Question

If the direction ratios of two lines are given by $l+m+n=0,mn-2ln+lm=0,$ then the angle between the line is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

Given lines are $l+m+n=0\Rightarrow l=-(m+n)$ ...$\left(1\right)$

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

$\Rightarrow mn+2(m+n)n-(m+n)m=0$ (from $\left(1\right)$)

$\Rightarrow mn+2mn+2n^2-m^2-nm=0\Rightarrow 2{\left(\frac{n}{m}\right)}^2+\left(\frac{2n}{m}\right)-1=0$

This is a quadratic equation in $\left(\frac{n}{m}\right)$

$\therefore \frac{n_1{n}_2}{m_1{m}_2}=-\frac{1}{2}$ ...(3)

$[$ where $\frac{n_1}{m_1},\frac{n_2}{m_2}$ are the roots of the equation $]$

From equation $\left(1\right)$ $m=-(n+1)$

On putting in equation $\left(2\right)$, we get

$-(n+l)n-2ln-l(n+l)=0\Rightarrow l^2+4ln+n^2=0$

$\Rightarrow {\left(\frac{l}{n}\right)}^2+\frac{4l}{n}+1=0\Rightarrow \frac{l_1{l}_2}{n_1{n}_2}=1$ ...(4)

$[$ where $\frac{l_1}{n_1},\frac{l_2}{n_2}$ are the roots of the equation $]$

$\therefore$ From equation $\left(3\right)$ and $\left(4\right)$

$l_1{l}_2=-2m_1{m}_2=n_1{n}_2\Rightarrow \frac{l_1{l}_2}{1}=\frac{m_1{m}_2}{-2}=\frac{n_1{n}_2}{1}=k$

Now $l_1{l}_2+m_1{m}_2+n_1{n}_2=k-2k+k=0$

$\therefore \cos \theta \Rightarrow \theta ={90}^0=\frac{\pi }{2}$