Let θ be the angle between the planes P₁: r⃗ ·(î+ĵ+2 k̂)=9 and P₂: r⃗ ·(2 î-ĵ+k̂)=15. Let L be the line that meets P₂ at the point (4,-2,5) and makes an angle θ with the normal of P₂. If α is the angle between L and P₂, then (tan ² θ)(cot ² α) is equal to ____________.

Solution

Maths sir · 4 steps

A line at angle θ to the normal sits at 90° − θ to the plane, so α is just the complement of θ.

1. Angle between planes. The angle between two planes is the angle between their normals n⃗₁ = (1, 1, 2) and n⃗₂ = (2, -1, 1). Both have length √6.

   cos θ = (n⃗₁ · n⃗₂)/(|n⃗₁| |n⃗₂|) = (2 - 1 + 2)/(√6×√6) = 1/2

2. Find theta. So θ = 60° and tan² θ = 3.
3. Line and plane. The normal is perpendicular to P₂. A line making 60° with the normal therefore makes α = 90° - 60° = 30° with the plane. That gives cot² α = (√3)² = 3. The point (4, -2, 5) only fixes where the line meets P₂. It does not change any angle.
4. So. (tan² θ)(cot² α) = 3 × 3 = 9.

The common slip: 1. Treating the angle with the normal as the angle with the plane gives α = 60°, so cot² α = 1/3 and the product becomes 1, not 9.

The idea to keep. Angle with the normal and angle with the plane always add up to 90°. So cot α = tan θ here, and the product is just tan⁴ θ.

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