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If two lines $L_1$ and $L_2$ in space, are defined by
$$
\begin{aligned}
& L_1=\{x=\sqrt{\lambda} y+(\sqrt{\lambda}-1), \\
& \quad z=(\sqrt{\lambda}-1) y+\sqrt{\lambda}\} \text { and } \\
& L_2=\{x=\sqrt{\mu} y+(1-\sqrt{\mu}),
\end{aligned}
$$
$$
z=(1-\sqrt{\mu}) y+\sqrt{\mu}\}
$$
then $L_1$ is perpendicular to $L_2$, for all nonnegative reals $\lambda$ and $\mu$, such that :
Options:
$$
\begin{aligned}
& L_1=\{x=\sqrt{\lambda} y+(\sqrt{\lambda}-1), \\
& \quad z=(\sqrt{\lambda}-1) y+\sqrt{\lambda}\} \text { and } \\
& L_2=\{x=\sqrt{\mu} y+(1-\sqrt{\mu}),
\end{aligned}
$$
$$
z=(1-\sqrt{\mu}) y+\sqrt{\mu}\}
$$
then $L_1$ is perpendicular to $L_2$, for all nonnegative reals $\lambda$ and $\mu$, such that :
Solution:
2383 Upvotes
Verified Answer
The correct answer is:
$\lambda=\mu$
$\lambda=\mu$
For $\mathrm{L}_1$,
$$
\begin{aligned}
& x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}} \\
& z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}
\end{aligned}
$$
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
$$
\begin{aligned}
& \sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0 \\
& \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu} \\
& \Rightarrow \lambda=\mu
\end{aligned}
$$
$$
\begin{aligned}
& x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}} \\
& z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}
\end{aligned}
$$
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
$$
\begin{aligned}
& \sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0 \\
& \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu} \\
& \Rightarrow \lambda=\mu
\end{aligned}
$$
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