Question

If the direction cosines of two lines satisfy the equations $l+m+n=0$ and $2lm+2\ln -mn=0$, then the acute angle between those two lines is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{2\pi }{5}$

Answer 1

Answer: (B)

We have,
$l+m+n=0$ and $2lm+2\ln -mn=0$
$\therefore 2(-m-n)m+2(-m-n)n-mn=0$
$[\because l=-(m+n)]$
$\Rightarrow -2m^2-2mn-2mn-2n^2-mn=0$
$\Rightarrow -2m^2-5mn-2n^2=0$
$\Rightarrow 2m^2+5mn+2n^2=0$
$\Rightarrow 2m^2+4mn+mn+2n^2=0$
$\Rightarrow 2m(m+2n)+n(m+2n)=0$
$\Rightarrow (2m+n)(m+2n)=0$
$\Rightarrow m=\frac{-n}{2},-2n$
$\Rightarrow l=\frac{-n}{2},n$
$\therefore$ DR's of two lines are $<\frac{-n}{2},\frac{-n}{2},n>$ and $⟨n_i-2n,n>$ i.e. $<-1,-1,2>$ and $<1,-2,1>$
$\therefore$ Angle between these two lines
$={\cos }^{-1}\left[\frac{-1\times 1+(-1)\times (-2)+2\times 1}{\sqrt{(-1{)}^2+(-1{)}^2+(2{)}^2}\sqrt{(1{)}^2+(-2{)}^2+(1{)}^2}}\right]$
$={\cos }^{-1}\left[\frac{-1+2+2}{\sqrt{1+1+4}\sqrt{1+4+1}}\right]$
$={\cos }^{-1}\left(\frac{3}{6}\right)={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$
$\left[\because \cos \frac{\pi }{3}=\frac{1}{2}\right]$