Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Substituting $\mathrm{n}=-l-\mathrm{m}$ in $$\begin{array}{rl}& 2l^2+2{\mathrm{m}}^2-{\mathrm{n}}^2=0,\text{ we get }\\ & 2l^2+2{\mathrm{m}}^2-(-l-\mathrm{m}{)}^2=0\\ & \Rightarrow l^2+{\mathrm{m}}^2-2l\mathrm{m}=0\\ & \Rightarrow (l-\mathrm{m}{)}^2=0\\ & \Rightarrow l=\mathrm{m}\end{array}$$ 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. $$\begin{array}{ll}\therefore & \cos \theta =\pm 1\\ & \Rightarrow \theta =0^{\circ }\text{ or }180^{\circ }\end{array}$$