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JEE Main & Advanced AIEEE Solved Paper-2011

  • question_answer
    The vectors \[\vec{a}\] and \[\vec{b}\] are not perpendicular and \[\vec{c}\] and \[\vec{d}\] are two vectors satisfying \[\vec{b}\times \vec{c}=\vec{b}\times \vec{d}\]and \[\vec{a}.\,\vec{d}=0\]. Then the vector \[\vec{d}\] is equal to   AIEEE  Solved  Paper-2011

    A) \[\vec{c}-\left( \frac{\vec{a}.\,\vec{c}}{a.\,\vec{b}} \right)\vec{b}\]                          

    B) \[\vec{b}-\left( \frac{\vec{b}.\,\vec{c}}{\vec{a}.\,\vec{b}} \right)\vec{c}\]

    C) \[\vec{b}+\left( \frac{\vec{a}.\,\vec{c}}{\vec{a}.\,\vec{b}} \right)\vec{b}\]                            

    D) \[\vec{b}+\left( \frac{\vec{b}.\,\vec{c}}{\vec{a}.\,\vec{b}} \right)\vec{c}\]

    Correct Answer: A

    Solution :

                 We have \[\vec{b}\times \vec{c}=\vec{b}\times \vec{d}\] \[\Rightarrow \,\,\vec{a}\times (\vec{b}\times \vec{c})=\vec{a}\times (\vec{b}\times \vec{d})\] \[\Rightarrow \,\,\,(\vec{a}.\,\vec{c})\vec{b}-(\vec{a}.\,\vec{b})\vec{c}=(\vec{a}.\,\vec{d})\vec{b}-(\vec{a}.\,\vec{b})\vec{d}\] \[\Rightarrow \,\,\,(\vec{a}.\,\vec{b})\vec{d}=-(\vec{a}.\,\vec{c})\vec{b}+(\vec{a}.\,\vec{b})\vec{c}\] \[\Rightarrow \,\,\,\vec{d}=-\frac{(\vec{a}.\,\vec{c})}{(\vec{a}.\,\vec{b})}\vec{b}+\vec{c}\]


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