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. $\lambda \ne \mu$
• D. $\sqrt{\lambda }+\sqrt{\mu }=1$

Answer 1

Answer: (B)

Line $L_1$ is parallel to vector
$v_1\text{ (say) }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -\sqrt{\lambda }& 0\\ 0& (\sqrt{\lambda }-1)& -1\end{array}\right|=(\sqrt{\lambda })\hat{i}-\hat{j}+(\sqrt{\lambda }-1)\hat{k}$
Also, line $L_2$ is parallel to vector
$v_2$ (say) $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -\sqrt{\mu }& 0\\ 0& 1-\sqrt{\mu }& -1\end{array}\right|=(\sqrt{\mu })\hat{i}-\hat{j}+(1-\sqrt{\mu })\hat{k}$
As, $v_1\cdot v_2=0\Rightarrow \sqrt{\lambda }+\sqrt{\mu }=0\Rightarrow \lambda =\mu =0$