---
title: "Derive parallel projection matrix"
subject: "Computer Graphics"
chapter: 4
key: "semester_5/computer_graphics:parallel-projection-matrix-derive"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/parallel-projection-matrix-derive
site: BCT Engineering
---

# Derive parallel projection matrix

Computer Graphics · Chapter 4: Three-Dimensional Graphics

- Asked in: 2073 Shrawan, 6 marks; 2079 Baisakh, 4 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/parallel-projection-matrix-derive
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-4

## Question

Derive the transformation matrix for parallel projection.

## Short answer

For projectors direction d=(dx,dy,dz) onto plane z=zvp: xp = x − (dx/dz)(z−zvp), yp = y − (dy/dz)(z−zvp). Matrix rows: (1,0,−dx/dz, zvp·dx/dz) and (0,1,−dy/dz, zvp·dy/dz) with zp = zvp. Orthographic: dx=dy=0; oblique: −dx/dz = L cos φ, −dy/dz = L sin φ.

## Answer

In **parallel projection**, all projectors are parallel to a given direction d = (dx, dy, dz). The view plane is z = zvp.

**Derivation:** a point P(x, y, z) is projected along the direction d. The projector through P is

(x', y', z') = (x, y, z) + t·(dx, dy, dz)

It meets the view plane when z' = zvp:

t = (zvp − z)/dz

Therefore

- xp = x + dx·(zvp − z)/dz = x − (dx/dz)(z − zvp)
- yp = y − (dy/dz)(z − zvp)
- zp = zvp

**Matrix:**

```text
| xp |   | 1 0 -dx/dz  zvp·dx/dz | | x |
| yp | = | 0 1 -dy/dz  zvp·dy/dz | | y |
| zp |   | 0 0  0      zvp       | | z |
| 1  |   | 0 0  0      1         | | 1 |
```

**Special cases**
1. **Orthographic projection** (projectors perpendicular to the plane, d = (0, 0, 1)): dx = dy = 0, so xp = x, yp = y, zp = zvp. For zvp = 0:

```text
| 1 0 0 0 |
| 0 1 0 0 |
| 0 0 0 0 |
| 0 0 0 1 |
```

2. **Oblique projection** (d not perpendicular): with −dx/dz = L1cosφ and −dy/dz = L1sinφ, the matrix becomes the standard oblique matrix with entries L1cosφ and L1sinφ in the third column.

Parallel lines remain parallel and the size of faces parallel to the view plane is unchanged.

## Check yourself

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

1. In parallel projection along d = (dx,dy,dz), the projector through P is:
   A. (x,y,z) + t(zvp,zvp,zvp)
   B. (x,y,z) + t(dx,dy,dz)
   C. t(x,y,z)
   D. (x,y,z) - zvp
   Answer: B. All projectors have the same direction d. (Key point: Projector equation)

2. Intersecting with the plane z = zvp gives t equal to:
   A. (zvp - z)/dz
   B. (z - zvp)/dx
   C. zvp/dz
   D. dz/(zvp - z)
   Answer: A. Setting z + t*dz = zvp gives t = (zvp - z)/dz. (Key point: Parameter t)

3. The projected x coordinate is:
   A. xp = x + dz*z
   B. xp = x*dx/dz
   C. xp = x - (dx/dz)(z - zvp)
   D. xp = x + zvp
   Answer: C. Substituting t gives xp = x + dx(zvp - z)/dz. (Key point: Projected x)

4. When d = (0,0,1) the projection becomes:
   A. Perspective with vanishing points
   B. xp = z, yp = z
   C. Oblique with L1 = 1
   D. xp = x, yp = y (orthographic)
   Answer: D. dx = dy = 0 so only z is dropped; for zvp = 0 the matrix is diag(1,1,0,1). (Key point: Orthographic case)

## Related in Computer Graphics

- [Derive perspective projection matrix](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aperspective-projection-matrix-derive) — Computer Graphics, chapter 4, asked 3×
- [Derive oblique projection matrix](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aoblique-projection-matrix-derive) — Computer Graphics, chapter 4, asked 3×
- [Parallel versus oblique projection](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aparallel-vs-oblique-projection) — Computer Graphics, chapter 4
- [Need of perspective projection](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Awhy-perspective-projection-world-to-viewing) — Computer Graphics, chapter 4

## Related in other subjects

- [Derive bar element stiffness matrix](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Abar-element-stiffness-derivation) — Finite Element Method, chapter 3
- [Derive beam element stiffness matrix](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Abeam-element-stiffness-derivation) — Finite Element Method, chapter 3
- [Derive CST element stiffness matrix](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Acst-stiffness-matrix-derivation) — Finite Element Method, chapter 6