Question

Find the angle between the two straight lines whose direction cosines are given by $2l+2m-n=0$ and $mn+nl+lm=0$.

Answer 1

Let the angle between the lines be $\alpha$

Given Direction cosines of two lines are $<l,m,n>$

Also given that $2l+2m-n=0⟹n=2l+2m\cdots \left(1\right)$

$mn+nl+lm=0\cdots \left(2\right)$

$⟹m(2l+2m)+l(2l+2m)+lm=0$ $(\because \text{from}(1\left)\right)$

$⟹2m^2+5lm+2l^2=0$

$⟹2m^2+4lm+lm+2l^2=0$

$⟹2m(m+2l)+l(m+2l)=0$

$⟹(2m+l)(m+2l)=0$

$⟹m=-\frac{l}{2},-2l$

$n=2l+2m=2l+2(-\frac{l}{2}),2l+2(-2l)=2l-l,2l-4l=l,-2l$

So the direction cosines of two lines are $<l,-\frac{l}{2},l>$ and $<l,-2l,-2l>$

So the direction ratios will be $<2,-1,2>$ and $<1,-2,-2>$

As we know that

If $\theta$ is the angle between the two lines having direction ratios as $<a,b,c>$ and $<d,e,f>$

then $\cos \theta =\frac{ad+be+cf}{\sqrt{a^2+b^2+c^2}\sqrt{d^2+e^2+f^2}}$

Hence $\cos \alpha =\frac{\left(2\right)\left(1\right)+(-1)(-2)+2(-2)}{\sqrt{2^2+(-1{)}^2+2^2}\sqrt{1^2+(-2{)}^2+(-2{)}^2}}=\frac{2+2-4}{3\times 3}=0$

$⟹\alpha =\frac{\pi }{2}$

So the angle between the two lines is $\frac{\pi }{2}$