Question

If two lines whose direction cosines given by the relations $3l+m+5n=0$ and $6mn-2nl+5lm=0$ and the angle between them is $\theta$, then $\left|12\cos \theta \right|$

Answer 1

$3l+m+5n=0-----\left(i\right)6mn-2nl+5lm=0------\left(ii\right)so,from6n\left(equ\right(i)..........m=-3l-5nsub{}^{\prime }{m}^{\prime }inequ(ii)\Rightarrow 6n(-3l-5n)-2nl+5l(-3l-5n)=0\Rightarrow -18nl-30n^2-2nl-15l^2-25nl=0\Rightarrow 18nl+30n^2+15l^2+27nl=0\Rightarrow 45nl+30n^2+15l^2=0\Rightarrow 2n^2+3nl+l^{}=0(2n+l\left)\right(l+n)=0\therefore l=-n,l=-2nIfl=-n,fromequ(i)\Rightarrow -3n+m+5n=0\Rightarrow m+2n=0,n=\frac{-m}{2}then,\frac{l}{1}=\frac{-m}{2}=\frac{-n}{1}ifl=-2n,,fromequ(ii)\Rightarrow 6mn-2n(-2n)+5(-2n)m=0\Rightarrow 6mn+4n^2-10mn=0\Rightarrow 4n^2-4mn=0,n-m=0\therefore n=m\frac{l}{-2}=\frac{m}{1}=\frac{n}{1}\cos \theta =\frac{\left|-2+2-1\right|}{\sqrt{1+4+1}\sqrt{4+1+1}}=\frac{1}{6}6={\cos }^{-1}(\frac{1}{6})\therefore \left|12\cos \theta \right|=12\left(\frac{1}{6}\right)=2$