Perpendicular Lines.

Question

If two lines $L_1$ and $L_2$ in space, are defined by: $$L_1=\{x=\sqrt{\lambda}y+(\sqrt{\lambda}-1)\\z=(\sqrt{\lambda}-1)y+\sqrt{\lambda}\}\text{ and }\\L_2=\{x=\sqrt{\mu}y+(1-\sqrt{\mu})\\z=(1-\sqrt{\mu})y+\sqrt{\mu}\}$$ then $L_1$ is perpendicular to $L_2$, forall non-negative reals $\lambda$ and $\mu$, such that: It can be easily seen: $$L_1:\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y}{1}=\frac{z-\sqrt{\lambda}}{(\sqrt{\lambda}-1)}\\ L_2:\frac{x-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y}{1}=\frac{z-\sqrt{\mu}}{(1-\sqrt{\mu})}\\ $$ So dot product (of those) must be zero: $$\sqrt{\lambda}\sqrt{\mu}+1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0\\ \sqrt\lambda+\sqrt\mu=0\\ \lambda=\mu=0$$ Options given are: $$\lambda=\mu,\lambda\ne\mu,\sqrt\lambda+\sqrt\mu=1,\lambda+\mu=0$$ It seems three of them are correct but only one is actually correct.

Answer 1

Let me see: the lines are

$$L_1 \;:\;\;\left\{\;\left(\;x=\sqrt\lambda\,y+\sqrt\lambda-1\,,\,\,y\,,\,\,z=(\sqrt\lambda-1)y+\sqrt\lambda\;\right)\right\}=$$

$$=\left(\sqrt\lambda-1\,,\,\,0\,,\,\,\sqrt\lambda\right)+t\left(\sqrt\lambda\,,\,\,1\,,\,\,\sqrt\lambda-1\right)\;,\;\;t\in\Bbb R$$

$$L_2\;:\;\;\left\{\;\left(\;x=\sqrt\mu\,y+1-\sqrt\lambda\,,\,\,y\,,\,\,z=(1-\sqrt\mu)y+\sqrt\mu\;\right)\right\}=$$

$$=\left(1-\sqrt\mu\,,\,0\,,\,\,\sqrt\mu\right)+t\left(\sqrt\mu\,,\,1\,,\,1-\sqrt\mu\right)$$

Thus, $\;L_1\perp L_2\iff\;$ their direction vectors are perpendicular, iff

$$0=\left(\sqrt\lambda\,,\,\,1\,,\,\,\sqrt\lambda-1\right)\cdot\left(\sqrt\mu\,,\,1\,,\,1-\sqrt\mu\right)=\sqrt{\lambda\mu}+1+\sqrt\lambda-\sqrt{\lambda\mu}-1+\sqrt\mu\iff$$

$$\iff\sqrt\lambda=-\sqrt\mu$$