Le Forgeron <jgr### [at] freefr> wrote:
> Le 05.11.2007 21:30, Grassblade nous fit lire :
> > Le Forgeron <jgr### [at] freefr> wrote:
> >> Le 04.11.2007 17:16, alphaQuad nous fit lire :
> >>> the task:
> >>> Area = abs(   (xB*yA - xA*yB)  + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2
>
> >> Let's make a numerical application with A(0,-1) B(1,0) and C(0,1)
> >>
> >>  abs ( (-1 - 0) + (0 - 1) + ( 0 +1 ) ) /2 ---> abs ( -1 ) /2 --> 1/2
> >> (wrong)
> > I get ( (-1 - 0) + (0 - 1) + ( 0 + 0 ) ) /2= -1, which is correct.
> >
> Ok, my fault.
>
> So, I let you with alphaQuad trying to generalize that to 3D.
>
> And I will keep using length(crossproduct(B-A,C-A))/2.
>
> --
> The superior man understands what is right;
> the inferior man understands what will sell.
> -- Confucius

I didn't say anything against cross product, did I? It's a useful tool, but it
doesn't work on four or more dimensions. What if I go nuts and want a mesh in
4D?
But to take you up to the task, let's start with the big formula on wikipedia.
With no loss of generality, we can consider A to be in the origin by simply
taking the difference between B and A and C and A. Consequently two of the
determinants are nil, and the square root cancels out with the square. Sarrus'
rule yields: (xB*yC+yB*zC+zB*xC-xB*zC-yB*xC-zB*yC)/2, or
(xB*(yC-zC)+yB*(zC-xC)+zB*(xC-yC))/2. You can reintroduce the coordinates of A
by substitution if you want. I bet that's exactly the same result as the cross
product. :-p

Post a reply to this message