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JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let L_1: (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and L_2:...

Let

\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

and

\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

L_1

and

L_2

?

Options

A $\left(\frac{14}{3},-3, \frac{22}{3}\right)$
B $\left(-\frac{5}{3},-7,1\right)$
C $\left(2,3, \frac{1}{3}\right)$
D $\left(\frac{8}{3},-1, \frac{1}{3}\right)$

Correct answer

A. $\left(\frac{14}{3},-3, \frac{22}{3}\right)$

Step-by-step solution

\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}

\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

,

\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} \\ & \Rightarrow \frac{2 \lambda-3 \mu-1}{-1}=\frac{3 \lambda-4 \mu-2}{2}=\frac{4 \lambda-5 \mu-2}{-1} \\ & \Rightarrow \lambda=\frac{1}{3} \mu=\frac{-1}{6} \\ & \Rightarrow 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

P Q\langle 1,-2,1\rangle \therefore \quad \text{ Line } \frac{y-\frac{5}{3}}

{1}=

\frac{y-3}{-2}=\frac{y-\frac{13}{3}}${1}

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