Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{L}}_1\equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda \\ & \mathrm{P}(\lambda +2,5\lambda +4,\lambda +2)\\ & {\mathrm{L}}_2\equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2}\\ & \mathrm{P}(2\mu +3,3\mu +2,2\mu +3)\\ & \lambda +2=2\mu +33\mu +2=5\lambda +4\\ & \lambda =2\mu +13\mu =5\lambda +2\\ & 3\mu =5(2\mu +1)+2\\ & 3\mu =10\mu +7\\ & \mu =-1\lambda =-1\end{array}$
Both satisfies (P) $\begin{array}{rl}& \mathrm{P}(1,-1,1)\\ & {\mathrm{L}}_3\equiv \frac{\mathrm{x}}{1/4}=\frac{\mathrm{y}}{1/2}=\frac{\mathrm{z}}{1}\\ & {\mathrm{L}}_3=\frac{\mathrm{x}}{\mathrm{l}}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{4}=\mathrm{k}\end{array}$
Coordinates of $\mathrm{Q}(\mathrm{k},2\mathrm{k},4\mathrm{k})$ $\mathrm{DR}$ 's $\mathrm{PQ}=⟨\mathrm{k}-1,2\mathrm{k}+1,4\mathrm{k}-1>$ $\mathrm{PQ}\perp$ to ${\mathrm{L}}_3$ $\begin{array}{rl}& (\mathrm{k}-1)+2(2\mathrm{k}+1)+4(4\mathrm{k}-1)=0\\ & \mathrm{k}-1+4\mathrm{k}+2+16\mathrm{k}-4=0\\ & \mathrm{k}=\frac{1}{7}\\ & \mathrm{Q}\left(\frac{1}{7},\frac{2}{7},\frac{4}{7}\right)\\ & \mathrm{PQ}=\sqrt{{\left(1-\frac{1}{7}\right)}^2+{\left(-1-\frac{2}{7}\right)}^2+{\left(1-\frac{4}{7}\right)}^2}\\ & =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}\\ & PQ=\frac{3\sqrt{14}}{7}\end{array}$