JEE Main 2020 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 8 January 2020, Shift 1, Maths Q65

Question 65 of 75 in this shift, Maths question 15 of 25, Section A.

The shortest distance between the lines

\dfrac{x-3}{3}=\dfrac{y-8}{-1}=\dfrac{z-3}{1}

and

\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}

is :

(1) $2\sqrt{30}$
(2) $3\sqrt{30}$ Official answer
(3) $3$
(4) $\frac{7}{2}\sqrt{30}$

Official answer

Option 2

NTA final key (Jan 2020).

Chapter: Three Dimensional Geometry
Topic: Skew Lines
Idea tested: Cartesian Shortest Distance

Same idea in other shifts

Asked 5× in all

6 Apr 2026, Shift 2 · Q14The shortest distance between the lines $\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$ and $\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$ is :EasySingle correct
7 Apr 2025, Shift 1 · Q14If the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and…MediumSingle correct
9 Apr 2024, Shift 1 · Q16The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is :EasySingle correct
31 Jan 2024, Shift 2 · Q17The shortest distance, between lines $L_1$ and $L_2$ , where $L_1:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line,…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.