Question

The direction cosines of a line 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. $-3$

Answer 1

Answer: (B)

$2$

Given relations between the direction cosines are $\lambda (l+m)=n$ ------$\left(1\right)$

and $lm+mn+ln=0$ -------$\left(2\right)$

Eliminating $n$ from $\left(1\right)$ and $\left(2\right)$ gives,

$lm+(l+m)\lambda (l+m)=0$

$\Rightarrow lm+\lambda (l^2+m^2+2lm)=0$

Dividing equation with $m^2$ gives,

$\Rightarrow \lambda {\left(\frac{l}{m}\right)}^2+(2\lambda +1)\left(\frac{l}{m}\right)+\lambda =0$Pproduct of roots $\frac{l_1{l}_2}{m_1{m}_2}=1$ -----$\left(3\right)$

Now eliminating $l$ from $\left(1\right)$ and $\left(2\right)$ gives,

$mn+(n+m)\left(\frac{n}{\lambda }-m\right)=0$

$\Rightarrow \lambda mn+(n+m)(n-\lambda m)=0$

$\Rightarrow \lambda m^2-mn-n^2=0$

Dividign eqaution with $n^2$ gives,,

$\lambda {\left(\frac{m}{n}\right)}^2-\frac{m}{n}-1=0$

Product of roots $\frac{m_1{m}_2}{n_1{n}_2}=-\frac{1}{\lambda }$ ------$\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$, we get

$\frac{l_1{l}_2}{1}=\frac{m_1{m}_2}{1}=\frac{n_1{n}_2}{-\lambda }$

The two lines are perpendicular to each other if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$

$\Rightarrow 1+1-\lambda =0$

$\therefore \lambda =2$

Hence, option B is correct.