The acute angle between the two lines whose direction ratios are given by $l+m-n=0$ and $l^2+m^2-n^2=0$, is

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Aicharya · Accepted Answer

Given that $ l+m-n=0 $ and $ l^2+m^2-n^2=0 $ From equaion (i) $l=-(m-n)$ Putting in equation (ii), we get $ \begin{aligned} \Rightarrow & (m-n)^2+m^2-n^2 & =0 \ \Rightarrow & m^2+n^2-2 m n+m^2-n^2 & =0 \ \Rightarrow & 2 m^2-2 m n & =0 \ & 2 m(m-n) & =0 \ \Rightarrow & m=0, m & =n \end{aligned} $ From equation (i), if $m=0$, then $ l=n \Rightarrow l: m: n=1: 0: 1 $ If $m=n$, then $l=0 \Rightarrow l: m: n=0: 1: 1$ If $	heta$ is the acute angle, then $ \begin{aligned} \cos 	heta & =\left|\frac{1.0+0.1+1.1}{\sqrt{1+1} \sqrt{1+1}}ight|=\frac{1}{2}=\cos \frac{\pi}{3} \ \Rightarrow \quad 	heta & =\frac{\pi}{3} \end{aligned} $

The acute angle between the two lines whose direction ratios are given by $l+m-n=0$ and $l^2+m^2-n^2=0$, is

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