JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q14

MathematicsSingle correct+4 / -1Question 14 of 75

The shortest distance between the lines $\dfrac{x-4}1=\dfrac{y-3}2=\dfrac{z-2}{-3}$ and $\dfrac{x+2}2=\dfrac{y-6}4=\dfrac{z-5}{-5}$ is :

\dfrac{5\sqrt6}6

2\sqrt5

3\sqrt5

4\sqrt5

Answer and worked solution

Official answer: C

Setup. $A(4,3,2)$ , $\vec b_1=(1,2,-3)$ ; $B(-2,6,5)$ , $\vec b_2=(2,4,-5)$ .

Cross product. $\vec b_1\times\vec b_2=(2,-1,0)$ , with length $\sqrt5$ .

Distance. $\vec{AB}=(-6,3,3)$ , and $\vec{AB}\cdot(2,-1,0)=-15$ :
$d=\frac{15}{\sqrt5}=3\sqrt5.$

(C)

Topics:Skew lines

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 14