Back-face visibility of BED and ABCD

Vertices: A(1,0,1), B(1,0,−1), C(−1,0,1), D(−1,0,−1), apex E(0,2,0). Observer P(5,5,5). Test: a face is a back face if V · N ≤ 0, where N is the outward normal and V = P − (point on face) is the vector towards the observer.

Surface ABCD (base, plane y = 0)

The base is at the bottom of the pyramid, so its outward normal points down: N = (0, −1, 0).

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

V · N = 0·4 + (−1)·5 + 0·4 = −5

V · N < 0, so ABCD is a back face (not visible).

Surface BED

B(1,0,−1), E(0,2,0), D(−1,0,−1) is the face on the side z = −1.

BE = E − B = (−1, 2, 1), BD = D − B = (−2, 0, 0)

BE × BD = (2·0 − 1·0, 1·(−2) − (−1)·0, (−1)·0 − 2·(−2)) = (0, −2, 4)

This points inward (towards +z, −y); the outward normal is the reverse: N = (0, 2, −4). Check: this face is on the z = −1 side of the pyramid, so its outward normal must have a negative z-component.

V = P − B = (4, 5, 6)

V · N = 0·4 + 2·5 + (−4)·6 = 10 − 24 = −14

V · N < 0, so BED is a back face (not visible).

Answer: both ABCD and BED are back faces for the observer at (5, 5, 5), so neither is visible. (For comparison, the face AEC on the z = +1 side has outward normal (0, 2, 4), and V·N = 10 + 16 = 26 > 0, so it is visible.)