Question

Find the angle between the two lines whose direction cosines satisfy $l-5m-3n=0,7l^2+5m^2-3n^2=0$

Answer 1

$l-5m-3n=0\Rightarrow l=5m+3n$

$\Rightarrow 7l^2+5m^2-3n^2=0$

$\Rightarrow 7{(5m+3n)}^2+5m^2-3n^2=0$

$\Rightarrow 7(25m^2+9n^2+30mn)+5m^2-3n^2=0$

$\Rightarrow 175m^2+63n^2+210mn+5m^2-3n^2=0$

$\Rightarrow 180m^2+60n^2+210mn=0$

$\Rightarrow 6m^2+2n^2+7mn=0$

$\Rightarrow 6m^2+7mn+2n^2=0$

$\Rightarrow 6m^2+4mn+3mn+2n^2=0$

$\Rightarrow 2m(3m+2n)+n(3m+2n)=0$

$\Rightarrow (3m+2n)(2m+n)=0$

$\Rightarrow m=\frac{-2n}{3},\frac{n}{2}$

$\Rightarrow m=\frac{-2n}{3}\Rightarrow l=5m+3n=\frac{-10n}{3}+3n=\frac{-10n+9n}{3}=\frac{-n}{3}$

Direction ratio's are $-1,-2,3$ for $n=3$ .........$\left(1\right)$

$\Rightarrow m=\frac{n}{2}\Rightarrow l=5m+3n=\frac{5n}{2}+3n=\frac{5n+6n}{2}=\frac{11n}{2}$

Direction ratio's are $1,-1,2$ for $n=2$ .........$\left(2\right)$

$\cos \theta =\sqrt{\frac{-1+2+6}{14+6}}=\frac{7}{12}$

$\therefore \theta ={\cos }^{-1}\sqrt{\frac{7}{12}}$