The acute angle between the planes P₁ and P₂, when P₁ and P₂ are the planes passing through the intersection of the planes 5x + 8y + 13z - 29 = 0 and 8x - 7y + z - 20 = 0 and the points (2, 1, 3) and (0, 1, 2), respectively, is :

Solution

Maths sir · 4 steps

Write the family of planes through the line, let each point fix its own λ, then compare the two normals.

1. Family of planes. Every plane through the line of intersection has the form (5x + 8y + 13z - 29) + λ(8x - 7y + z - 20) = 0. Each given point fixes one value of λ.
2. Plane P1. At (2, 1, 3) the first bracket is 28 and the second is -8, so 28 - 8λ = 0 and λ = 7/2. This gives 66x - 33y + 33z - 198 = 0, that is, 2x - y + z - 6 = 0, with normal (2, -1, 1).
3. Plane P2. At (0, 1, 2) the first bracket is 5 and the second is -25, so 5 - 25λ = 0 and λ = 1/5. This gives 33x + 33y + 66z - 165 = 0, that is, x + y + 2z - 5 = 0, with normal (1, 1, 2).

   cos θ = (|(2)(1) + (-1)(1) + (1)(2)|)/(√6 √6) = 3/6 = 1/2

4. So. cos θ = 1/2, so the acute angle between P₁ and P₂ is θ = π/3, option (a).

The idea to keep. Plane through the intersection of two planes: write P + λ Q = 0 and let the given point fix λ. The angle between planes is the angle between their normals, with a modulus for the acute one.

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