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 \[\sqrt{\lambda }\sqrt{\mu }+1\times 1+(\sqrt{\lambda }-1)(1-\sqrt{\mu })=0\] \[\Rightarrow \]\[\sqrt{\lambda }+\sqrt{\mu }=0\Rightarrow \sqrt{\lambda }=-\sqrt{\lambda }=-\sqrt{\mu }\] \[\Rightarrow \]\[\lambda =\mu \]You need to login to perform this action.
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