- 2075 Chaitra · 5 marks
Back-face test with given normals
Short answer
For viewing direction V=(1,-1,1), V·N1=(2,-3,4) is 9 >0, so surface 1 is a back face. V·N2=(1,1,-2) is -2 <0, so surface 2 is a front face.
Determine whether two surfaces of an object with normals 2i - 3j + 4k and i + j - 2k respectively, viewed from a direction given by i - j + k, are backface or frontface.
Test: with the viewing direction V (from the eye towards the object), a surface with normal N is a back face if V · N > 0, and a front face if V · N < 0 (V · N = 0 means the surface is seen edge-on).
V = i − j + k = (1, −1, 1)
Surface 1: N1 = 2i − 3j + 4k = (2, −3, 4)
V · N1 = 1·2 + (−1)(−3) + 1·4 = 2 + 3 + 4 = 9 > 0 → back face (hidden).
Surface 2: N2 = i + j − 2k = (1, 1, −2)
V · N2 = 1·1 + (−1)(1) + 1·(−2) = 1 − 1 − 2 = −2 < 0 → front face (visible).
Answer: surface 1 is a back face; surface 2 is a front face.
(Note: this uses V as the direction in which the viewer looks, from the eye to the object. If V were taken as the vector pointing towards the viewer, the condition would be reversed and the two results would swap.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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