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3D Geometry question

2025 · 28 Jan · Shift 1 · Q38
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  5. /2025 · 28 Jan · Shift 1 · Q38

3D Geometry question

2025 · 28 Jan · Shift 1 · Q38

JEE MainMathematics3D GeometryMCQ+4 / −1
If the image of the point (4,4,3)(4,4,3)(4,4,3) in the line x−12=y−21=z−13\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}2x−1​=1y−2​=3z−1​ is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to
  1. A
    12
  2. B
    9
  3. C
    7
  4. D
    8
View written solutionFree

Correct answer: B

  1. Interpret the problem

We need the image of the point P(4,4,3)P(4,4,3)P(4,4,3) in the line

x−12=y−21=z−13.\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}.2x−1​=1y−2​=3z−1​.

Here, “image in the line” means the reflection of the point across the given line in 3D.

  1. Write the line in parametric form

Let

x−12=y−21=z−13=t.\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}=t.2x−1​=1y−2​=3z−1​=t.

Then the line is

x=1+2t,y=2+t,z=1+3t.x=1+2t,\quad y=2+t,\quad z=1+3t.x=1+2t,y=2+t,z=1+3t.

So a point on the line is

A(1,2,1)A(1,2,1)A(1,2,1)

and its direction vector is

d⃗=(2,1,3).\vec d=(2,1,3).d=(2,1,3).
  1. Find the foot of the perpendicular from PPP to the line

Let the foot of the perpendicular be F=A+td⃗F=A+t\vec dF=A+td. Then

F=(1+2t, 2+t, 1+3t).F=(1+2t,\,2+t,\,1+3t).F=(1+2t,2+t,1+3t).

Since PF⊥d⃗PF \perp \vec dPF⊥d, we use

(PF⃗)⋅d⃗=0.(\vec{PF})\cdot \vec d=0.(PF)⋅d=0.

Now,

PF⃗=F−P=(1+2t−4, 2+t−4, 1+3t−3)=(−3+2t, −2+t, −2+3t).\vec{PF}=F-P=(1+2t-4,\,2+t-4,\,1+3t-3)=(-3+2t,\,-2+t,\,-2+3t).PF=F−P=(1+2t−4,2+t−4,1+3t−3)=(−3+2t,−2+t,−2+3t).

Thus,

(−3+2t, −2+t, −2+3t)⋅(2,1,3)=0.(-3+2t,\,-2+t,\,-2+3t)\cdot(2,1,3)=0.(−3+2t,−2+t,−2+3t)⋅(2,1,3)=0.

Compute:

2(−3+2t)+(−2+t)+3(−2+3t)=0.2(-3+2t)+(-2+t)+3(-2+3t)=0.2(−3+2t)+(−2+t)+3(−2+3t)=0. −6+4t−2+t−6+9t=0.-6+4t-2+t-6+9t=0.−6+4t−2+t−6+9t=0. 14t−14=0.14t-14=0.14t−14=0. t=1.t=1.t=1.

Therefore,

F=(1+2, 2+1, 1+3)=(3,3,4).F=(1+2,\,2+1,\,1+3)=(3,3,4).F=(1+2,2+1,1+3)=(3,3,4).
  1. Use midpoint property for reflection

If P′(α,β,γ)P'(\alpha,\beta,\gamma)P′(α,β,γ) is the reflection of P(4,4,3)P(4,4,3)P(4,4,3) in the line, then the foot FFF is the midpoint of PP′PP'PP′.

So,

P′=2F−P.P'=2F-P.P′=2F−P.

Hence,

P′=2(3,3,4)−(4,4,3)=(6,6,8)−(4,4,3)=(2,2,5).P'=2(3,3,4)-(4,4,3)=(6,6,8)-(4,4,3)=(2,2,5).P′=2(3,3,4)−(4,4,3)=(6,6,8)−(4,4,3)=(2,2,5).

Thus,

α=2,β=2,γ=5.\alpha=2,\quad \beta=2,\quad \gamma=5.α=2,β=2,γ=5.
  1. Find the required sum
α+β+γ=2+2+5=9.\alpha+\beta+\gamma=2+2+5=9.α+β+γ=2+2+5=9.
  1. Check options

The correct option is:

9\boxed{9}9​

which is Option B.

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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