Consider the line [L] passing through the points (1,2,3) and (2,3,5).

Question

Consider the line $$\mathrm{L}$$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $$\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$$ from the line $$\mathrm{L}$$ along the line $$\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$$ is equal to

Accepted Answer

$$L: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu$$

Measured along $$L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda$$

Any point on $$L_1:(\mu+1, \mu+2,2 \mu+3)$$

Any point on $$L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)$$

Now

$$\begin{aligned}
& \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \
& \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \
& \mu=\frac{2}{3} \
\end{aligned}$$

$$\begin{aligned}
& \text { Point on } L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \
& d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \
& d=\sqrt{4+1+4} \
& d=3
\end{aligned}$$

Solutions · 1