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Q.
The angle between the lines whose direction cosines are connected by the relations $l+m+n=0$ and $2 l m+2 n l-m n=0$, is
Three Dimensional Geometry
Solution:
Given, $1+m+n=0$....(i)
$2 l m+2 n l-m n =0$.....(ii)
$m =-1-n$[from Eq.(i)]
On substituting the value of $m$ in Eq. (ii), we get
$2 l(-1-n)+2 n l-(-1-n) n=0$
$\rightarrow-2 i^2-2 i n+2 n l+i n+n^2=0$
$\rightarrow n^2+l n-2 l^2=0$
$\rightarrow n^2+2 l n-l n-2 l^2=0 $
$\rightarrow n(n+2 l)-l(n+2 l)=0$
$\rightarrow (n+2 l)(n-l) \rightarrow n=1, n=-2 l$
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} $
$\therefore \theta =\cos ^{-1}\left(-\frac{1}{2}\right)-\pi-\cos ^{-1}\left(\frac{1}{2}\right)-\pi-\frac{\pi}{3}-\frac{2 \pi}{3}$