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

Consider the line $L$ 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. 6
  2. 5
  3. 4
  4. 3

Solution

$\begin{aligned} & \frac{x-1}{2-1}=\frac{y-2}{3-2}=\frac{z-3}{5-3} \\ & \Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda\end{aligned}$

$\begin{aligned} & \mathrm{B}(1+\lambda, 2+\lambda, 3+2 \lambda) \\ & \text { D.R. of } \mathrm{AB}= < \frac{3 \lambda-8}{3}, \frac{3 \lambda-5}{3}, \frac{6 \lambda-10}{3}> \\ & \mathrm{B}\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \frac{3 \lambda-8}{3 \lambda-5}=\frac{2}{1} \Rightarrow 3 \lambda-8=6 \lambda-10 \\ & 3 \lambda=2 \\ & \lambda=\frac{2}{3} \\ & \mathrm{AB}=\frac{\sqrt{36+9+36}}{3}=\frac{9}{3}=3\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 2)

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