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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Three Dimensional Geometry

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

Consider the line LL passing through the points (1,2,3)and(2,3,5)(1,2,3) and (2,3,5). The distance of the point (113,113,193)\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right) from the line L\mathrm{L} along the line 3x−112=3y−111=3z−192\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2} is equal to

Options

  1. A6
  2. B5
  3. C4
  4. D3

Correct answer

D. 3

Step-by-step solution

x−12−1=y−23−2=z−35−3⇒x−11=y−21=z−32=λ\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}B(1+λ,2+λ,3+2λ) D.R. of  AB=<3λ−83,3λ−53,6λ−103>B(53,83,133)3λ−83λ−5=21⇒3λ−8=6λ−103λ=2λ=23AB=36+9+363=93=3\begin{aligned} & \mathrm{B}(1+\lambda, 2+\lambda, 3+2 \lambda) \\ & \text{ D.R.\text{ 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}$
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