JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 4 April 2025, Shift 2, Maths Q14

Question 14 of 75 in this shift, Maths question 14 of 25, Section A.

Let the values of p, for which the shortest distance between the lines

\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}

and

\vec{r}=\left(p\hat{i}+2\hat{j}+\hat{k}\right)+\lambda\left(2\hat{i}+3\hat{j}+4\hat{k}\right)

\frac{1}{\sqrt{6}}

, be a, b,

(a<b)

. Then the length of the latus rectum of the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

is :

(1) $\frac{3}{2}$
(2) $\frac{2}{3}$ Official answer
(3)18
(4)9

Official answer

Option 2

NTA final key.

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

Same idea in other shifts

Asked 5× in all

4 Apr 2026, Shift 2 · Q13The shortest distance between the lines…EasySingle correct
2 Apr 2025, Shift 2 · Q15The line $L_1$ is parallel to the vector $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ and passes through the point $(7, 6, 2)$ and the line…EasySingle correct
8 Apr 2024, Shift 1 · Q14If the shortest distance between the lines $L_1:\vec{r}=(2+\lambda)\hat{i}+(1-3\lambda)\hat{j}+(3+4\lambda)\hat{k}$ ,…MediumSingle correct
24 Jun 2022, Shift 1 · Q30If the shortest distance between the lines $\vec{r}=(-\hat{i}+3\hat{k})+\lambda(\hat{i}-a\hat{j})$ and…MediumNumerical value

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.