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JEE MAIN 2026 · Three Dimensional Geometry

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 15

Updated
Single correct+4 / -1Answer matches NTA’s final key

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

Choose one

Solution

Solution

Foot of the perpendicular. A point of the line is (4+t,9+2t,5+t)(4 + t, 9 + 2t, 5 + t). Matching (1,μ,2)(1, \mu, 2): t=−3t = -3, so μ=3\mu = 3.

The vector from the foot to (λ,2,3)(\lambda, 2, 3) is (λ−1,−1,1)(\lambda - 1, -1, 1), and it is perpendicular to (1,2,1)(1, 2, 1):
(λ−1)−2+1=0⇒λ=2(\lambda - 1) - 2 + 1 = 0 \Rightarrow \lambda = 2.

Distance between the parallel lines. Both have direction d⃗=(2,3,6)\vec d = (2, 3, 6), ∣d⃗∣=7|\vec d| = 7. Points: A(1,2,−4)A(1, 2, -4) and B(2,3,−5)B(2, 3, -5), so AB⃗=(1,1,−1)\vec{AB} = (1, 1, -1).

AB⃗×d⃗=(1⋅6−(−1)⋅3,  −1⋅2−1⋅6,  1⋅3−1⋅2)=(9,−8,1),\vec{AB} \times \vec d = \big(1\cdot 6 - (-1)\cdot 3,\; -1\cdot 2 - 1\cdot 6,\; 1\cdot 3 - 1 \cdot 2\big) = (9, -8, 1),

distance=∣AB⃗×d⃗∣∣d⃗∣=81+64+17=1467.\text{distance} = \frac{|\vec{AB} \times \vec d|}{|\vec d|} = \frac{\sqrt{81 + 64 + 1}}{7} = \frac{\sqrt{146}}{7}.

Answer: 1467\frac{\sqrt{146}}{7}.

Answer checked against the official answer key. The solution is our own working.

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