Mathematics - Point Question with Solution | TestHub

For ${\mathrm{L}}_1$, $$\begin{array}{rl}& x=\sqrt{\lambda }y+(\sqrt{\lambda }-1)\Rightarrow y=\frac{x-(\sqrt{\lambda }-1)}{\sqrt{\lambda }}\\ & z=(\sqrt{\lambda }-1)y+\sqrt{\lambda }\Rightarrow y=\frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1}\end{array}$$ From (i) and (ii) $$\frac{x-(\sqrt{\lambda }-1)}{\sqrt{\lambda }}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1}$$ The equation (A) is the equation of line ${\mathrm{L}}_1$. Similarly equation of line ${\mathrm{L}}_2$ is $$\frac{x-(1-\sqrt{\mu })}{\sqrt{\mu }}=\frac{y-0}{1}=\frac{z-\sqrt{\mu }}{1-\sqrt{\mu }}$$ Since ${\mathrm{L}}_1\perp {\mathrm{L}}_2$, therefore $$\begin{array}{rl}& \sqrt{\lambda }\sqrt{\mu }+1\times 1+(\sqrt{\lambda }-1)(1-\sqrt{\mu })=0\\ & \Rightarrow \sqrt{\lambda }+\sqrt{\mu }=0\Rightarrow \sqrt{\lambda }=-\sqrt{\mu }\\ & \Rightarrow \lambda =\mu \end{array}$$