3D Composite Transformation

Theory of Computation & Computer Graphics — Three-Dimensional Transformation, NEC licence examination syllabus (Nepal Engineering Council).

3D Composite Transformation

Same "combine matrices, apply once" philosophy from 2D — now juggling more matrices.

Worked example — scaling about a fixed point P(a,b,c) in 3D: Step 1: Translate by (−a,−b,−c) — move P to origin Step 2: Scale by (sx,sy,sz) — now scaling is about origin Step 3: Translate by (+a,+b,+c) — move P back Combined matrix = T(a,b,c) · S(sx,sy,sz) · T(−a,−b,−c)

T · R is not the same as R · T With rz = 40 °, tx = 2, x' = 3.53, y' = 1.29.

Set tx to 0 and the cube rotates about the origin. Add tx and it rotates first, then slides — which is what T·R means, read right to left.

Exactly the same 3-step recipe from 2D composite transformation — translate to origin, do the operation, translate back — just now with 4×4 matrices instead of 3×3. If you're comfortable with the 2D version, this is the same skill applied one dimension further.

💡 Practice a full worked numerical: scale or rotate a 3D point about a specific fixed point (not the origin), showing all three matrices multiplied together in the correct order.

Why the order cannot be swapped

Matrix multiplication is not commutative, so T·R and R·T are different transformations. Rotating then translating moves the object to a new place; translating then rotating swings the already-moved object around the origin on a long arc.

Read a composite matrix product right to left: in T(a,b,c)·S(sx,sy,sz)·T(−a,−b,−c), the rightmost matrix acts first. That is the step most often reversed in exam answers, and it produces an answer that is not merely inaccurate but describes a completely different motion.

💡 A quick sanity check on any composite answer: the fixed point must come back to itself. Substitute P(a,b,c) into your combined matrix — if it does not map to exactly P, the order is wrong somewhere.

Why bother combining at all

Because the combined matrix is computed once and then applied to every vertex. A model with fifty thousand vertices costs one matrix multiplication plus fifty thousand vector multiplications, rather than three separate passes over the whole mesh. Composition is not tidiness — it is the optimisation that makes the pipeline affordable.

A 3D composite, worked

Scale by 2 about the fixed point P(2, 2, 2), applied to the point (4, 4, 4).

Step 1 — translate P to the origin: subtract (2,2,2) (4,4,4) → (2, 2, 2) Step 2 — scale by 2 about the origin (2,2,2) → (4, 4, 4) Step 3 — translate back: add (2,2,2) (4,4,4) → (6, 6, 6) Combined: M = T(2,2,2) · S(2,2,2) · T(−2,−2,−2)

💡 The same self-check as in 2D: the fixed point must map to itself. Putting (2,2,2) through the three steps gives (0,0,0) → (0,0,0) → (2,2,2). If your fixed point moves, the order is wrong — and in 3D, where the matrices are 4×4, that check is far quicker than re-multiplying them.

Why order matters, concretely

Rotate-then-translate and translate-then-rotate give genuinely different results, and it is worth seeing why rather than memorising the rule.

🔀 The same two operations, two answers

T · RRotate about the origin first, then slide. The object turns in place, then moves.

R · TSlide first, then rotate about the origin. The object is now far from the origin, so rotating swings it along a long arc to somewhere else entirely.

Read right to left: the matrix nearest the point acts first. That single habit prevents almost every composite-transformation error.

Syllabus points

Combining 3D transformations (numerical)

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