JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Three Dimensional Geometry

Let P be the plane passing through the intersection of the planes

r \cdot (\hat{i} + 3 \hat{j} - \hat{k}) = 5

and

r \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 3

, and the point

(2, 1, - 2)

. Let the position vectors of the points

X

and

Y

\hat{i} - 2 \hat{j} + 4 \hat{k}

and

5 \hat{i} - \hat{j} + 2 \hat{k}

respectively. Then the points

Answer: (C)

Solution

The family of planes through the intersection is

(x + 3 y - z - 5) + λ (2 x - y + z - 3) = 0

.

Substituting

(2, 1, - 2)

(2 + 3 + 2 - 5) + λ (4 - 1 - 2 - 3) = 0 \Rightarrow 2 - 2 λ = 0 \Rightarrow λ = 1

.

The plane P is:

(x + 3 y - z - 5) + (2 x - y + z - 3) = 0 \Rightarrow 3 x + 2 y - 8 = 0

.

For point X

(1, - 2, 4)

3 (1) + 2 (- 2) - 8 = 3 - 4 - 8 = - 9 < 0

.

For point Y

(5, - 1, 2)

3 (5) + 2 (- 1) - 8 = 15 - 2 - 8 = 5 > 0

.

Since the values have opposite signs, X and Y are on opposite sides of P.

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option