Question

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

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $4$

Answer 1

The correct option is B $2$

By Eliminating n, we get
$\lambda (l+m{)}^2+lm=0\Rightarrow \frac{\lambda l^2}{m^2}+(2\lambda +1)\frac{l}{m}+\lambda =0\Rightarrow \frac{l_l{l}_2}{m_1{m}_2}=1$
Where $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$ are the roots of this equation, Further eliminating m, we get
$\lambda l^2-ln-n^2=0\Rightarrow \frac{l_1{l}_2}{n_1{n}_2}=-\frac{1}{\lambda }$
Since, the lines with direction cosines $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ are perpendicular.
$l_1{l}_2+m_1{m}_2+n_1{n}_2=0$
$\Rightarrow 1+1-\lambda =0\Rightarrow \lambda =2$