d}}]=0\Rightarrow \mathbf{b}\,.\,(\mathbf{c}\times \mathbf{\hat{d}})=0\]                   
    • \[\Rightarrow \left| \begin{matrix}    0 & 1 & -1  \\    -1 & 0 & 1  \\    {{a}_{1}} & {{a}_{2}} & {{a}_{3}}  \\ \end{matrix} \right|=-1(-{{a}_{3}}-{{a}_{1}})-1(-{{a}_{2}})\]                   
    • \[\therefore \,{{a}_{1}}+{{a}_{2}}+{{a}_{3}}=0\Rightarrow {{a}_{1}}-{{a}_{2}}+0{{a}_{3}}=0\],    {from (ii)}                   
    • \[\therefore \,\frac{{{a}_{1}}}{0+1}=\frac{{{a}_{2}}}{1-0}=\frac{{{a}_{3}}}{-1-1}\Rightarrow \frac{{{a}_{1}}}{1}=\frac{{{a}_{2}}}{1}=\frac{{{a}_{3}}}{-2}=\lambda \], (say)                   
    • \[\Rightarrow {{a}_{1}}=\lambda ,\] \[{{a}_{2}}=\lambda ,\] \[{{a}_{3}}=-2\lambda \]                   
    • \[\therefore \,{{\lambda }^{2}}+{{\lambda }^{2}}+4{{\lambda }^{2}}=1\] ,    {from (i)}                   
    • \[\Rightarrow 6{{\lambda }^{2}}=1\Rightarrow \lambda =\pm \frac{1}{\sqrt{6}};\] \[\therefore \,\mathbf{\hat{d}}=\pm \frac{\mathbf{i}+\mathbf{j}-2\mathbf{k}}{\sqrt{6}}\].


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