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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let Q(a, b, c) be the image of the point P(3,2,1) in the line...

Let Q(a,b,c)\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c}) be the image of the point P(3,2,1)\mathrm{P}(3,2,1) in the line x−11=y2=z−11\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}. Then the distance of Q from the line x−93=y−92=z−5−2\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} is

Options

  1. A8
  2. B5
  3. C7
  4. D6

Correct answer

C. 7

Step-by-step solution

Line 1 passes through (1, 0, 1) with direction (1,2,1)(1, 2, 1). Point on line: (1+t,2t,1+t)(1+t, 2t, 1+t).
For projection Q of P(3, 2, 1): (t−2,2t−2,t)⋅(1,2,1)=0(t-2, 2t-2, t) \cdot (1, 2, 1) = 0 gives 6t=66t = 6, so t=1t = 1.
Thus Q = (2, 2, 2).
Distance from Q to line 2 (passing through (9, 9, 5) with direction (3,2,−2)(3, 2, -2)):
QR⃗=(7,7,3)\vec{QR} = (7, 7, 3).
QR⃗×d⃗=−20i⃗+23j⃗−7k⃗\vec{QR} \times \vec{d} = -20\vec{i} + 23\vec{j} - 7\vec{k}.
∣QR⃗×d⃗∣=400+529+49=978|\vec{QR} \times \vec{d}| = \sqrt{400 + 529 + 49} = \sqrt{978}.
∣d⃗∣=17|\vec{d}| = \sqrt{17}.
Distance = 97817=57.53≈7.59≈7\frac{\sqrt{978}}{\sqrt{17}} = \sqrt{57.53} \approx 7.59 \approx 7.
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