Question

The angle between the lines whose direction cosines are connected by the relations $l+m+n=0$ and $2lm+2nl-mn=0$, is

• A. $\frac{\pi }{6}$
• B. $\frac{2\pi }{3}$
• C. $\pi$
• D. None of these

Answer 1

Answer: (B)

Given, $1+m+n=0$....(i)
$2lm+2nl-mn=0$.....(ii)
$m=-1-n$[from Eq.(i)]
On substituting the value of $m$ in Eq. (ii), we get
$2l(-1-n)+2nl-(-1-n)n=0$
$\to -2i^2-2in+2nl+in+n^2=0$
$\to n^2+ln-2l^2=0$
$\to n^2+2ln-ln-2l^2=0$
$\to n(n+2l)-l(n+2l)=0$
$\to (n+2l)(n-l)\to n=1,n=-2l$
When $n=1,m=-21$
When, $n=-21,m=1$
Therefore, direction ratios of two lines are
$1,1,-2$ and $1,-2,1$
Now, $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{1-2-2}{\sqrt{1^2+1^2+(-2{)}^2}\sqrt{1^2+(-2{)}^2+1^2}}=\frac{-3}{\sqrt{6}\sqrt{6}}$
$=\frac{-3}{6}-\frac{-1}{2}$
$\therefore \theta ={\cos }^{-1}\left(-\frac{1}{2}\right)-\pi -{\cos }^{-1}\left(\frac{1}{2}\right)-\pi -\frac{\pi }{3}-\frac{2\pi }{3}$