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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Three Dimensional Geometry

The line L_1 is parallel to the vector a =-3 i +2 j +4 k and passes...

The line L1L_1 is parallel to the vector a⃗=−3i^+2j^+4k^\vec{a}=-3 \hat{i}+2 \hat{j}+4 \hat{k} and passes through the point (7,6,2)(7,6,2) and the line L2L_2 is parallel to the vector b⃗=2i^+j^+3k^\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k} and passes through the point (5,3,4)(5,3,4). The shortest distance between the lines L1L_1 and L2L_2 is :

Options

  1. A2338\frac{23}{\sqrt{38}}
  2. B2157\frac{21}{\sqrt{57}}
  3. C2357\frac{23}{\sqrt{57}}
  4. D2138\frac{21}{\sqrt{38}}

Correct answer

A. 2338\frac{23}{\sqrt{38}}

Step-by-step solution

L1:(7i^+6j^+2k)+λ(−3i^+2j^+4k)L2:(5i^+3j^+4k)+λ(2i^+j^+3k)\begin{aligned} & L_1:(7 \hat{i}+6 \hat{j}+2 k)+\lambda(-3 \hat{i}+2 \hat{j}+4 k) \\ & L_2:(5 \hat{i}+3 \hat{j}+4 k)+\lambda(2 \hat{i}+\hat{j}+3 k)\end{aligned}
Distance between skew lines
=(2i^+3j^−2k^)⋅(2i^+17j^−7k^)342=69342=69338=2338\begin{aligned} & =\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+17 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})}{\sqrt{342}} \\ & =\frac{69}{\sqrt{342}}=\frac{69}{3 \sqrt{38}}=\frac{23}{\sqrt{38}}\end{aligned}
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