Angle between surface normals

The normal to a surface φ(x,y,z) = 0 at a point is ∇φ. Take φ = xy − z².

∇φ = y i + x j − 2z k

At P(1,4,2): n₁ = 4 i + 1 j − 4 k. At Q(−3,−3,3): n₂ = −3 i − 3 j − 6 k.

(Both points lie on the surface: 1·4 = 2² and (−3)(−3) = 3².)

n₁·n₂ = (4)(−3) + (1)(−3) + (−4)(−6) = −12 − 3 + 24 = 9
|n₁| = √(16 + 1 + 16) = √33
|n₂| = √(9 + 9 + 36) = √54
cos θ = 9/(√33 · √54) = 9/√1782 = 9/42.21 = 0.2132
θ = cos⁻¹(0.2132) ≈ 77.69°

Answer: θ = cos⁻¹(9/√1782) ≈ 77.7°.