The fastest way to obtain the two vectors perpendicular to a known vector? (it sounds like the inverse of the cross product :) )
precisely however i'd need the 4 points of the cross resulting from these 2 vectors (p1,p2,p3,p4)
thx
Not sure I understand - there's an infinite number of vectors perpendicular to a single vector (imagine the cross in your pic rotating).
but they lie on the same plane
In theory I should choose a random value and do a cross product with the other vector generated by the new point.
Ah, so you don't care about the orientation of the vectors?
Then I guess you can calc a normal to the vector, and then calc the cross prod of these two vectors to find the 3rd vector.
the only thing I have is the green vector
I understand that. :)
What I'm saying is:
- calculate a normal to the green vector (sorry, I don't know how to do this off the top of my head).
- calc cross prod of green vector and normal to get 3rd vector.
Easy way, create a pivot, align it's Y axis to the vector you know, then use tformpoint to find -X, +Z, +X, -Z (-1,0,0 - 0,0,1 - 1,0,0 - 0,0,-1)
The hard way, in psuedo code:
vecY = 1,2,3 ;your known vector
normalize vecY
;use any vector that isn't vecY
vecZ = 0,1,0
if vecZ = vecY or vecZ = -VecY then vecZ = 0,0,1 ;can't crossProduct parallel vectors
;find a perpendicular vector, vecX, from vecY & vecZ
vecX = crossProduct(vecY,vecZ)
;orthoganalize vecZ (this just rotates vecZ around the vecX axis until it's perpendicular to vecY)
vecZ = crossProduct(vecX,vecY)
normalize vecX
normalize vecZ
;in clockwise order around vector vecY, 4 points lying on the plane X,Z would be
-vecX
vecZ
vecX
-vecZ
nice the first solution :D
thanx!