The angle between the lines, whose direction cosines $l, \mathrm{~m}, \mathrm{n}$ satisfy the equations…
The angle between the lines, whose direction cosines $l, \mathrm{~m}, \mathrm{n}$ satisfy the equations $l+\mathrm{m}+\mathrm{n}=0$ and $2 l^2+2 \mathrm{~m}^2-\mathrm{n}^2=0$, is
$60^{\circ}$
$180^{\circ}$
$90^{\circ}$
$30^{\circ}$
Solution
Substituting $\mathrm{n}=-l-\mathrm{m}$ in
$\begin{aligned}
& 2 l^2+2 \mathrm{~m}^2-\mathrm{n}^2=0, \text { we get } \\
& 2 l^2+2 \mathrm{~m}^2-(-l-\mathrm{m})^2=0 \\
& \Rightarrow l^2+\mathrm{m}^2-2 l \mathrm{~m}=0 \\
& \Rightarrow(l-\mathrm{m})^2=0 \\
& \Rightarrow l=\mathrm{m}
\end{aligned}$
If $l=\mathrm{m}$, then $\mathrm{n}=-2 \mathrm{~m}$
$\Rightarrow \frac{l}{1}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{-2}$
The direction ratios of both the lines are same.
$\begin{array}{ll}
\therefore \quad & \cos \theta= \pm 1 \\
& \Rightarrow \theta=0^{\circ} \text { or } 180^{\circ}
\end{array}$