RE: [Algorithms] Where is the enemy?

[Steve W. <Ste...@im...>]

Come to think of it...the vector can be eliminated since it's in both
equations...just this test need be done:

if Px + Py + Pz > P1x + P1y + P1z then it's to the right
if Px + Py + Pz < P1x + P1y + P1z then it's to the left
if Px + Py + Pz = P1x + P1y + P1z then it's straight ahead

R&R

> From: Aldo . [mailto:al...@ho...]
> How can I decide wich side must I turn?
> 
> I've tried cross product, but it doesn't worked fine.
> 
> Any sugestions?
> 

Hmm, how about using the equation of a plane which has at it's normal a
vector V pointing at a right angle to the up and forward vectors and
positive to the right:

VxP1x + VyP1y + VzP1z = C1

Where C1 would be an index to compare with C in

VxPx + VyPy + VzPz = C

that tells you if point P is to the right (C>C1) or left (C<C1).

R&R