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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    If two lines \[{{\operatorname{L}}_{1}}\]and \[{{\operatorname{L}}_{2}}\] in space, are defined by                 \[{{\operatorname{L}}_{2}}=\{x=\sqrt{\lambda }y+(\sqrt{\lambda }-1)\}\]                 \[z=\left( \sqrt{\lambda }-1 \right)y+\sqrt{\lambda }\}\]and                 \[{{L}_{2}}=\{x=\sqrt{\mu }y+(1-\sqrt{\mu }),\]                 \[z=(1-\sqrt{\mu })y+\sqrt{\mu }\},\]                 then\[{{L}_{1}}\] is perpendicular to \[{{L}_{2}},\] for all non-negative reals\[\lambda \] and \[\mu \] such that:     JEE Main Online Paper ( Held On 23  April 2013 )

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

    B)                                          \[\lambda \ne \mu \]

    C)                                          \[\lambda +\mu =0\]                  

    D)                                          \[\lambda =\mu \]

    Correct Answer: D

    Solution :

                      For \[{{L}_{1}},\] \[x=\sqrt{\lambda y}+(\sqrt{\lambda }-1)\Rightarrow y=\frac{x-(\sqrt{\lambda }-1)}{\sqrt{\lambda }}\]?.(i) \[z=(\sqrt{\lambda }+1)y+\sqrt{\lambda }\Rightarrow y=\frac{z-(\sqrt{\lambda }}{\sqrt{\lambda }-1}\]?.(ii) From (i) and (ii) \[\frac{x-(\sqrt{\lambda }-1)}{\sqrt{\lambda }}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1}\]                         ? The equation   is the equation of line \[{{L}_{1}}.\] Similarly equation of line \[{{L}_{2}}\]is \[\frac{x-(1-\sqrt{\mu })}{\sqrt{\mu }}=\frac{y-0}{1}=\frac{z-\sqrt{\mu }}{1-\sqrt{\mu }}\]                           ? Since \[{{L}_{1}}\bot {{L}_{2}},\] therefore \[\sqrt{\lambda }\sqrt{\mu }+1\times 1+(\sqrt{\lambda }-1)(1-\sqrt{\mu })=0\] \[\Rightarrow \]\[\sqrt{\lambda }+\sqrt{\mu }=0\Rightarrow \sqrt{\lambda }=-\sqrt{\lambda }=-\sqrt{\mu }\] \[\Rightarrow \]\[\lambda =\mu \]


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