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} \bot L_{2}, $ therefore
$\sqrt{\lambda}\sqrt{\mu}+1\times1+\left(\sqrt{\lambda }-1\right)\left(1-\sqrt{\mu }\right)=0$
$\Rightarrow \sqrt{\lambda }\sqrt{\mu }=0 \Rightarrow \sqrt{\lambda }=\sqrt{\mu}$
$\Rightarrow \lambda=\mu$