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
$L_{2} = \left\{x = \sqrt{\mu}y+\left(1-\sqrt{\mu }\right), z = \left(1-\sqrt{\mu }\right)y+\sqrt{\mu }\right\}$
then $L_{1}$ is perpendicular to $L_{2}$, for all non-negative reals $\lambda$ and $\mu$, such that :

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

Answer 1

Answer: (D)

For $L_1$,
$x = \sqrt{\lambda} y +\left(\sqrt{\lambda } -1\right) \Rightarrow y = \frac{x-\left(\sqrt{\lambda }-1\right)}{\sqrt{\lambda }}\quad...\left(i\right)$
$z= \left(\sqrt{\lambda }-1\right) y+ \sqrt{\lambda } \Rightarrow y = \frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1}\quad ...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\frac{x-\left(\sqrt{\lambda }-1\right)}{\sqrt{\lambda }} = \frac{y-0}{1} = \frac{z-\sqrt{\lambda }}{\sqrt{\lambda }-1} \quad...\left(A\right)$
The equation $\left(A\right)$ is the equation of line $L_{1}$.
Similarly equation of line $L_{2}$ is
$\frac{x-\left(1 - \sqrt{\mu }\right)}{\sqrt{\mu }} = \frac{y-0}{1} = \frac{z-\sqrt{\mu }}{1 - \sqrt{\lambda }} \quad ...\left(B\right)$
Since $L_{1} \bot L_{2 }$, therefore
$\sqrt{\lambda }\sqrt{\mu}+1\times1+\left(\sqrt{\lambda }-1\right)\left(1-\sqrt{\mu }\right) = 0$
$\Rightarrow \sqrt{\lambda } + \sqrt{\mu } = 0 \Rightarrow \sqrt{\lambda } = -\sqrt{\mu } \Rightarrow \lambda = \mu$