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

Question

Consider the line $L$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $(\frac{11}{3},\frac{11}{3},\frac{19}{3})$ from the line $L$ along the line $\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}$ 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 ${L}_{2}(\frac{2}{3}\lambda +\frac{11}{3},\frac{\lambda }{3}+\frac{11}{3},\frac{2}{3}\lambda +\frac{19}{3})$

Now

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

$\begin{gathered} L=(\frac{5}{3},\frac{8}{3},\frac{13}{3}) \ d=\sqrt{{(\frac{11}{3}-\frac{5}{3})}^{2}+{(\frac{8}{3}-\frac{11}{3})}^{2}+{(\frac{19}{3}-\frac{13}{3})}^{2}} \ d=\sqrt{4+1+4} \ d=3 \end{gathered}$

Solutions · 1