Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The given equations are $3l+m+5n=0$ ...(i) and $6\mathrm{mn}-2\mathrm{n}l+5l\mathrm{m}=0$ ...(ii) From (i), we have $m=-3l-5n$. Putting $\mathrm{m}=-3l-5\mathrm{n}$ in (ii), we get $6(-3l-5n)n-2nl+5l(-3l-5n)=0$ $\Rightarrow (n+l)(2n+l)=0$ $\Rightarrow$ either $l=-\mathrm{n}$ or $l=-2\mathrm{n}$. If $l=-n$, then putting $l=-n$ in (i), we obtain $m=-2n$. If $l=-2\mathrm{n}$, then putting $l=-2\mathrm{n}$ in (i), we obtain $\mathrm{m}=\mathrm{n}$. Thus, the direction ratios of two lines are $-\mathrm{n},-$ $2\mathrm{n},\mathrm{n}$ and $-2\mathrm{n},\mathrm{n},\mathrm{n}$ i.e., $1,2,-1$ and $-2,1,1$. Hence, the direction cosines are $\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}$ or $\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}$. The angle $\theta$ between the lines is given by $\begin{array}{rl}& \cos \theta =\frac{1}{\sqrt{6}}\times \frac{-2}{\sqrt{6}}+\frac{2}{\sqrt{6}}\times \frac{1}{\sqrt{6}}+\frac{-1}{\sqrt{6}}\times \frac{1}{\sqrt{6}}=\frac{-1}{6}\\ & \Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{6}\right)\end{array}$