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JEE Advanced 2025 Paper 1 · Mathematics · Three Dimensional Geometry · hard

The line L: x/3=y/2=z/1 meets the plane Pi: x+2y+2z=9 at a point Q. The line L' is the mirror image (reflection) of L in the plane Pi. The point at…

Single correct option

The line L: x3=y2=z1L:\ \dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{1} meets the plane Π: x+2y+2z=9\Pi:\ x+2y+2z=9 at a point QQ. The line L′L' is the mirror image (reflection) of LL in the plane Π\Pi. The point at which L′L' meets the xyxy-plane is
  1. A. (103,43,0)\left(\dfrac{10}{3},\dfrac43,0\right)
  2. B. (83,103,0)\left(\dfrac{8}{3},\dfrac{10}{3},0\right)
  3. C. (113,23,0)\left(\dfrac{11}{3},\dfrac{2}{3},0\right)
  4. D. (73,53,0)\left(\dfrac{7}{3},\dfrac{5}{3},0\right)
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Answer: (A) (103,43,0)\left(\dfrac{10}{3},\dfrac43,0\right)

Put (3t,2t,t)(3t,2t,t) in Π\Pi: 3t+4t+2t=93t+4t+2t=9, so t=1t=1 and Q=(3,2,1)Q=(3,2,1). The reflected line passes through QQ. Its direction is the reflection of d⃗=(3,2,1)\vec d=(3,2,1) in the plane with normal n⃗=(1,2,2)\vec n=(1,2,2): d⃗′=d⃗−2(d⃗⋅n⃗)∣n⃗∣2n⃗\vec d'=\vec d-\dfrac{2(\vec d\cdot\vec n)}{|\vec n|^2}\vec n. Here d⃗⋅n⃗=3+4+2=9\vec d\cdot\vec n=3+4+2=9 and ∣n⃗∣2=9|\vec n|^2=9, so d⃗′=(3,2,1)−2(1,2,2)=(1,−2,−3)\vec d'=(3,2,1)-2(1,2,2)=(1,-2,-3). As a check, the image of the origin (a point of LL) is 2n⃗=(2,4,4)2\vec n=(2,4,4) and (2,4,4)−Q=(−1,2,3)(2,4,4)-Q=(-1,2,3), parallel to d⃗′\vec d'. So L′: (3+s, 2−2s, 1−3s)L':\ (3+s,\ 2-2s,\ 1-3s). Setting z=0z=0 gives s=13s=\tfrac13, hence the point (103,43,0)\left(\tfrac{10}{3},\tfrac{4}{3},0\right). The other options come from using d⃗\vec d itself or a sign slip in d⃗′\vec d'.
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Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake