gRG: X = [- 50/7, 50/7, 15] + r·([-10, 10, 35] - [- 50/7, 50/7, 15]) = [- 20/7·r - 50/7, 20/7·r + 50/7, 20·r + 15]
Davon suchen wir jetzt genau den Punkt der die Entfernung 5 von der Ebene hat.
d = ABS(2·x - 2·y - z + 75)/√(2^2 + 2^2 + 1^2) = 5
d = ABS(2·(- 20/7·r - 50/7) - 2·(20/7·r + 50/7) - (20·r + 15) + 75)/√(2^2 + 2^2 + 1^2) = 5 --> r = 23/44
S = [- 50/7, 50/7, 15] + 23/44·([-10, 10, 35] - [- 50/7, 50/7, 15]) = [- 95/11, 95/11, 280/11] = [-8.636, 8.636, 25.455]