---
title: "Back-face test with given normals"
subject: "Computer Graphics"
chapter: 7
key: "semester_5/computer_graphics:back-face-normals-frontface"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/back-face-normals-frontface
site: BCT Engineering
---

# Back-face test with given normals

Computer Graphics · Chapter 7: Visible Surface Determination

- Asked in: 2075 Chaitra, 5 marks
- Video: [Backface Culling Algorithm](https://www.youtube.com/watch?v=wpNef1Nu4pA) — Jamie King
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/back-face-normals-frontface
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-7

## Question

Determine whether two surfaces of an object with normals 2i - 3j + 4k and i + j - 2k respectively, viewed from a direction given by i - j + k, are backface or frontface.

## Short answer

For viewing direction V=(1,-1,1), V·N1=(2,-3,4) is 9 >0, so surface 1 is a back face. V·N2=(1,1,-2) is -2 <0, so surface 2 is a front face.

## Answer

**Test:** with the viewing direction V (from the eye towards the object), a surface with normal N is a back face if **V · N > 0**, and a front face if V · N < 0 (V · N = 0 means the surface is seen edge-on).

V = i − j + k = (1, −1, 1)

**Surface 1:** N1 = 2i − 3j + 4k = (2, −3, 4)

V · N1 = 1·2 + (−1)(−3) + 1·4 = 2 + 3 + 4 = **9 > 0** → **back face** (hidden).

**Surface 2:** N2 = i + j − 2k = (1, 1, −2)

V · N2 = 1·1 + (−1)(1) + 1·(−2) = 1 − 1 − 2 = **−2 < 0** → **front face** (visible).

**Answer:** surface 1 is a back face; surface 2 is a front face.

(Note: this uses V as the direction in which the viewer looks, from the eye to the object. If V were taken as the vector pointing towards the viewer, the condition would be reversed and the two results would swap.)

## Check yourself

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

1. Here V is the direction the viewer looks (eye to object). When is a surface a back face?
   A. V . N > 0
   B. V . N < 0
   C. V . N = 0
   D. V . N < -1
   Answer: A. The viewing direction and normal pointing the same way means the surface faces away. (Key point: Convention)

2. V = (1,-1,1), N1 = (2,-3,4). What is V . N1?
   A. -9
   B. 9
   C. 3
   D. 5
   Answer: B. 2 + 3 + 4 = 9. (Key point: Surface 1 dot product)

3. V . N2 for N2 = (1,1,-2) is -2. What is the conclusion?
   A. Back face (hidden)
   B. Edge-on
   C. Front face (visible)
   D. Cannot be decided
   Answer: C. V . N < 0 means it faces the viewer. (Key point: Surface 2 result)

4. If V were taken as the vector pointing towards the viewer instead, what would change?
   A. Nothing changes
   B. Both become back faces
   C. Only surface 1 changes to edge-on
   D. The condition reverses and the two results swap
   Answer: D. Reversing V flips the sign of every dot product. (Key point: Convention dependence)

## Related in Computer Graphics

- [Back-face detection algorithm](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aback-face-detection-method) — Computer Graphics, chapter 7, asked 8×
- [Limitations of back-face detection](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aback-face-limitations-overcome) — Computer Graphics, chapter 7, asked 2×
- [Back-face visibility of BED and ABCD](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aback-face-bed-abcd-observer) — Computer Graphics, chapter 7

## Related in other subjects

- [Reciprocal system of 2i−3j+k, i+2j−k, 3i−j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k) — Mathematics, chapter 5, asked 4×
- [Reciprocal system of 2i+3j−k, i−j−2k, −i+2j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k-2) — Mathematics, chapter 5, asked 2×
- [Angle between normals to surfaces](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aangle-between-normals-surfaces) — Mathematics, chapter 5, asked 2×
- [Interface two 4K x 8 ROM with 3:8 decoder](https://www.bctengineer.com/question.md?key=semester_4%2Fmicroprocessor%3Adecoder-two-4k-rom-38-decoder) — Microprocessor, chapter 4