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QJEEM25Ma61d7: JEE Main 2025 Mathematics - Class 12 JEEM25Ma61d7 Solution: Let and be two lines. Then which of the following points lies…

Let L1:x−12=y−23=z−34\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and L2:x−23=y−44=z−55\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5} be two lines. Then which of the following points lies on the line of the shortest distance between L1\mathrm{L}_1 and L2\mathrm{L}_2 ?

Multiple Choice Options

  • A (143,−3,223)\left(\frac{14}{3},-3, \frac{22}{3}\right)
  • B (2,3,13)\left(2,3, \frac{1}{3}\right)
  • C (83,−1,13)\left(\frac{8}{3},-1, \frac{1}{3}\right)
  • D (−53,−7,1)\left(-\frac{5}{3},-7,1\right)

Step-by-Step Verified Solution

amp;L1:x−12=y−23=z−34amp;L2:x−23=y−44=z−55\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\\\ & L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\end{aligned}

JEE Main 2025 (Online) 22nd January Morning Shift Mathematics - 3D Geometry Question 12 English Explanation

amp;P(2λ+1,3λ+2,4λ+3)amp;Q(3μ+2,4μ+4,5μ+5)\begin{aligned} & P(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & Q(3 \mu+2,4 \mu+4,5 \mu+5) \end{aligned}

Dr's of P Q<2 \lambda-3 \mu-1,3 \lambda-4 \mu-2 ,

amp;4λ−5μ−2gt;amp;PQ=∣i^amp;j^amp;k^2amp;3amp;43amp;4amp;5∣=−i^+2j^−k^\begin{aligned} & 4 \lambda-5 \mu-2> \\ & P Q=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|=-\hat{i}+2 \hat{j}-\hat{k} \end{aligned}

amp;⇒2λ−3μ−1−1=3λ−4μ−22=4λ−5μ−2−1amp;⇒λ=13μ=−16amp;⇒P(53,3,133)Q(32,103,256)\begin{aligned} & \Rightarrow \quad \frac{2 \lambda-3 \mu-1}{-1}=\frac{3 \lambda-4 \mu-2}{2}=\frac{4 \lambda-5 \mu-2}{-1} \\ & \Rightarrow \quad \lambda=\frac{1}{3} \mu=\frac{-1}{6} \\ & \Rightarrow \quad P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right) \end{aligned}

Dr's PQ⟨1,−2,1⟩P Q\langle 1,-2,1\rangle

∴\therefore Line

y−531=y−3−2=y−1331\frac{y-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{y-\frac{13}{3}}{1}

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