Question

An angle between the lines whose direction cosines are given by the equations, $l+3m+5n=0$ and $5lm-2mn+6nl=0$, is?

Answer 1

$l+3m+5n=0\Rightarrow l=-3m-5n5lm-2mn+6nl=05(-3m-5n)m-2mn+6n(-3m-5n)=0-15m^2-25nm-2mn-18mn-30n^2=015m^2+30n^2+45mn=0m^2+mn+2mn+2n^2=0m(m+n)+2n(m+n)=0(m+2n)(m+n)=0\therefore m=2norm=-n\therefore \frac{m}{-2}=\frac{n}{1}or\frac{m}{-1}=\frac{n}{1}$

Case(i) $\frac{m}{-2}=\frac{n}{1}=k\Rightarrow m=-2k,n=k\therefore l=-3m-5n=k\therefore \frac{m}{-2}=\frac{l}{1}=\frac{n}{1}$ $\therefore$directions $=(1,-2,1)$

Case(ii)$\frac{m}{-1}=\frac{n}{1}\Rightarrow m=-k,n=k\therefore l=-3m-5n=-2k$ $\therefore$ Directions $=(-2,-1,1)$

Angle between angles.

$\cos \theta =\frac{1\times -2+(-2)(-1)+1}{\sqrt{(1^2+2^2+1^2)(1^2+1^2+2^2)}}=\frac{1}{\sqrt{6}.\sqrt{6}}\therefore \theta ={\cos }^{-1}\left(\frac{1}{6}\right)$