If the direction ratio of two lines are given by $l+m+n=0, m n-2 l n+l m=0, \quad$ then the angle between the lines is

Question

Aicharya · Accepted Answer

From equation (i) $ l=-(m+n) $ Putting in equation (ii), we get $ \begin{aligned} & & m n+2(m+n) n-(m+n) m & =0 \\ \Rightarrow & & m n+2 m n+2 n^2-m^2-n m & =0 \\ \Rightarrow & & 2 n^2-m^2+2 m n & =0 \\ \Rightarrow & & 2\left(\frac{n}{m}\right)^2+\frac{2 n}{m}-1 & =0 \end{aligned} $

[where $\frac{n_1}{m_1}, \frac{n_2}{m_2}$ are the roots of the equation] From equaiton (i) $ m=-(n+l) $ Putting in equation (ii), we get $ \begin{aligned} & & -(n+l) n-2 \ln -l(n+l) & =0 \\ \Rightarrow & & n^2+\ln +2 \ln +\ln +l^2 & =0 \\ \Rightarrow & & l^2+3 \ln +n^2 & =0 \\ \Rightarrow & & \left(\frac{l}{n}\right)^2+\frac{3 l}{n}+1 & =0 \end{aligned} $

[where $\frac{l_1}{n_1}, \frac{l_2}{n_2}$ are the roots of the equation] From equations (iii) and (iv), we get

If the direction ratio of two lines are given by $l+m+n=0, m n-2 l n+l m=0, \quad$ then the angle between…

Solution