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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    Let \[\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\]and \[\vec{b}=\hat{i}+2\hat{j}-2\hat{k}\]be two vectors. If a vector perpendicular to both the vectors \[\vec{a}+\vec{b}\]and \[\vec{a}-\vec{b}\] has the magnitude 12 then one such vector is[JEE Main Held on 12-4-2019 Morning]        

    A) \[4(2\hat{i}+2\hat{j}-\hat{k})\]        

    B) \[4(-2\hat{i}-2\hat{j}+\hat{k})\]

    C) \[4(2\hat{i}-2\hat{j}-\hat{k})\]                      

    D) \[4(2\hat{i}+2\hat{j}+\hat{k})\]

    Correct Answer: C

    Solution :

    \[(\vec{a}+\vec{b})\times (\vec{a}-\vec{b})\]             \[=2\,(\vec{b}\times \vec{a})\]           \[=2\,\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & 2 & -2  \\    3 & 2 & 2  \\ \end{matrix} \right|\]\[=2(8\hat{i}-8\hat{j}+4\hat{k})\] Required vector\[=\pm 12\frac{(2\hat{i}-2\hat{j}-\hat{k})}{3}\]                         \[=\pm 4(2\hat{i}-2\hat{j}-\hat{k})\]


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