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3D Geometry question

2020 · Shift 2 · Q28
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3D Geometry question

2020 · Shift 2 · Q28

JEE AdvancedMathematics3D GeometryMultiple correct+4 / −2
Let α\alphaα 2 + β\betaβ 2 + γ\gammaγ 2 eee 0 and α\alphaα+γ\gammaγ= 1. Suppose the point (3, 2, −-− 1) is the mirror image of the point (1, 0, −-− 1) with respect to the plane α\alphaα x + β\betaβ y + γ\gammaγ z = δ\deltaδ. Then which of the following statements is/are TRUE?
  1. A
    α\alphaα+β\betaβ = 2
  2. B
    δ−γ\delta -\gammaδ−γ = 3
  3. C
    δ\deltaδ+β\betaβ = 4
  4. D
    α\alphaα+β\betaβ+γ\gammaγ=δ\deltaδ
View written solutionFree

Correct answer: A, B, C

  1. Use the mirror-image property of a plane

If point P′(3,2,−1)P'(3,2,-1)P′(3,2,−1) is the mirror image of P(1,0,−1)P(1,0,-1)P(1,0,−1) with respect to the plane

αx+βy+γz=δ,\alpha x+\beta y+\gamma z=\delta,αx+βy+γz=δ,

then the plane is the perpendicular bisector of the segment joining PPP and P′P'P′. Therefore:

  • the midpoint of PP′PP'PP′ lies on the plane,
  • the normal vector of the plane is parallel to PP′→\overrightarrow{PP'}PP′.
  1. Find the midpoint

Given

P=(1,0,−1),P′=(3,2,−1).P=(1,0,-1),\qquad P'=(3,2,-1).P=(1,0,−1),P′=(3,2,−1).

Midpoint:

M=(1+32,0+22,−1+(−1)2)=(2,1,−1).M=\left(\frac{1+3}{2},\frac{0+2}{2},\frac{-1+(-1)}{2}\right)=(2,1,-1).M=(21+3​,20+2​,2−1+(−1)​)=(2,1,−1).

Since the plane passes through MMM,

2α+β−γ=δ.(1)2\alpha+\beta-\gamma=\delta. \tag{1}2α+β−γ=δ.(1)
  1. Use the normal vector condition

Direction vector of PP′PP'PP′ is

PP′→=(3−1,2−0,−1−(−1))=(2,2,0).\overrightarrow{PP'}=(3-1,2-0,-1-(-1))=(2,2,0).PP′=(3−1,2−0,−1−(−1))=(2,2,0).

The normal vector of the plane is (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ), so it must be parallel to (2,2,0)(2,2,0)(2,2,0). Hence,

(α,β,γ)=λ(2,2,0)(\alpha,\beta,\gamma)=\lambda(2,2,0)(α,β,γ)=λ(2,2,0)

for some nonzero λ\lambdaλ. Thus,

α=2λ,β=2λ,γ=0.\alpha=2\lambda,\quad \beta=2\lambda,\quad \gamma=0.α=2λ,β=2λ,γ=0.

Also given:

α+γ=1.\alpha+\gamma=1.α+γ=1.

So,

2λ+0=1  ⟹  λ=12.2\lambda+0=1\implies \lambda=\frac12.2λ+0=1⟹λ=21​.

Therefore,

α=1,β=1,γ=0.\alpha=1,\quad \beta=1,\quad \gamma=0.α=1,β=1,γ=0.

Now from (1),

δ=2α+β−γ=2(1)+1−0=3.\delta=2\alpha+\beta-\gamma=2(1)+1-0=3.δ=2α+β−γ=2(1)+1−0=3.

So the plane is

x+y=3.x+y=3.x+y=3.
  1. Check each option

Option A: α+β=2\alpha+\beta=2α+β=2

α+β=1+1=2.\alpha+\beta=1+1=2.α+β=1+1=2.

So A is true.

Option B: δ−γ=3\delta-\gamma=3δ−γ=3

δ−γ=3−0=3.\delta-\gamma=3-0=3.δ−γ=3−0=3.

So B is true.

Option C: δ+β=4\delta+\beta=4δ+β=4

δ+β=3+1=4.\delta+\beta=3+1=4.δ+β=3+1=4.

So C is true.

Option D: α+β+γ=δ\alpha+\beta+\gamma=\deltaα+β+γ=δ

α+β+γ=1+1+0=2≠3=δ.\alpha+\beta+\gamma=1+1+0=2 \neq 3=\delta.α+β+γ=1+1+0=2=3=δ.

So D is false.

  1. Final answer

The true statements are:

A, B, C\boxed{A,\ B,\ C}A, B, C​
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