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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 15

Let the foot of perpendicular from the point (λ,2,3)(\lambda,2,3) on the line x−41=y−92=z−51\dfrac{x-4}1=\dfrac{y-9}2=\dfrac{z-5}1 be the point (1,μ,2)(1,\mu,2). Then the distance between the lines x−12=y−23=z+46\dfrac{x-1}2=\dfrac{y-2}3=\dfrac{z+4}6 and x−λ2=y−μ3=z+56\dfrac{x-\lambda}2=\dfrac{y-\mu}3=\dfrac{z+5}6 is equal to :

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Answer and worked solution

Official answer: C

Foot. (1,μ,2)(1,\mu,2) on the line: z−5=−3z-5=-3 gives parameter −3-3, so μ=3\mu=3.

Perpendicular. (λ−1,−1,1)⋅(1,2,1)=0(\lambda-1,-1,1)\cdot(1,2,1)=0 gives λ=2\lambda=2.

Parallel lines. Through (1,2,−4)(1,2,-4) and (2,3,−5)(2,3,-5) with direction (2,3,6)(2,3,6) (length 7). The difference is (1,1,−1)(1,1,-1) and
(1,1,−1)×(2,3,6)=(9,−8,1),d=1467.(1,1,-1)\times(2,3,6)=(9,-8,1),\qquad d=\frac{\sqrt{146}}7.

(C)

Topics:Angles and distancesEquation of a line in space
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