JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    Given two vectors \[\widehat{\mathbf{i}}-\widehat{\mathbf{j}}\] and \[\widehat{\mathbf{i}}+2\widehat{\mathbf{j}}\] the unit vector coplanar with the two vectors and perpendicular to first is:

    A) \[\pm \frac{1}{\sqrt{2}}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}})\]                               

    B) \[\frac{1}{\sqrt{5}}(2\widehat{\mathbf{i}}+\widehat{\mathbf{j}})\]

    C) \[\pm \frac{1}{\sqrt{2}}(\widehat{\mathbf{i}}+\widehat{\mathbf{k}})\]                              

    D)  none of these

    Correct Answer: A

    Solution :

    The required vector is along the vector                 \[\overset{\to }{\mathop{\mathbf{a}}}\,\times (\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,)=(\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,)\overset{\to }{\mathop{\mathbf{a}}}\,-(\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{a}}}\,)\overset{\to }{\mathop{\mathbf{a}}}\,\]                 \[=-(\widehat{\mathbf{i}}-\widehat{\mathbf{j}})-2(\widehat{\mathbf{i}}+2\widehat{\mathbf{j}})\]                 \[=-3\widehat{\mathbf{i}}-3\widehat{\mathbf{j}}\] Hence, required vector are given by                 \[\pm \frac{(-3\widehat{\mathbf{i}}-3\widehat{\mathbf{j}})}{\sqrt{9+9}}=\pm \frac{1}{\sqrt{2}}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}})\]


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