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 2x + y + 3z = 1 is :

Solution

Maths sir · 3 steps

Points equidistant from A and B lie on their perpendicular bisector plane, so AB itself is P's normal.

1. Normal of P. Every point of P is equally far from A(-4, 2, 1) and B(2, -2, 3), so P is the perpendicular bisector plane of AB. Its normal is along AB⃗ = (6, -4, 2), which is the same direction as (3, -2, 1).
2. Angle between normals. The angle between two planes is the angle between their normals. The second plane has normal (2, 1, 3). Both normals have length √14.

   cos θ = (|3 × 2 + (-2) × 1 + 1 × 3|)/(√14 √14) = 7/14 = 1/2

3. So. cos θ = 1/2 gives the acute angle θ = π/3, which is option (c).

The idea to keep. The locus of points equidistant from two points is a plane with normal along the segment joining them. For the angle between planes, use cos of the angle between normals. You do not need the midpoint for this.

Answer from the NTA final answer key. Solution by PrepWiser.