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Q.
The direction cosines $l, m, n$ of two lines are satisfying $3 l+m+5 n=0$ and $6 m n-2 n l+5 l m=0$. If $\theta$ is the angle between those lines then $|\cos \theta|=$
TS EAMCET 2020
Solution:
We have
$3 l+m+5 n=0$ and $6 m n-2 n l+5 l m=0$
$\Rightarrow 6 n(-3 l-5 n)-2 n l+5 l(-3 l-5 n)=0$
$\Rightarrow -18 n l-30 n^{2}-2 n l-15 l^{2}-25 n l=0$
$\Rightarrow -15 l^{2}-30 n^{2}-45 n l=0$
$\Rightarrow l^{2}+3 n l+2 n^{2}=0$
$\Rightarrow (l+n)(l+2 n)=0$
$ \Rightarrow l=-n,-2 n$
When $I=-n$
$-3 n+m+5 n=0$
$\Rightarrow m=-2 n$
So, $d^{\prime} r$ are
$-n,-2 n, n$ or $1,2,-1$
When $l=-2 n$
$-6 n+m+5 n=0$
$ \Rightarrow m=n$
So, $d' r$ are
$-2 n, n, n$
$-2,1,1$
$\therefore \cos \theta=\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6}$
$\therefore |\cos \theta|=\frac{1}{6}$