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Angle between surfaces x²+y²+z²=9 and z=x²+y²−3

Short answer

Normals from gradients: sphere ∇f = (4,−2,4), paraboloid ∇g = (4,−2,−1). cosθ = (16+4−4)/(6√21) = 16/(6√21) = 8/(3√21). The angle is θ = cos⁻¹(8/(3√21)) ≈ 54.4°.

Find the angle between the surfaces x²+y²+z² = 9 and z = x²+y²−3 at the point (2,−1,2).

The angle between two surfaces at a common point is the angle between their normals there. For f(x, y, z) = 0 the normal is ∇f.

Check the point

  • Sphere: 4 + 1 + 4 = 9 ✓.
  • Paraboloid z = x² + y² − 3: 4 + 1 − 3 = 2 = z ✓.

Normals

Surface 1: f = x² + y² + z² − 9, ∇f = (2x, 2y, 2z). At (2, −1, 2): n₁ = (4, −2, 4).

Surface 2: g = x² + y² − z − 3, ∇g = (2x, 2y, −1). At (2, −1, 2): n₂ = (4, −2, −1).

Angle

n₁ · n₂ = 16 + 4 − 4 = 16
|n₁| = √(16 + 4 + 16) = 6
|n₂| = √(16 + 4 + 1) = √21
cos θ = 16 / (6√21) = 8 / (3√21) = 8√21 / 63 ≈ 0.5819
θ ≈ 54.4°

Answer: θ = cos⁻¹(8/(3√21)) ≈ 54.4°.

✓ Answer checked by Claude Opus 5.5 (high effort)

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