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JEE Main & Advanced Mathematics Vector Algebra Question Bank Vector or Cross product of two vectors and its application

  • question_answer
    A unit vector perpendicular to the plane determined by the points \[P\,(1,\,\,-1,\,\,2),\,\,Q\,(2,\,\,0,\,-1)\] and \[R\,(0,\,\,2,\,\,1)\] is           [IIT 1994]

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

    B)                 \[\frac{2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]

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

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

    Correct Answer: B

    Solution :

               A vector perpendicular to the plane determined by the points \[P(1,\,-1,\,2);\] \[Q(2,\,0,\,-1)\] and \[R(0,\,2,\,1)\] is given by                     \[\overrightarrow{QR}\times \overrightarrow{PR}\Rightarrow (-2\mathbf{i}+2\mathbf{j}+2\mathbf{k})\times (-\mathbf{i}+3\mathbf{j}-\mathbf{k})\]                                 Therefore, unit vector \[=\frac{2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{4+1+1}}=\frac{2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}.\]


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