Question

If two lines $L_1$ and $L_2$ in space, are defined by
$L_1=\left\{x=\sqrt{\lambda }y+\left(\sqrt{\lambda }-1\right),z=+\left(\sqrt{\lambda }-1\right)+y+\sqrt{\lambda }\right\}$ and $L_2=\left\{x=\sqrt{\mu }y+\left(1-\sqrt{\mu }\right)\right\},z=\left(1-\sqrt{\mu }\right)y+\sqrt{\mu }$
then $L_1$ is perpendicular to $L_2$, for all nonnegative reals $\lambda$ and $\mu$, such that :

• A. $\sqrt{\lambda }+\sqrt{\mu }=1$
• B. $\lambda \ne \mu$
• C. $\lambda +\mu =0$
• D. $\lambda =\mu$

Answer 1

Answer: (D)

For $L_1.$
$x=\sqrt{\lambda }y+\left(\sqrt{\lambda }-1\right)\Rightarrow y=\frac{x-\left(\sqrt{\lambda }\right)-1}{\sqrt{\lambda }}...\left(i\right)$
$z=\left(\sqrt{\lambda }-1\right)y+\sqrt{\lambda }\Rightarrow y=\frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1}...\left(ii\right)$
From $\left(i\right)and\left(ii\right)$
$\frac{x-\left(\sqrt{\lambda }-1\right)}{\sqrt{\lambda }}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1}...\left(A\right)$
The equation $\left(A\right)$ is the equation of line $L_1.$ Similarly equation of line $L_2$ is
$\frac{x-\left(1-\sqrt{\mu }\right)}{\sqrt{\mu }}=\frac{y-0}{1}=\frac{z-\sqrt{\mu }}{1-\sqrt{\mu }}...\left(B\right)$
Since $L_1\perp L_2,$ therefore
$\sqrt{\lambda }\sqrt{\mu }+1\times 1+\left(\sqrt{\lambda }-1\right)\left(1-\sqrt{\mu }\right)=0$
$\Rightarrow \sqrt{\lambda }\sqrt{\mu }=0\Rightarrow \sqrt{\lambda }=\sqrt{\mu }$
$\Rightarrow \lambda =\mu$