Question

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

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

Answer 1

Answer: (B)

Line $L _1$ is parallel to vector
$v_1 \text { (say) }=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & -\sqrt{\lambda} &0 \\0 & (\sqrt{\lambda}-1) & -1\end{vmatrix}=(\sqrt{\lambda}) \hat{i}-\hat{j}+(\sqrt{\lambda}-1) \hat{k}$
Also, line $L _2$ is parallel to vector
${v}_2$ (say) $=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -\sqrt{\mu} & 0 \\ 0 & 1-\sqrt{\mu} & -1\end{vmatrix}=(\sqrt{\mu}) \hat{i}-\hat{j}+(1-\sqrt{\mu}) \hat{k}$
As, ${v}_1 \cdot {v}_2=0 \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \lambda=\mu=0$