Question

If $\theta$ is the acute angle between the two lines whose direction cosines are given by $3l+m+5n=0$ and $6mn-2ln+5lm=0,$ then the value of $\sec \theta$ is equal to

Answer 1

Given, $3l+m+5n=0\dots \left(1\right)$
$6mn-2ln+5lm=0\dots \left(2\right)$
From $\left(1\right),$ $m=-3l-5n$
From $\left(2\right),$
$6n(-3l-5n)-2ln+5l(-3l-5n)=0$
$\Rightarrow -18ln-30n^2-2ln-15l^2-25ln=0$
$\Rightarrow -30n^2-45ln-15l^2=0$
$\Rightarrow 2n^2+3ln+l^2=0$
$\Rightarrow (2n+l)(n+l)=0$
$\Rightarrow l=-n$ or $l=-2n$
$\Rightarrow m=-2n$ or $m=n$
$\therefore (l,m,n)=(-1,-2,1)$ and $(-2,1,1)$
$\cos \theta =\frac{|2-2+1|}{\sqrt{6}\cdot \sqrt{6}}=\frac{1}{6}$ $(\because \theta$ is acute $)$
$\Rightarrow \sec \theta =6$