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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    The vector \[\vec{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}\] lies in the plane of the vectors \[\vec{b}=\hat{i}+\hat{j}\] and \[\vec{c}=\hat{j}+\hat{k}\] and bisects the angle between \[\vec{b}\] and \[\vec{c}\]. Then which one of the following gives possible values of \[\alpha \] and\[\beta \]?         AIEEE  Solved  Paper-2008

    A) \[\alpha =2,\,\,\beta =1\]                           

    B) \[\alpha =1,\,\,\beta =1\]           

    C)        \[\alpha =2,\,\,\beta =2\]           

    D)        \[\alpha =1,\,\,\beta =2\]

    Correct Answer: B

    Solution :

                    \[\because \]  \[\vec{a},\vec{b}\] and \[\vec{c}\] are coplanar.                 \[\Rightarrow \]               \[[\vec{a}\,\,\vec{b}\,\,\vec{c}]=0\Rightarrow \alpha +\beta =2\]                            ? (i) Also \[\vec{a}\] bisects the angle between \[\vec{b}\] and \[\vec{c}\] \[\Rightarrow \,\vec{a}=\lambda \left( \hat{b}+\hat{c} \right)\Rightarrow \vec{a}=\lambda \left( \frac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{2}} \right)\]        ? (ii) Comparing (ii) with \[\vec{a}=\alpha \,\hat{i}+2\hat{j}+\beta \hat{k}\], we get \[\lambda =\sqrt{2},\,\,\therefore \alpha =1\] and \[\beta =1\], which also satisfies (i)


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