JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    If two lines  \[{{L}_{1}}\] and  \[{{L}_{2}}\] in space are defined by\[{{L}_{1}}=\left\{ x=\sqrt{\lambda }y+(\sqrt{\lambda }-1),\,z=(\sqrt{\lambda }-1)y+\sqrt{\lambda } \right\}\]   and \[{{L}_{2}}=\left\{ x=\sqrt{\mu }y+(1-\sqrt{\mu }),\,z=(1-\sqrt{\mu })y+\sqrt{\mu } \right\}\]then \[{{L}_{1}}\] is perpendicular to \[{{L}_{2}}\] for all non-negative reals \[\lambda \] and \[\mu \], such that

    A)  \[\lambda =\mu \]                                        

    B)  \[\lambda +\mu =0\]

    C)  \[\lambda \ne \mu \]                                   

    D)  \[\sqrt{\lambda }+\sqrt{\mu }=1\]

    Correct Answer: B

    Solution :

    Line \[{{L}_{1}}\] is parallel to vector \[\overset{\to }{\mathop{{{v}_{1}}}}\,(say)=\left| \begin{matrix}    \widehat{i} & \widehat{j} & \widehat{k}  \\    1 & -\sqrt{\lambda } & 0  \\    0 & (\sqrt{\lambda }-1) & -1  \\ \end{matrix} \right|=(\sqrt{\lambda })\widehat{i}-\widehat{j}+(\sqrt{\lambda }-1)\widehat{k}\]            Also, line \[{{L}_{2}}\]is parallel to vector \[\overset{\to }{\mathop{{{v}_{2}}}}\,(say)=\left| \begin{matrix}    \widehat{i} & \widehat{j} & \widehat{k}  \\    1 & -\sqrt{\mu } & 0  \\    0 & 1-\sqrt{\mu } & -1  \\ \end{matrix} \right|=(\sqrt{\mu })\widehat{i}-\widehat{j}+(1-\sqrt{\mu })\widehat{k}\]As, \[{{v}_{1}}.{{v}_{2}}=0\Rightarrow \sqrt{\lambda }+\sqrt{\mu }=0\Rightarrow \lambda =\mu =0\]


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