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

Question

Let $$\mathrm{P}$$ 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 $$\mathrm{P}$$ from the line $4 x=2 y=z$ is

Accepted Answer

$$\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \
& L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}
\end{aligned}$$

Point of intersection of $L_1$ and $L_2$ is $(-1,1,-1)$

Distance of point $P$ from $L_3: 4 x=2 y=z$

$$L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}$$

Any point on $L_3$ be

$$\begin{aligned}
& \left(\frac{\lambda}{4}, \frac{\lambda}{2}, \lambda\right) \
& P R:\left\langle\frac{\lambda}{4}+1, \frac{\lambda}{2}-1, \lambda+1\right\rangle \
& \because P R \perp\left\langle\frac{1}{4}, \frac{1}{2}, 1\right\rangle
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow\left(\frac{\lambda}{4}+1\right) \frac{1}{4}+\frac{1}{2}\left(\frac{\lambda}{2}-1\right)+\lambda+1=0 \
& \quad \frac{\lambda}{16}+\frac{1}{4}+\frac{\lambda}{4}-\frac{1}{2}+\lambda+1=0 \
& \Rightarrow \quad \lambda=\frac{-4}{7} \
& \therefore \quad R\left(\frac{-1}{7}, \frac{-2}{7}, \frac{-4}{7}\right) \
& \text { Now } R P: \sqrt{\left(\frac{-1}{7}+1\right)^2+\left(\frac{-2}{7}-1\right)^2+\left(\frac{-4}{7}+1\right)^2} \
& \quad=\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3 \sqrt{14}}{7}
\end{aligned}$$

Solutions · 1