Mathematics - Vector Question with Solution | TestHub

${\mathrm{L}}_1:\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda$ ${\mathrm{L}}_2:\frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu$ dr of line MN will be $<3+\lambda -\frac{\mu }{2},-1-\lambda -\frac{\mu }{2},2+\lambda -\mu >\&$ it will be proportional to $<3,1,2>$
$\Rightarrow \lambda =-\frac{4}{3}\&\mu =-\frac{2}{3}$ $\therefore$ Coordinate of $\mathrm{M}$ will be $<\left(\frac{2}{3},\frac{4}{3},-\frac{1}{3}\right)$ and equation of required line will be. $\frac{\mathrm{x}-\frac{2}{3}}{3}=\frac{\mathrm{y}-\frac{4}{3}}{1}=\frac{\mathrm{z}+\frac{1}{3}}{2}=\mathrm{k}$ So any point on this line will be $\begin{array}{rl}& \left(\frac{2}{3}+3\mathrm{k},\frac{4}{3}+\mathrm{k},-\frac{1}{3}+2\mathrm{k}\right)\\ & \because \frac{2}{3}+3\mathrm{k}=-\frac{1}{3}\Rightarrow \mathrm{k}=-\frac{1}{3}\end{array}$ $\therefore$ Point lie on the line for $\mathrm{k}=-\frac{1}{3}\text{ is }\left(-\frac{1}{3},1,-1\right)$