Let P be the point of intersection of the lines (x-2)/(1)=(y-4)/(5)=(z-2)/(1) and (x-3)/(2…

Question

Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of $P$ from the line $4x=2y=z$ is

Accepted Answer

$\begin{gathered} {L}_{1}:\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \ {L}_{2}:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2} \end{gathered}$

Point of intersection of ${L}_{1}$ and ${L}_{2}$ is $(-1,1,-1)$

Distance of point $P$ from ${L}_{3}:4x=2y=z$

${L}_{3}:\frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}$

Any point on ${L}_{3}$ be

$\begin{gathered} (\frac{\lambda }{4},\frac{\lambda }{2},\lambda ) \ PR:\langle \frac{\lambda }{4}+1,\frac{\lambda }{2}-1,\lambda +1angle \ \because PR\perp \langle \frac{1}{4},\frac{1}{2},1angle \end{gathered}$

$\begin{gathered} \Rightarrow (\frac{\lambda }{4}+1)\frac{1}{4}+\frac{1}{2}(\frac{\lambda }{2}-1)+\lambda +1=0 \ \, \frac{\lambda }{16}+\frac{1}{4}+\frac{\lambda }{4}-\frac{1}{2}+\lambda +1=0 \ \Rightarrow \, \lambda =\frac{-4}{7} \ 	herefore \, R(\frac{-1}{7},\frac{-2}{7},\frac{-4}{7}) \ RP:\sqrt{{(\frac{-1}{7}+1)}^{2}+{(\frac{-2}{7}-1)}^{2}+{(\frac{-4}{7}+1)}^{2}} \ \, =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3\sqrt{14}}{7} \end{gathered}$

Solutions · 1