If two lines L₁ and L₂ in space, are defined by
$\begin{array}{l}{\mathrm{L}}_1=\{\mathrm{x}=\sqrt{\mathrm{\lambda }}\mathrm{y}+(\sqrt{\mathrm{\lambda }}-1),\mathrm{z}=(\sqrt{\mathrm{\lambda }}-1)\mathrm{y}+\sqrt{\mathrm{\lambda }}\}\text{ and }\\ {\mathrm{L}}_2=\{\mathrm{x}=\sqrt{\mathrm{\mu }}\mathrm{y}+(1-\sqrt{\mathrm{\mu }}),\mathrm{z}=(1-\sqrt{\mathrm{\mu }})\mathrm{y}+\sqrt{\mathrm{\mu }}\}\end{array}$
Then L₁ is perpendicular to L₂ , for all non-negative reals $\mathrm{\lambda }$ and $\mathrm{\mu }$ such that:

For L₁
$\begin{array}{l}\mathrm{x}=\sqrt{\mathrm{\lambda }}\mathrm{y}+(\sqrt{\mathrm{\lambda }}-1)\Rightarrow \mathrm{y}=\frac{\mathrm{x}-(\mathrm{\lambda }-1)}{\sqrt{\mathrm{\lambda }}}\to \left(i\right)\\ \mathrm{z}=(\sqrt{\mathrm{\lambda }}-1)\mathrm{y}+\sqrt{\mathrm{\lambda }}\Rightarrow \mathrm{y}=\frac{\mathrm{z}-\sqrt{\mathrm{\lambda }}}{\sqrt{\mathrm{\lambda }}-1}\to \left(ii\right)\end{array}$
From (i) and (ii)$\frac{\mathrm{x}-(\sqrt{\mathrm{\lambda }}-1)}{\sqrt{\mathrm{\lambda }}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\sqrt{\mathrm{\lambda }}}{\sqrt{\mathrm{\lambda }}-1}$
The equation (i) and (ii) is the equation of line L₁
Similarly equation of line L₂ is $\frac{\mathrm{x}-(1-\sqrt{\mathrm{\mu }})}{\sqrt{\mathrm{\mu }}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\sqrt{\mathrm{\mu }}}{1-\sqrt{\mathrm{\mu }}}$
Since ${\mathrm{L}}_1\perp {\mathrm{L}}_2$, therefore
$\begin{array}{r}\sqrt{\mathrm{\lambda }}\sqrt{\mathrm{\mu }}+1\times 1+(\sqrt{\mathrm{\lambda }}-1)(1-\sqrt{\mathrm{\mu }})=0\\ \Rightarrow \sqrt{\mathrm{\lambda }}+\sqrt{\mathrm{\mu }}=0\Rightarrow \sqrt{\mathrm{\lambda }}=-\sqrt{\mathrm{\mu }}\Rightarrow \mathrm{\lambda }=\mathrm{\mu }\end{array}$