Start free

Mathematics · 3D Geometry

JEE Main 2024 — 31 January, Shift 2 — Question 81

The shortest distance between lines L1L_1 and L2L_2 where x−12=y+1−3=z+42\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+4}{2} and L2L_2 is the line passing through the points A(−4,4,3), B(−1,6,3)A(-4,4,3),\ B(-1,6,3) and perpendicular to the line x−3−2=y3=z−11\dfrac{x-3}{-2}=\dfrac{y}{3}=\dfrac{z-1}{1} is

  1. Option A:

    121221\dfrac{121}{\sqrt{221}}

  2. Option B:

    24117\dfrac{24}{\sqrt{117}}

  3. Option C:

    141221\dfrac{141}{\sqrt{221}}

    Correct
  4. Option D:

    42117 \dfrac{42}{\sqrt{117}}

Answer: C

Step-by-step solution

L2: x+43=y−42=z−30L_2:\ \dfrac{x+4}{3}=\dfrac{y-4}{2}=\dfrac{z-3}{0}

$\text{S.D.}= \dfrac{\left|

x_2-x_1 & y_2-y_1 & z_2-z_1\\ 2 & -3 & 2\\ 3 & 2 & 0 \end{matrix}$$ \right|}{|\vec{n}_1 \times \vec{n}_2|}$ $= \dfrac{\left| $$\begin{matrix} 5 & -5 & -7\\ 2 & -3 & 2\\ 3 & 2 & 0 \end{matrix}$$ \right|}{|\vec{n}_1 \times \vec{n}_2|}$ $= \dfrac{141}{|-4\hat{i}+6\hat{j}+13\hat{k}|}$ $= \dfrac{141}{\sqrt{16+36+169}}$ $= \dfrac{141}{\sqrt{221}}$

Answer key and solution verified before publishing.

Practise 3D Geometry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them

More 3D Geometry questions from this paper

The shortest distance between lines L 1 and L 2 where… | JEE Main 2024 PYQ with Solution · DhiX AI