---
title: "Spatial orientation / normal of a polygon"
subject: "Computer Graphics"
chapter: 6
key: "semester_5/computer_graphics:spatial-orientation-polygon-calculation"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/spatial-orientation-polygon-calculation
site: BCT Engineering
---

# Spatial orientation / normal of a polygon

Computer Graphics · Chapter 6: Surface Modeling

- Asked in: 2078 Kartik, 3 marks; 2071 Shrawan, 4 marks; 2071 Chaitra, 2 marks; 2075 Chaitra, 3 marks; 2075 Asoj, 3 marks (asked 5 times)
- Video: [Normal Vectors and Their Applications in Computer Graphics](https://www.youtube.com/watch?v=wgrKs6ItJUs) — Manifolds in Maryland
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/spatial-orientation-polygon-calculation
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-6

## Question

How do you calculate the spatial orientation of a polygon surface (the surface normal / plane coefficients A, B, C)?

## Short answer

For three non-collinear vertices V1,V2,V3 in CCW order, normal N=(V2−V1)×(V3−V1)=(A,B,C). D=−N·V1. Plane equation Ax+By+Cz+D=0. The outward normal is used for back-face test and shading.

## Answer

The **spatial orientation** of a polygon surface is given by its plane equation and especially its normal vector N = (A, B, C).

Plane equation: **Ax + By + Cz + D = 0**

For three non-collinear vertices (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) taken in counter-clockwise order (seen from the outside):

- A = y1(z2 − z3) + y2(z3 − z1) + y3(z1 − z2)
- B = z1(x2 − x3) + z2(x3 − x1) + z3(x1 − x2)
- C = x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)
- D = −x1(y2z3 − y3z2) − x2(y3z1 − y1z3) − x3(y1z2 − y2z1)

Equivalent method: normal N = (V2 − V1) × (V3 − V1) = (A, B, C), and D = −N · V1.

**Meaning**
- N = (A, B, C) is perpendicular to the surface. The unit normal is N/|N|.
- A point (x, y, z) is inside (behind) the surface if Ax + By + Cz + D < 0, and outside (in front) if > 0.
- Back-face test: if viewing direction V · N > 0, the surface faces away.
- Used in shading (angle between N and light direction).

Example: A(1,0,0), B(0,1,0), C(0,0,1) gives N = (1,1,1), D = −1, plane x + y + z − 1 = 0.

## Check yourself

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

1. The plane equation of a polygon surface is:
   A. Ax + By + C = 0
   B. Ax^2 + By^2 + Cz^2 = D
   C. x + y + z = N
   D. Ax + By + Cz + D = 0 with normal N = (A,B,C)
   Answer: D. (A,B,C) is perpendicular to the surface. (Key point: Plane equation)

2. The normal is computed as:
   A. N = V1 + V2 + V3
   B. N = (V2 - V1) . (V3 - V1)
   C. N = (V2 - V1) x (V3 - V1)
   D. N = V1 x V2 x V3
   Answer: C. Cross product of two edge vectors is perpendicular to both; D = -N.V1. (Key point: Normal via cross product)

3. If Ax+By+Cz+D < 0 for a point, the point is:
   A. Outside (in front)
   B. Inside (behind) the surface
   C. On the surface
   D. Undefined
   Answer: B. Greater than 0 is outside; less than 0 is inside. (Key point: Inside/outside)

4. For A(1,0,0), B(0,1,0), C(0,0,1), D equals:
   A. 1
   B. 0
   C. -3
   D. -1
   Answer: D. N=(1,1,1) and D = -N.V1 = -1. (Key point: Example)

## Related in Computer Graphics

- [Spatial orientation and polygon tables](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aspatial-orientation-surface-significance) — Computer Graphics, chapter 6
- [Polygon table and normal of S1](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Apolygon-table-surface-normal-s1) — Computer Graphics, chapter 6

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- [Plane table principles; orientation by back sighting](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aorientation-back-sighting) — Surveying I, chapter 6, asked 4×
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