JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2020 • Paper 2

Reflection of a Point in Plane Geometry

Let α,β,γ,δ\alpha, \beta, \gamma, \delta be real numbers satisfying α2+β2+γ2≠0\alpha^2 + \beta^2 + \gamma^2 \neq 0 along with α+γ=1\alpha + \gamma = 1. Assume that the point (3,2,−1)(3, 2, -1) is the mirror reflection of (1,0,−1)(1, 0, -1) relative to the plane αx+βy+γz=δ\alpha x + \beta y + \gamma z = \delta. Which of the following assertions is/are TRUE?

Options

A

α+β=2\alpha + \beta = 2

Correct
B

δ−γ=3\delta - \gamma = 3

Correct
C

δ+β=4\delta + \beta = 4

Correct
D

α+β+γ=δ\alpha + \beta + \gamma = \delta

Topics & Concepts

Three Dimensional GeometryPlane in 3D

Step-by-Step Solution

To determine which of the given options are correct, we analyze the properties of a mirror reflection of a point relative to a plane in three-dimensional space.

Let the given points be A(1,0,−1)A(1, 0, -1) and its mirror image B(3,2,−1)B(3, 2, -1) with respect to the plane: P:αx+βy+γz=δP: \alpha x + \beta y + \gamma z = \delta

Step 1: Relation between the Normal Vector and the Line Joining the Points

The segment joining AA and BB must be perpendicular to the plane PP. Thus, the direction vector of the line segment ABAB, denoted as AB⃗\vec{AB}, is parallel to the normal vector to the plane, n⃗=αi^+βj^+γk^\vec{n} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}.

Calculating AB⃗\vec{AB}: AB⃗=(3−1)i^+(2−0)j^+(−1−(−1))k^=2i^+2j^+0k^\vec{AB} = (3 - 1)\hat{i} + (2 - 0)\hat{j} + (-1 - (-1))\hat{k} = 2\hat{i} + 2\hat{j} + 0\hat{k}

Since n⃗\vec{n} is parallel to AB⃗\vec{AB}, there exists a non-zero scalar kk such that: (α,β,γ)=k(2,2,0)(\alpha, \beta, \gamma) = k(2, 2, 0)

From this, we get: γ=0\gamma = 0 α=2kandβ=2k  ⟹  α=β\alpha = 2k \quad \text{and} \quad \beta = 2k \implies \alpha = \beta

Step 2: Determining α,β,\alpha, \beta, and γ\gamma

We are given that α+γ=1\alpha + \gamma = 1. Substituting γ=0\gamma = 0: α+0=1  ⟹  α=1\alpha + 0 = 1 \implies \alpha = 1

Since α=β\alpha = \beta: β=1\beta = 1

Thus, we have: α=1,β=1,γ=0\alpha = 1, \quad \beta = 1, \quad \gamma = 0

Step 3: Determining δ\delta

The midpoint MM of the segment ABAB must lie on the plane PP. The coordinates of MM are: 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)

Substituting M(2,1,−1)M(2, 1, -1) into the equation of the plane αx+βy+γz=δ\alpha x + \beta y + \gamma z = \delta: 1(2)+1(1)+0(−1)=δ1(2) + 1(1) + 0(-1) = \delta 2+1+0=δ  ⟹  δ=32 + 1 + 0 = \delta \implies \delta = 3

Step 4: Verification of Options

Using the values α=1\alpha = 1, β=1\beta = 1, γ=0\gamma = 0, and δ=3\delta = 3:

  • Option (A): α+β=1+1=2\alpha + \beta = 1 + 1 = 2
    (True)

  • Option (B): δ−γ=3−0=3\delta - \gamma = 3 - 0 = 3
    (True)

  • Option (C): δ+β=3+1=4\delta + \beta = 3 + 1 = 4
    (True)

  • Option (D): α+β+γ=1+1+0=2≠δ\alpha + \beta + \gamma = 1 + 1 + 0 = 2 \neq \delta (since δ=3\delta = 3)
    (False)

Conclusion

The correct assertions are A, B, and C.

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