Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

Let α, β , γ , δ be real numbers such that α² + β² + γ² + δ² ≠ 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 αx + βy + γz = δ. Then which of the following statements is/are TRUE ?

solution : here let R is the midpoint of points P (1, 0, -1) and Q(3, 2, -1)

from midpoint formula, R = [(1 + 3)/2, (0 + 2)/2 , (- 1-1)/2 ] = (2, 1, -1) it lies on the plane of equation, αx + βy + γz = δ

so, 2α + β - γ = δ ......(1)

normal vector to the plane = (3 - 1)i + (2 - 0)j + (-1 + 1)k = 2i + 2j

so, α/2 = β/2 = γ/0 = k (let)

⇒α = 2k, β = 2k and γ = 0

given α + γ = 1 ⇒2k + 0 = 1 , k = 1/2

so, α = 2k = 2 × 1/2 = 1 , β = 2 × 1/2 = 1 and γ = 0

putting α, β, γ in equation (1) we get, δ = 2(1) + 1 - 0 = 3

now let's check all options

α + β = 1 + 1 = 2, option (A) is correct.

δ - γ = 3 - 0 = 3, option (B) is also correct.

δ + β = 3 + 1 = 4, option (C) is also correct.

α + β + γ = 1 + 1 + 0 = 2 ≠ δ , option (D) is incorrect.

Therefore options (A), (B) and (C) are correct choices.

Answer 2

AnswEr:-

Let ,

• R be the midpoint if PQ.

Therefore,

• R(2,1,-1) and it lies on a plane .

Given,

• Equation of plane is $\sf \alpha a +\beta y +\gamma z = \delta$

Therefore,

$\implies \sf \: 2 \alpha + \beta - \gamma = \delta \: ....(1)$

• Normal vector to the plane is ,

$\implies \sf \: \: \bar{n} \: = 2i + 2j$

Therefore,

$\bigstar$  Option A,B,C are correct .