---
title: "Shear unit cube w.r.t. yz plane"
subject: "Computer Graphics"
chapter: 4
key: "semester_5/computer_graphics:cube-shear-yz-constant-2"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/cube-shear-yz-constant-2
site: BCT Engineering
---

# Shear unit cube w.r.t. yz plane

Computer Graphics · Chapter 4: Three-Dimensional Graphics

- Asked in: 2072 Chaitra, 8 marks; 2078 Bhadra, 7 marks; 2067 Mangsir (old course), 10 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/cube-shear-yz-constant-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-4

## Question

A unit length cube with diagonal passing through (0,0,0) and (1,1,1) is sheared with respect to yz plane with the shear constants = 2 in both directions. Obtain the coordinates of all the corners of the cube after shearing.

## Short answer

Shearing with respect to the yz plane fixes x and displaces y and z by 2x. Original unit cube corners (0,0,0) to (1,1,1) become (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,2,2), (1,2,3), (1,3,2), (1,3,3).

## Answer

**Assumption:** shear "with respect to the yz plane" keeps the x-coordinate fixed (the yz plane, x = 0, does not move). The y and z coordinates are displaced in proportion to x with shear constants sh_y = sh_z = 2. The unit cube has corners from (0,0,0) to (1,1,1).

Equations: x' = x, y' = y + 2x, z' = z + 2x

```text
| x' |   | 1 0 0 0 | | x |
| y' | = | 2 1 0 0 | | y |
| z' |   | 2 0 1 0 | | z |
| 1  |   | 0 0 0 1 | | 1 |
```

| Corner | (x, y, z) | y + 2x | z + 2x | New corner |
|---|---|---|---|---|
| A | (0,0,0) | 0 | 0 | (0,0,0) |
| B | (0,0,1) | 0 | 1 | (0,0,1) |
| C | (0,1,0) | 1 | 0 | (0,1,0) |
| D | (0,1,1) | 1 | 1 | (0,1,1) |
| E | (1,0,0) | 2 | 2 | (1,2,2) |
| F | (1,0,1) | 2 | 3 | (1,2,3) |
| G | (1,1,0) | 3 | 2 | (1,3,2) |
| H | (1,1,1) | 3 | 3 | (1,3,3) |

The face x = 0 stays in place and the face x = 1 slides by 2 units in both y and z.

**Answer:** (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,2,2), (1,2,3), (1,3,2), (1,3,3).

## Check yourself

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

1. Shear with respect to the yz plane (shear constants in y and z) keeps which coordinate fixed?
   A. x
   B. y
   C. z
   D. None; all three change
   Answer: A. The yz plane is x = 0, so x is unchanged and y, z are displaced in proportion to x. (Key point: Fixed coordinate)

2. With constants 2 in both directions, the transformation is:
   A. x' = x + 2y, y' = y, z' = z + 2y
   B. x' = x, y' = y + 2x, z' = z + 2x
   C. x' = 2x, y' = 2y, z' = 2z
   D. x' = x, y' = 2y, z' = 2z
   Answer: B. y and z are displaced by 2 times x while x stays the same. (Key point: Shear equations)

3. What is the new position of corner (1,0,1)?
   A. (1,2,1)
   B. (1,0,3)
   C. (1,2,3)
   D. (3,0,3)
   Answer: C. y' = 0 + 2(1) = 2 and z' = 1 + 2(1) = 3, x stays 1. (Key point: Corner calculation)

4. What happens to the faces of the cube?
   A. Face x = 1 stays; face x = 0 slides by 2
   B. Both faces slide equally
   C. Face x = 0 stays; face x = 1 slides by 2 in x only
   D. Face x = 0 stays; face x = 1 slides by 2 in both y and z
   Answer: D. Points with x = 0 are unchanged, while those with x = 1 are displaced by 2 in y and z. (Key point: Effect on faces)

## Close variants set in other exams

- [Shear cube w.r.t. ZX plane, constant 3](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acube-shear-zx-constant-3) — Computer Graphics, chapter 4, asked 4×