Find the angle between the two straight lines whose direction cosines are given by $l+m+n=0,l^2+m^2-n^2=0$ 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.
 A: You must add a third equation, from the definition of direction cosines:
$$
l^2+m^2+n^2=1,
$$
to get all possible solutions:
$$
l=0,\quad m=\mp1/\sqrt2,\quad n=\pm1/\sqrt2
\quad\text{or}\quad
l=\mp1/\sqrt2,\quad m=0,\quad n=\pm1/\sqrt2.
$$
These represent FOUR directions but only TWO lines, because directions in each pair are opposite. Hence to compute the angle you must take one triple from the first solution and one triple from the second one, to get:
$$
\cos\theta=\pm{1\over2}.
$$
A: 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 $(l,m,n)=(0,1,-1)$
$$\cos\theta=\dfrac{\vec{b_1}\cdot\vec{b_2}}{\vec{|b_1|}\vec{|b_2|}}=\dfrac{1}{\sqrt{2}\sqrt{2}}=\dfrac12\implies\theta=\dfrac{\pi}{3}$$
