JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Three Dimensional Geometry

Let the points on the plane P be equidistant from the points

(- 4, 2, 1)

and

(2, - 2, 3)

. Then the acute angle between the plane P and the plane

2 x + y + 3 z = 1

Answer: (C)

Solution

The plane P is the perpendicular bisector of the segment joining

A = (- 4, 2, 1)

and

B = (2, - 2, 3)

. Its normal vector is

A B = (6, - 4, 2) = 2 (3, - 2, 1)

. The midpoint is

(- 1, 0, 2)

. Plane P:

3 (x + 1) - 2 y + (z - 2) = 0 \Rightarrow 3 x - 2 y + z + 1 = 0

. The given plane is

P^{'} : 2 x + y + 3 z - 1 = 0

. The angle

θ

between them satisfies

cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣} = \frac{∣6 - 2 + 3∣}{14 \cdot 14} = \frac{7}{14} = \frac{1}{2}

. Thus

θ = cos^{- 1} (\frac{1}{2}) = \frac{π}{3}

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option