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

Solution:

The given equations are 

$l+3m+5n=0$                                 …(i)

$5lm-2mn+6nl=0$                         …(ii)   

From (i) $l=-3m-5n$

Putting this value of $I$ in (ii), we have 

$5(-3m-5m)m-2mn+6n(-3m-5n)=0$

$\begin{array}{l}\Rightarrow -15m^2-30n^2-45mn=0\Rightarrow m^2+2n^2+3mn=0\\ \Rightarrow m^2+3mn+2n^2=0\Rightarrow m(m+2n)+n(m+2n)=0\end{array}$

$\Rightarrow (m+n)(m+2n)=0\Rightarrow$either $m=-n$ or $m=-2n$

For, $m=-n,l=-2n$; for $;m=-2n,l=n$

$\therefore$ Direction ratios of two lines are 

$⟨-2n,-n,n⟩$ and $⟨n,-2n,n⟩$

i.e. $⟨-2,-1,1⟩$ and $⟨1,-2,1⟩$

$\therefore$ The required angle is

$\begin{array}{l}\cos \theta =\frac{-2\cdot 1+2\cdot 1+1\cdot 1}{\sqrt{4+1+1}\sqrt{1+4+1}}\\ \Rightarrow \cos \theta =\frac{1}{\sqrt{6}\cdot \sqrt{6}}=\frac{1}{6}\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{6}\right)\end{array}$