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

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 : 

(1) 3

(2) 5

(3) 4

(4) 6

Answer 1

Correct option is : (1)  3 

\(\frac{x-1}{2-1}=\frac{y-2}{3-2}=\frac{z-3}{5-3}\)

\(\Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda\) 

\(\mathrm{B}(1+\lambda, 2+\lambda, 3+2 \lambda)\)

D.R. of \(\mathrm{AB}=<\frac{3 \lambda-8}{3}, \frac{3 \lambda-5}{3}, \frac{6 \lambda-10}{3}>\)

В \(\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \frac{3 \lambda-8}{3 \lambda-5}=\frac{2}{1} \Rightarrow 3 \lambda-8=6 \lambda-10\)

\(3 \lambda=2\)

\(\lambda=\frac{2}{3}\)

\(\mathrm{AB}=\frac{\sqrt{36+9+36}}{3}=\frac{9}{3}=3\)