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Visibility of triangle ABC from (5,5,5)

Short answer

Compute normal N=(1,1,1) and viewer vector V=(4,5,5). Dot product is 14 >0, so the face is a front face and visible from (5,5,5).

Given a triangular object ABC having A(1,0,0), B(0,1,0) and C(0,0,1), if the observer is at point (5,5,5), is the face ABC visible?

Vertices A(1,0,0), B(0,1,0), C(0,0,1); observer at P(5,5,5).

Step 1: normal of ABC (vertex order A → B → C, so the normal points away from the origin)

AB = B − A = (−1, 1, 0), AC = C − A = (−1, 0, 1)

N = AB × AC = (1·1 − 0·0, 0·(−1) − (−1)·1, (−1)·0 − 1·(−1)) = (1, 1, 1)

Step 2: plane equation

x + y + z + D = 0; at A: 1 + D = 0, so D = −1. Plane: x + y + z − 1 = 0 (A = B = C = 1).

Step 3: vector to the observer

V = P − A = (5−1, 5−0, 5−0) = (4, 5, 5)

Step 4: test

V · N = 4·1 + 5·1 + 5·1 = 14 > 0

Since V · N > 0, the face points towards the observer, so it is a front face and is visible.

Check using the plane: substituting the observer's position gives 5 + 5 + 5 − 1 = 14 > 0, so the eye is on the outside (positive) side of the plane, and the face is visible.

Answer: face ABC is visible from (5, 5, 5).

✓ Answer checked by Claude Opus 5.5 (high effort)

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