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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/computer-graphics/2/a/i/solution

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Past exam of the computer science course of the University of Oxford / 2015 / Computer graphics / 2 / a / i / Solution by

Codex  0 2026-10-31

Represent a point by the column of homogeneous coordinates

(X, Y, Z, 1)^{T}

. The actions of the four matrices are:

Matrix	Geometric action and parameters
$G (θ)$	A proper rotation about the $x$ -axis: $(X, Y, Z) \mapsto (X, Y cos θ - Z sin θ, Y sin θ + Z cos θ)$ . The angle $θ$ is in radians, with the usual right-hand sense.
$H (x, y, z)$	A translation by $(x, y, z)$ : $(X, Y, Z) \mapsto (X + x, Y + y, Z + z)$ .
$J (λ)$	Uniform scaling by $λ$ : $(X, Y, Z) \mapsto (λ X, λY, λ Z)$ . Positive $λ$ gives a uniform dilation; $λ = 0$ collapses the object to the origin; negative $λ$ additionally reverses its direction through the origin.
$K (f)$	Perspective projection from the origin onto the plane $Z = f$ , where $f \neq = 0$ is the focal length in the chosen coordinate units.

The last operation produces

(X, Y, Z, Z / f)^{T}

. It requires a perspective divide: divide the first three components by the last. For

Z \neq = 0

K (f) : (X, Y, Z) ⟼ (\frac{f X}{Z}, \frac{f Y}{Z}, f) .

(1)

This is why the last component of homogeneous coordinates cannot simply be discarded. Points with

Z = 0

have no finite projected position, and a visible scene normally uses

Z > 0

under this forward-axis convention. The first three transformations leave the last component equal to one and need no nontrivial divide. The displayed

K

retains no depth variation after division; it describes projection onto the image plane, rather than a complete depth-buffer mapping.

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