Question

If the direction cosines of two lines satisfy the equations $l+m+n=0$ and $2\, l m+2 \ln -m n=0$, 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 $2 l m+2 \ln -m n=0$
$\therefore 2(-m-n) m+2(-m-n) n-m n=0$
$[\because l=-(m+ n)]$
$\Rightarrow -2 m^{2}-2 m n-2 m n-2 n^{2}-m n=0$
$\Rightarrow -2 m^{2}-5 m n-2 n^{2}=0$
$\Rightarrow 2 m^{2}+5 m n+2 n^{2}=0$
$\Rightarrow 2 m^{2}+4 m n+ m n+2 n^{2}=0$
$\Rightarrow 2 m(m+2 n)+n(m+2 n)=0$
$\Rightarrow (2 m+n)(m+2 n)=0$
$\Rightarrow m=\frac{-n}{2},-2 n$
$\Rightarrow l=\frac{-n}{2}, n$
$\therefore $ DR's of two lines are $<\frac{-n}{2}, \frac{-n}{2}, n>$ and $\left\langle n_{i}-2 n, n>\right.$ i.e. $<-1,-1,2>$ and $<1,-2,1>$
$\therefore $ Angle between these two lines
$=\cos ^{-1} \left[\frac{-1 \times 1+(-1) \times(-2)+2 \times 1}{\sqrt{(-1)^{2}+(-1)^{2}+(2)^{2}} \sqrt{(1)^{2}+(-2)^{2}+(1)^{2}}}\right]$
$=\cos ^{-1}\left[\frac{-1+2+2}{\sqrt{1+1+4} \sqrt{1+4+1}}\right]$
$=\cos ^{-1}\left(\frac{3}{6}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
$\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right]$