Angle between normals to surfaces

The normal to a surface f(x, y, z) = 0 at a point is along ∇f. The angle between two surfaces at a point is the angle between their normals there, found from cos θ = (n₁ · n₂)/(|n₁||n₂|).

Surface 1: f = x log z − y² + 1 = 0

∇f = (log z, −2y, x/z). At (1, 1, 1): log 1 = 0, so n₁ = (0, −2, 1). (The point lies on the surface: 0 − 1 + 1 = 0 ✓.)

Surface 2: g = x²y + z − 2 = 0

∇g = (2xy, x², 1). At (1, 1, 1): n₂ = (2, 1, 1). (1 + 1 − 2 = 0 ✓.)

Angle

n₁ · n₂ = 0·2 + (−2)(1) + (1)(1) = −1
|n₁| = √5,   |n₂| = √6
cos θ = −1/√30

θ = cos⁻¹(−1/√30) ≈ 100.5° between these two normal directions. The acute angle between the surfaces (taking the normals in the other sense) is cos⁻¹(1/√30) ≈ 79.5°.

Answer: the angle between the normals is cos⁻¹(1/√30) ≈ 79.5° (acute angle; the obtuse value from the signs of the chosen normals is cos⁻¹(−1/√30) ≈ 100.5°).