The direction cosines of a line satisfy the relations $\mathrm{\lambda }(\mathrm{l}+\mathrm{m})=\mathrm{n}$ and $\mathrm{mn}+\mathrm{nl}+\mathrm{lm}=0\text{. }$the value of  $\mathrm{\lambda }$ for which the two lines are perpendicular to each other, is

Eliminating n, we get 

$\begin{array}{l}\mathrm{\lambda }(\mathrm{l}+\mathrm{m}{)}^2+\mathrm{lm}=0\\ \Rightarrow \frac{{\mathrm{\lambda l}}^2}{{\mathrm{m}}^2}+(2\mathrm{\lambda }+1)\frac{\mathrm{l}}{\mathrm{m}}+\mathrm{\lambda }=0\\ \Rightarrow \frac{{\mathrm{l}}_1{\mathrm{l}}_2}{{\mathrm{m}}_1{\mathrm{m}}_2}=1\left(\mathrm{product}\mathrm{of}\mathrm{roots} \frac{{\mathrm{l}}_1}{{\mathrm{m}}_1}\text{ and }\frac{{\mathrm{l}}_2}{{\mathrm{m}}_2}\right)\end{array}$

where $\frac{{\mathrm{l}}_1}{{\mathrm{m}}_1}$and $\frac{{\mathrm{l}}_2}{{\mathrm{m}}_2}$ are the roots of this equation, further eliminating m, we get 

$\begin{array}{r}{\mathrm{\lambda l}}^2-\ln -{\mathrm{n}}^2=0\\ \Rightarrow \frac{{\mathrm{l}}_1{\mathrm{l}}_2}{{\mathrm{n}}_1{\mathrm{n}}_2}=-\frac{1}{\mathrm{\lambda }}\end{array}$

Since the lines with direction cosines (l₁,m₁,n₁ and (1₂, m₂, n₂) are perpendicular, 

we have  ${\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2=0$

$\begin{array}{l}\mathrm{or} 1+1-\mathrm{\lambda }=0\\ \mathrm{or} \mathrm{\lambda }=2\end{array}$