---
title: "Angle between surface normals"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:angle-between-surface-normals"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/angle-between-surface-normals
site: BCT Engineering
---

# Angle between surface normals

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2067 Asar
- Video: [Find the Angle Between Normals | xy = z² at (4,1,2) & (3,3,-3) | VTU Maths 2026 ](https://www.youtube.com/watch?v=0znZ3_wu5k4) — SK-Academy Of Mathematics 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/angle-between-surface-normals
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Find the angle between the normals to the surface xy = z² at the points (1,4,2) and (−3,−3,3).

## Short answer

Normals are gradients: n₁ = 4i + j − 4k, n₂ = −3i − 3j − 6k. Dot product is 9, magnitudes √33 and √54. Angle = cos⁻¹(9/√1782) ≈ 77.7°.

## Answer

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

```text
∇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.)

```text
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°.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The normal to xy = z^2 is found as:
   A. The curl of phi
   B. The divergence of phi
   C. The tangent vector to the surface
   D. The gradient of phi = xy - z^2
   Answer: D. For phi = 0 the normal is grad phi = y i + x j - 2z k. (Key point: Normal)

2. What is the normal at (1,4,2)?
   A. 4i + j - 4k
   B. i + 4j - 2k
   C. 4i + j + 4k
   D. -4i + j - 4k
   Answer: A. Substitute x=1, y=4, z=2 into y i + x j - 2z k. (Key point: Normal at P)

3. The angle is found from cos(theta) =
   A. n1 x n2 / (|n1||n2|)
   B. n1.n2 / (|n1||n2|) = 9/sqrt(1782)
   C. (n1.n2)/(|n1|+|n2|)
   D. |n1||n2| / n1.n2
   Answer: B. Dot product over product of magnitudes gives theta ≈ 77.7°. (Key point: Angle formula)

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