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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    Let \[\mathbf{a}=2\mathbf{i}+\mathbf{j}+\mathbf{k},\,\,\mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k}\]and a unit vector c be coplanar. If c is perpendicular to a, then c = [IIT 1999; Pb. CET 2003; DCE 2005]

    A) \[\frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k})\]

    B) \[\frac{1}{\sqrt{3}}(-\mathbf{i}-\mathbf{j}-\mathbf{k})\]

    C) \[\frac{1}{\sqrt{5}}\,(\mathbf{i}-2\mathbf{j})\]

    D) \[\frac{1}{\sqrt{3}}(\mathbf{i}-\mathbf{j}-\mathbf{k})\]

    Correct Answer: A

    Solution :

    • \[\mathbf{c}\] is coplanar with \[\mathbf{a},\,\mathbf{b}\]                   
    • \[\therefore \,\,\,\mathbf{c}=x\mathbf{a}+y\mathbf{b}\]                   
    • \[\Rightarrow \mathbf{c}=x(2\mathbf{i}+\mathbf{j}+\mathbf{k})+y(\mathbf{i}+2\mathbf{j}-\mathbf{k})\]                   
    • \[\Rightarrow \mathbf{c}=(2x+y)\mathbf{i}+(x+2y)\mathbf{j}+(x-y)\mathbf{k}\]                   
    • \[\because \,\,\,\mathbf{a}\,.\mathbf{c}=0\]                   
    • \[\therefore \,\,\,2(2x+y)+x+2y+x-y=0\]                   
    • \[\Rightarrow y=-2x\]                   
    • \[\mathbf{c}=-3x\mathbf{j}+3x\mathbf{k}=3x(-\mathbf{j}+\mathbf{k})\]                   
    • \[\because \,\,\,|\mathbf{c}|\,=1\]                     
    • \[\therefore \,\,\,9{{x}^{2}}+9{{x}^{2}}=1\]                   
    • \[\Rightarrow x=\pm \frac{1}{3\sqrt{2}}\Rightarrow \mathbf{c}=\frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k})\].


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