Lab M03 - Homogeneous Coordinates and Transformation Matrices

Task 1 - Why 3x3 Cannot Translate

This task shows concretely that no 3x3 matrix can add a fixed offset to every input vector, then demonstrates the 4x4 solution.

Suppose we want to translate every point by (tx, ty, tz) = (4, 5, 6). A 3x3 matrix M must satisfy M * v = v + (4, 5, 6) for every possible v. Consider the y component of the result:

y_out = m10 * vx + m11 * vy + m12 * vz

For this to equal vy + 5 for every input, we would need m11 = 1 and m10 * vx + m12 * vz = 5 for all vx and vz simultaneously. There are no constants m10 and m12 that satisfy this for every possible vx and vz. Translation is not a linear operation in 3D.

The 4x4 solution: add a fourth component w = 1 to every point. The translation matrix is:

T(4, 5, 6):
| 1  0  0  4 |
| 0  1  0  5 |
| 0  0  1  6 |
| 0  0  0  1 |

Apply T(4, 5, 6) to the point p = (1, 2, 3, 1):

| 1  0  0  4 |   | 1 |   | 1*1 + 0*2 + 0*3 + 4*1 |   | 5 |
| 0  1  0  5 | * | 2 | = | 0*1 + 1*2 + 0*3 + 5*1 | = | 7 |
| 0  0  1  6 |   | 3 |   | 0*1 + 0*2 + 1*3 + 6*1 |   | 9 |
| 0  0  0  1 |   | 1 |   | 0*1 + 0*2 + 0*3 + 1*1 |   | 1 |

Result: p_translated = (5, 7, 9, 1). The offset (4, 5, 6) was added correctly.

Now apply the same T(4, 5, 6) to the direction vector d = (0, 1, 0, 0):

| 1  0  0  4 |   | 0 |   | 1*0 + 0*1 + 0*0 + 4*0 |   | 0 |
| 0  1  0  5 | * | 1 | = | 0*0 + 1*1 + 0*0 + 5*0 | = | 1 |
| 0  0  1  6 |   | 0 |   | 0*0 + 0*1 + 1*0 + 6*0 |   | 0 |
| 0  0  0  1 |   | 0 |   | 0*0 + 0*1 + 0*0 + 1*0 |   | 0 |

Result: d_translated = (0, 1, 0, 0). The direction is unchanged. The w = 0 zeroed out the entire fourth column of the matrix, so translation has no effect on direction vectors. This is geometrically correct: directions represent orientation, not position.

Task 2 - The Y-Axis Rotation Matrix

The Y-axis rotation matrix R_y(t) rotates points around the world Y axis by angle t (in radians). Note the sign swap on the sin terms compared to the X and Z rotation matrices: the -sin(t) appears at row 2, column 0 (bottom-left of the 3x3 block) rather than at row 0, column 2.

R_y(t):
| cos(t)   0   sin(t)   0 |
|    0     1      0     0 |
| -sin(t)  0   cos(t)   0 |
|    0     0      0     1 |

At theta = 90 degrees: cos(90) = 0, sin(90) = 1. R_y(90) becomes:

R_y(90):
|  0   0   1   0 |
|  0   1   0   0 |
| -1   0   0   0 |
|  0   0   0   1 |

Verification 1: apply R_y(90) to the point (1, 0, 0, 1) (tip of the X axis).

|  0   0   1   0 |   | 1 |   | 0 |
|  0   1   0   0 | * | 0 | = | 0 |
| -1   0   0   0 |   | 0 |   |-1 |
|  0   0   0   1 |   | 1 |   | 1 |

Result: (0, 0, -1, 1). Rotating the X-axis tip 90 degrees around Y swings it to -Z. This follows the right-hand rule: curl the fingers of the right hand from X toward Z; the thumb points up (+Y); a positive rotation swings X toward -Z.

Verification 2: apply R_y(90) to the point (0, 0, 1, 1) (tip of the Z axis).

|  0   0   1   0 |   | 0 |   | 1 |
|  0   1   0   0 | * | 0 | = | 0 |
| -1   0   0   0 |   | 1 |   | 0 |
|  0   0   0   1 |   | 1 |   | 1 |

Result: (1, 0, 0, 1). The Z-axis tip swings to +X. Both results are consistent: a 90-degree Y rotation takes X to -Z and Z to +X, cycling the horizontal axes as expected.

The sign of the -sin(t) entry at row 2, column 0 encodes this direction. When rotating from X toward -Z (a positive angle around Y), the X component contributes negatively to the new Z component. This is exactly what -sin(t) at position [2][0] produces.

Task 3 - Compose and Apply by Hand

Goal: start at p = (0, 0, 0, 1) (the world origin). First translate by (2, 0, 0), then rotate 90 degrees around Y. The final position should be (0, 0, -2, 1).

In column-major convention, the transform applied first goes on the right in the multiplication. To translate first and then rotate: C = R_y(90) * T(2, 0, 0).

The two matrices:

T(2, 0, 0):             R_y(90):
| 1  0  0  2 |          |  0  0  1  0 |
| 0  1  0  0 |          |  0  1  0  0 |
| 0  0  1  0 |          | -1  0  0  0 |
| 0  0  0  1 |          |  0  0  0  1 |

Compute C = R_y(90) * T(2, 0, 0) by multiplying row by column. Each entry C[row][col] is the dot product of row row of R_y(90) with column col of T(2, 0, 0). Fill in the 16 entries below.

Check Answers

The composite matrix C = R_y(90) * T(2, 0, 0) should be:

C:
|  0   0   1   0 |
|  0   1   0   0 |
| -1   0   0  -2 |
|  0   0   0   1 |

Apply C to the starting point p = (0, 0, 0, 1):

|  0   0   1   0 |   | 0 |   | 0 |
|  0   1   0   0 | * | 0 | = | 0 |
| -1   0   0  -2 |   | 0 |   |-2 |
|  0   0   0   1 |   | 1 |   | 1 |

Result: (0, 0, -2, 1). Interpretation: the origin first moved to (2, 0, 0) by the translation, then a 90-degree Y rotation swung that point to (0, 0, -2). This matches the expected result.

The GLM equivalent (note: GLM calls are written in reverse order because each call left-multiplies the accumulated matrix):

// GLM applies right to left: rotate is called FIRST in code, translate SECOND
// This gives model = T(2,0,0) * I then model = R_y(90) * model = R_y(90) * T(2,0,0)
glm::mat4 model(1.0f);
model = glm::rotate(model, glm::radians(90.0f), glm::vec3(0.0f, 1.0f, 0.0f));
model = glm::translate(model, glm::vec3(2.0f, 0.0f, 0.0f));