JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    The unit vector perpendicular to each of the vectors\[3\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}\], and\[2\widehat{\mathbf{i}}-2\widehat{\mathbf{j}}+4\widehat{\mathbf{k}}\]

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

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

    C) \[\frac{\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}}}{\sqrt{3}}\]                

    D)  None of these

    Correct Answer: A

    Solution :

    Let\[\overset{\to }{\mathop{\mathbf{a}}}\,=3\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}\]and\[\overset{\to }{\mathop{\mathbf{b}}}\,=2\widehat{\mathbf{i}}-2\widehat{\mathbf{j}}+4\widehat{\mathbf{k}}\] Now,     \[\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,=\left| \begin{matrix}    \widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}}  \\    3 & 1 & 2  \\    2 & -2 & 4  \\ \end{matrix} \right|\]                 \[=\widehat{\mathbf{i}}(4+4)-\widehat{\mathbf{j}}(12-4)+\widehat{\mathbf{k}}(-6-2)\]                 \[=8\widehat{\mathbf{i}}-8\widehat{\mathbf{j}}-8\widehat{\mathbf{k}}\] \[\therefore \]Unit vector\[=\frac{\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,}{|\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,|}\]                         \[=\frac{8(\widehat{\mathbf{i}}-\widehat{\mathbf{j}}-\widehat{\mathbf{k}})}{8\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}}\]                         \[=\frac{\widehat{\mathbf{i}}-\widehat{\mathbf{j}}-\widehat{\mathbf{k}}}{\sqrt{3}}\]


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