This is basically what i am looking for... Thank you... =]
I'm glad that someone mentioned the rest of the functions for i am only just
beginning to understand the vrotate and they are all equally as confusing as
the next...



> > Could you please tell about vector cross & dot products too, then?  :o)
>
> The dot product is an scalar binary vector space inner product operator.
> Don't get scared, that only means that it is a function that takes TWO
> VECTORS and gives a NUMBER as a result.
>
> The dot product is very important because it lets you find the angle between
> two vectors, see if they are orthogonal (perpendicular), etc.
> It is also very important because it lets you project a vector into a
> plane... something like finding the "shadow" of a vector in a plane. It lets
> you do many things!
>
> If we have two 3d vectors: A=(a1,a2,a3), B=(b1,b2,b3), you can compute it's
> dot product in two alternative but equivalent ways:
>
> 1) A dot B = a1*b1 + a2*b2 + a3*b3
> 2) A dot B = |A|*|B|*cos(theta)
>
> where theta is the angle (in radians!) between A and B. And |A| is the
> euclidean norm (length) of A defined by
>
> |A|=sqrt(a1^2 + a2^2 + a3^2)
>
> So, if A and B are orthogonal (perpendicular), then cos(theta)=0 and then
> the dot product is equal to zero.
>
> Now, the cross product, is a function that takes TWO VECTORS and gives a
> VECTOR as a result. The most important characteristic from the cross product
> is that the resulting vector is orthogonal to the other two original
> vectors. And that's very useful for finding normal vectors, and other
> things.
>
> The computation is more cumbersome (although there is a trick to memorize it
> using a determinant, but I won't state it):
>
> if A cross B = C = [c1,c2,c3] then
>
> c1 = a2*b3 - a3*b2
> c2 = a3*b1 - a1*b3
> c3 = a1*b2 - a2*b1
>
> As you can prove easily, the resulting vector C is orthogonal to A and B
> (using dot product and a little of algebra)
>
> Well, I hope I made myself clear and help you out of this,
>
> Fernando.

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