- 2076 Asoj · 2+5 marks
Polygon table and normal of S1
Short answer
Construct polygon tables: vertices V1=A(1,2,3), V2=B(5,8,6), V3=C(2,5,8), V4=D(8,-3,2), V5=E(12,6,9); edges e1=AB,...,e6=BE; surfaces S1: e1,e2,e3, S2: e3,e4,e5,e6. For surface S1, compute AB=(4,6,3), AC=(1,3,5); cross product gives normal N = (21, -17, 6).
Represent the following surfaces by polygon table method and find the normal of surface S1. [Figure: points A(1,2,3), B(5,8,6), C(2,5,8), D(8,-3,2), E(12,6,9); edges e1 = AB, e2 = AC, e3 = BC, e4 = CD, e5 = DE, e6 = BE; surface S1 is the triangle ABC (edges e1, e2, e3) and surface S2 is the quadrilateral BEDC sharing edge e3 (edges e3, e4, e5, e6).]
Polygon tables
Vertices: V1 = A(1,2,3), V2 = B(5,8,6), V3 = C(2,5,8), V4 = D(8,−3,2), V5 = E(12,6,9).
Vertex table
| Vertex | x | y | z | Edges |
|---|---|---|---|---|
| A | 1 | 2 | 3 | e1, e2 |
| B | 5 | 8 | 6 | e1, e3, e6 |
| C | 2 | 5 | 8 | e2, e3, e4 |
| D | 8 | −3 | 2 | e4, e5 |
| E | 12 | 6 | 9 | e5, e6 |
Edge table
| Edge | Vertices | Surfaces |
|---|---|---|
| e1 | A, B | S1 |
| e2 | A, C | S1 |
| e3 | B, C | S1, S2 |
| e4 | C, D | S2 |
| e5 | D, E | S2 |
| e6 | B, E | S2 |
Surface table
| Surface | Edges |
|---|---|
| S1 | e1, e2, e3 |
| S2 | e3, e4, e5, e6 |
Normal of S1 (triangle ABC)
AB = B − A = (4, 6, 3), AC = C − A = (1, 3, 5)
N = AB × AC = (6·5 − 3·3, 3·1 − 4·5, 4·3 − 6·1) = (30 − 9, 3 − 20, 12 − 6) = (21, −17, 6)
Check with the plane formulas: A = y1(z2−z3) + y2(z3−z1) + y3(z1−z2) = 2(−2) + 8(5) + 5(−3) = 21; B = z1(x2−x3) + z2(x3−x1) + z3(x1−x2) = 3(3) + 6(1) + 8(−4) = −17; C = x1(y2−y3) + x2(y3−y1) + x3(y1−y2) = 3 + 15 − 12 = 6.
|N| = √(441 + 289 + 36) = √766 = 27.68, so the unit normal = (0.759, −0.614, 0.217).
Plane: 21x − 17y + 6z + D = 0; at A: 21 − 34 + 18 + D = 0 gives D = −5, so 21x − 17y + 6z − 5 = 0 (point C: 42 − 85 + 48 − 5 = 0, correct).
Answer: N(S1) = (21, −17, 6) (for the vertex order A → B → C).
✓ Answer checked by Claude Opus 5.5 (high effort)
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