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JEE Main & Advanced Physics Two Dimensional Motion Question Bank Self Evaluation Test - Motion in a Plane

  • question_answer
    Let two vectors \[\vec{A}=3\hat{i}+\hat{j}+2\hat{k}\] and\[\vec{B}=2\hat{i}-2\hat{j}+4\hat{k}\]. Consider the unit vector perpendicular to both A and B is

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

    B)        \[\frac{\hat{i}-\hat{j}-\hat{k}}{2\sqrt{3}}\]

    C) \[\frac{-\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}\]          

    D)        \[\frac{\hat{i}-\hat{j}-\hat{k}}{2\sqrt{3}}\]

    Correct Answer: A

    Solution :

    [a] Angle between \[\vec{A}\] and \[\vec{B}\] is given by \[\cos \theta =\frac{\text{\vec{A}}\,\text{.}\,\text{\vec{B}}}{\text{AB}}=\frac{3}{\sqrt{21}}\] The unit vector perpendicular to \[\vec{A}\] and \[\vec{B}\] is given by \[\hat{n}=\frac{\text{\vec{A}}\times \text{\vec{B}}}{|\text{\vec{A}}\times \text{\vec{B}}|}=\frac{(3\hat{i}+\hat{j}+2\hat{k})\times (2\hat{i}-2\hat{j}+4\hat{k})}{|(3\hat{i}+\hat{j}+2\hat{k})\times (2\hat{i}-2\hat{j}+4\hat{k})|}\] \[=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}\]


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