Consider the line L passing through the points (1,2,3) and (2,3,5) . The distance of the point (11/3, 11/3, 19/3 ) from the line L along the line 3 x-11/2=3 y-11/1=3 z-19/2 is equal to :

Mathematics · 3D Geometry

JEE Main 2024 — 9 April, Shift 2 — Question 2

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 L along the line $\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$ is equal to :

Option A:
3
Correct
Option B:
4
Option C:
5
Option D:
6

Answer: A

Step-by-step solution

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

$\Rightarrow \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-3}{2}=\lambda$

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

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

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

$3 \lambda=2$

$\lambda=\frac{2}{3}$

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

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Mathematics
Chapter: 3D Geometry
Topic: Intersection of lines, line & plane.

The square of the distance of the image of the point (6,1,5) in the line x-1/3=y/2=z-2/4 , from the origin is

Answer: A

Step-by-step solution

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