---
title: "Intensity at centroid of a triangle"
subject: "Computer Graphics"
chapter: 8
key: "semester_5/computer_graphics:total-intensity-triangle-centroid"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/total-intensity-triangle-centroid
site: BCT Engineering
---

# Intensity at centroid of a triangle

Computer Graphics · Chapter 8: Illumination and Surface Rendering methods

- Asked in: 2079 Baisakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/total-intensity-triangle-centroid
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-8

## Question

Find out the total intensity at the centroid of a triangle defined by A(2,1,1), B(0,1,1), C(0,0,1), when illuminated by a point light source of intensity IL = 0.6 at (2,2,6) using illumination model. The viewer is at (2,3,6). Assume ambient intensity Ia = 0.1 and parameters: ka = 0.5, kd = 0.8, ks = 0.7, take n = 10. [centroid: (x1 + x2 + x3)/3, (y1 + y2 + y3)/3, (z1 + z2 + z3)/3]

## Short answer

Centroid is (0.6667, 0.6667, 1), triangle lies in z=1 plane with normal (0,0,1). Using light at (2,2,6) and viewer at (2,3,6), Phong model gives ambient 0.05, diffuse 0.4491, specular 0.0069, total 0.506.

## Answer

**Model (Phong):** I = ka·Ia + IL [ kd (N·L) + ks (R·V)^n ]

**Step 1: centroid**
G = ((2 + 0 + 0)/3, (1 + 1 + 0)/3, (1 + 1 + 1)/3) = (0.6667, 0.6667, 1)

**Step 2: normal.** All three vertices have z = 1, so the triangle lies in the plane z = 1. N = (0, 0, 1) (towards the light and the viewer, both at z = 6).

**Step 3: vectors**
- L = light − G = (2 − 0.6667, 2 − 0.6667, 6 − 1) = (1.3333, 1.3333, 5); |L| = 5.3438 → L = (0.2495, 0.2495, 0.9357)
- V = viewer − G = (2 − 0.6667, 3 − 0.6667, 6 − 1) = (1.3333, 2.3333, 5); |V| = 5.6765 → V = (0.2349, 0.4111, 0.8808)

**Step 4: dot products**
- N·L = 0.9357
- R = 2(N·L)N − L = (−0.2495, −0.2495, 0.9357)
- R·V = −0.2495·0.2349 − 0.2495·0.4111 + 0.9357·0.8808 = −0.0586 − 0.1026 + 0.8242 = 0.6630
- (R·V)^10 = 0.6630^10 = 0.0164

**Step 5: components**

| Term | Calculation | Value |
|---|---|---|
| Ambient | ka·Ia = 0.5 × 0.1 | 0.0500 |
| Diffuse | IL·kd·(N·L) = 0.6 × 0.8 × 0.9357 | 0.4491 |
| Specular | IL·ks·(R·V)^n = 0.6 × 0.7 × 0.0164 | 0.0069 |
| **Total** | | **0.5060** |

**Answer: I ≈ 0.506.** (No distance attenuation is used since none is specified. If the halfway-vector form (N·H)^n is used instead, the specular term is about 0.166 and the total about 0.666.)

## Check yourself

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

1. What is the centroid of A(2,1,1), B(0,1,1), C(0,0,1)?
   A. (2, 1, 1)
   B. (0.667, 0.667, 1)
   C. (0.667, 0.333, 1)
   D. (1, 0.667, 0.667)
   Answer: B. Average each coordinate: x = 2/3, y = 2/3, z = 1. (Key point: Centroid)

2. Why is the normal N = (0, 0, 1)?
   A. The light is on the z-axis
   B. The viewer is at z = 6
   C. All three vertices have z = 1, so the triangle lies in the plane z = 1
   D. The centroid has z = 1
   Answer: C. A plane z = const has normal along z; the side toward the light and viewer is +z. (Key point: Normal)

3. How is L found for a point light at (2,2,6)?
   A. Use the light position directly
   B. Subtract the light position from the viewer position
   C. Use (0,0,1) because the light is infinitely far
   D. Subtract the centroid from the light position and normalise
   Answer: D. L = light - G, then normalised, since the source is a finite point. (Key point: Vector calculation)

4. What is the total intensity at the centroid?
   A. About 0.506
   B. About 0.666
   C. About 0.05
   D. About 1.2
   Answer: A. Ambient 0.05 + diffuse 0.449 + specular 0.007 = 0.506. (Key point: Final result)

## Related in Computer Graphics

- [Gouraud intensity at midpoint P](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Agouraud-intensity-midpoint-scan-line) — Computer Graphics, chapter 8
- [Phong illumination model expression](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aphong-illumination-model-derive) — Computer Graphics, chapter 8, asked 8×
- [Ambient, diffuse and specular light](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aambient-diffuse-specular-expressions) — Computer Graphics, chapter 8, asked 2×
- [General illumination model](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Ageneral-illumination-model) — Computer Graphics, chapter 8
- [Define terms; Phong intensity equation](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Adefine-terms-lambert-phong-equation) — Computer Graphics, chapter 8

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