JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Three Dimensional Geometry

Let

α, β, γ, δ

be real numbers such that

α^{2} + β^{2} + γ^{2} \neq = 0

and

α + γ = 1

. Suppose the point

(3, 2, - 1)

is the mirror image of the point

(1, 0, - 1)

with respect to the plane

α x + β y + γ z = δ

. Then which of the following statements is/are TRUE?

Answer: (A), (B), (C)

Solution

The midpoint

R

P (1, 0, - 1)

and

Q (3, 2, - 1)

(2, 1, - 1)

, which lies on the plane:

2 α + β - γ = δ

. The normal vector to the plane is

P Q = 2 \hat{i} + 2 \hat{j}

, so

\frac{α}{2} = \frac{β}{2} = \frac{γ}{0}

, giving

α = β

and

γ = 0

. From

α + γ = 1

, we get

α = 1

β = 1

γ = 0

. Substituting into the plane equation:

2 (1) + 1 - 0 = δ

, so

δ = 3

. Checking the options: (A)

α + β = 2

✓, (B)

δ - γ = 3

✓, (C)

δ + β = 4

✓, (D)

α + β + γ = 2 \neq = 3 = δ

✗.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options