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JEE Main 2016 · Mathematics · Three Dimensional Geometry · hard

The image of the point P(1,2,3) in the plane 2x+y-2z=7 is Q(α,β,γ). Then α+β+γ equals

Single correct option

The image of the point P(1,2,3)P(1,2,3) in the plane 2x+y−2z=72x+y-2z=7 is Q(α,β,γ)Q(\alpha,\beta,\gamma). Then α+β+γ\alpha+\beta+\gamma equals
  1. A. 77
  2. B. 88
  3. C. 66
  4. D. 99
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Answer: (B) 88

The normal is n⃗=(2,1,−2)\vec n=(2,1,-2) with ∣n⃗∣2=9|\vec n|^2=9. For PP: 2(1)+2−2(3)−7=−92(1)+2-2(3)-7=-9. The image is Q=P−2⋅(−9)9n⃗=P+2n⃗=(1+4, 2+2, 3−4)=(5,4,−1)Q=P-2\cdot\dfrac{(-9)}{9}\vec n=P+2\vec n=(1+4,\,2+2,\,3-4)=(5,4,-1). Check: the midpoint (3,3,1)(3,3,1) satisfies 6+3−2−7=06+3-2-7=0, so it lies on the plane, and PQPQ is parallel to n⃗\vec n. Hence α+β+γ=5+4−1=8\alpha+\beta+\gamma=5+4-1=8. The value 77 is the sum for the foot of the perpendicular (3,3,1)(3,3,1), not for the image.
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