[Math] Find the angle between the two straight lines whose direction cosines are given by $l+m+n=0,l^2+m^2-n^2=0$

Question

Find the angle between the two straight lines whose direction cosines are given by $$l+m+n=0(1),l^2+m^2-n^2=0(2)$$I attempt to solve it in regular process:
$l=-m-n$ and eliminating $l$ from (2) gives me $m=0$ or $m=-n$.
now how could i get other direction cosines to apply the formula$$cos\theta=l_1l_2+m_1m_2+n_1n_2$$Please anyone help me.
Thanks in advance.

Answer 1

Angle between two lines is $\cos\theta=\dfrac{\vec{b_1}\cdot\vec{b_2}}{\vec{|b_1|}\vec{|b_2|}}$

From the question we have $l+m+n=0.....(1)$

From $(1)$ we get $-(l+m)=n$

$l^2+m^2-n^2=0.....(2)$

Now, plug in $n=-(l+m)$ in $(2)$ and we get$$l^2+m^2-l^2-m^2-2ml=0$$Notice that either $l=0$ or $m=0$

Plug in $m=0$ in $(1)$ and we get

If $m=0$ then $l=-n$

Direction ratios $(l,m,n)=(1,0,-1)$

Now plug in $l=0$ in $(1)$ and we get $m=-n$

Direction ratios $(l,m,n)=(0,1,-1)$

$$\cos\theta=\dfrac{\vec{b_1}\cdot\vec{b_2}}{\vec{|b_1|}\vec{|b_2|}}=\dfrac{1}{\sqrt{2}\sqrt{2}}=\dfrac12\implies\theta=\dfrac{\pi}{3}$$