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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    If \[\mathbf{u}=2\,\mathbf{i}+2\mathbf{j}-\mathbf{k}\]and \[\mathbf{v}=6\,\mathbf{i}-3\,\mathbf{j}+2\,\mathbf{k},\] then a unit vector perpendicular to both u and v is [MP PET 1987]

    A) \[\mathbf{i}-10\mathbf{j}-18\mathbf{k}\]

    B) \[\frac{1}{\sqrt{17}}\,\left( \frac{1}{5}\mathbf{i}-2\mathbf{j}-\frac{18}{5}\mathbf{k} \right)\]

    C) \[\frac{1}{\sqrt{473}}\,(7\mathbf{i}-10\mathbf{j}-18\mathbf{k})\]

    D) None of these

    Correct Answer: B

    Solution :

    • \[\mathbf{u}=2\mathbf{i}+2\mathbf{j}-\mathbf{k}\] and \[\mathbf{v}=6\mathbf{i}-3\mathbf{j}+2\mathbf{k}\]                   
    • Let vector \[\mathbf{c}={{c}_{1}}\mathbf{i}+{{c}_{2}}\mathbf{j}+{{c}_{3}}\mathbf{k}\] is perpendicular to both \[\mathbf{u}\] and \[\mathbf{v},\] then \[\mathbf{c}\,.\,\mathbf{u}=0\]                   
    • \[\Rightarrow 2{{c}_{1}}+2{{c}_{2}}-{{c}_{3}}=0\]                           ...(i)                   
    • and \[\mathbf{c}\,.\,\mathbf{v}=0\Rightarrow 6{{c}_{1}}-3{{c}_{2}}+2{{c}_{3}}=0\] ...(ii)                   
    • Solving equation (i) and (ii) by cross multiplication                   
    • \[\frac{{{c}_{1}}}{4-3}=\frac{{{c}_{2}}}{-6-4}=\frac{{{c}_{3}}}{-6-12}=\lambda \],    (say)                   
    • \[\Rightarrow \frac{{{c}_{1}}}{1}=\frac{{{c}_{2}}}{-10}=\frac{{{c}_{3}}}{-18}=\lambda \]                   
    • \[\Rightarrow {{c}_{1}}=\lambda ,\] \[{{c}_{2}}=-10\lambda \] and \[{{c}_{3}}=-18\lambda \]                   
    • Thus \[\mathbf{c}=\lambda (\mathbf{i}-10\mathbf{j}-18\mathbf{k})\]                   
    • \[|\mathbf{c}|=\lambda \sqrt{1+100+324}=\lambda \sqrt{425}\]                   
    • Hence required unit vector is, \[\frac{\mathbf{c}}{|\mathbf{c}|}\]                   
    • \[=\frac{\lambda (\mathbf{i}-10\mathbf{j}-18\mathbf{k})}{\lambda \sqrt{425}}=\frac{1}{\sqrt{425}}(\mathbf{i}-10\mathbf{j}-18\mathbf{k})\] = \[\frac{1}{5\sqrt{17}}(\mathbf{i}-10\mathbf{j}-18\mathbf{k})=\frac{1}{\sqrt{17}}\left( \frac{1}{5}\mathbf{i}-2\mathbf{j}-\frac{18}{5}\mathbf{k} \right)\]                    Aliter : Required vector is \[\frac{\mathbf{u}\times \mathbf{v}}{|\mathbf{u}\times \mathbf{v}|}=\frac{\mathbf{i}-10\mathbf{j}-18\mathbf{k}}{\sqrt{425}}.\]


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