The direction cosines of two lines are given by $l+m+n=0$ and $2lm + 2nl -mn=0$. Find the angle between them.
My attempt:
Given: $l+m+n=0$ and $2lm+2nl-mn=0$
From first equation, $n=-(l+m)$ Putting this value of $n$ in second equation: $$2lm-2l(l+m) + m(l+m)=0$$ $$ m^2+ml-2l^2=0$$ $$(m-l)(m+2l)=0$$
For $l_1=m_1,n_1=-m_1-l_1=-2l_1$
For $m_2=-2l_2, n_2=-m_2-l_2=2l_2-l_2=l_2$
If $\theta$ be the angle between the lines,
$$\cos\theta=\dfrac{|l_1l_2+m_1m_2+n_1n_2|}{\sqrt{l_1^2+m_1^2+n_1^2}\cdot \sqrt{l_2^2+m_2^2+n_2^2}}$$
Replace the values of
$m_1,n_1$ in terms of $l_1$
$m_2,n_2$ in terms of $l_2$