Re: how to determine normal vector for points on a plane in ℤ³

[hermann@stamm-wilbrandt.de]

On 2024-02-07 18:15, Bill Allombert wrote:
> I assume you means 'the same affine plane', and the normal vector
> is not unique.
> All you need to do is to find a,b,c,d so that, for all your points,
>   a*x+b*y+c*z+d=0
>
> So just add a component 1 to your vectors and take the kernel
>
> ? matker(Mat(apply(x->concat(x,1),V~)))
> %15 = [0;0;1;1]
>
> So a=0, b=0 and c=1 so one of the normal vectors is [0,0,1]
>
> Cheers,
> Bill
Thanks for both answers.

Bill's method only adds a 1 to each vector
Loïc's method does vector subtraction for all vectors.

So I will go with Bill's method:

? P=[[1, 0, 2], [1, 1, 1], [-2, 0, -1], [0, 0, 1], [-1, 1, -1], [-1, -1, 
1], [0, 2, -1], [1, 2, 0], [-2, -1, 0], [0, 1, 0], [-1, 0, 0]];
? matker(Mat(apply(x->concat(x,1),P~)))[1..3,1]
%9 = [1, -1, -1]~
? P=[[-1, -3, -1], [0, -3, -1], [1, -3, -1], [-2, -2, -1], [-1, -2, -1], 
[0, -2, -1], [1, -2, -1], [2, -2, -1], [-3, -1, -1], [-2, -1, -1], [-1, 
-1, -1], [0, -1, -1], [1, -1, -1], [2, -1, -1], [3, -1, -1], [-3, 0, 
-1], [-2, 0, -1], [-1, 0, -1], [0, 0, -1], [1, 0, -1], [2, 0, -1], [3, 
0, -1], [-3, 1, -1], [-2, 1, -1], [-1, 1, -1], [0, 1, -1], [1, 1, -1], 
[2, 1, -1], [3, 1, -1], [-2, 2, -1], [-1, 2, -1], [0, 2, -1], [1, 2, 
-1], [2, 2, -1], [0, 3, -1], [1, 3, -1]];
? matker(Mat(apply(x->concat(x,1),P~)))[1..3,1]
%11 = [0, 0, 1]~
?


I will raise an issue against incorrect implementation in JSCAD:
https://discord.com/channels/775984095250612234/914627480512503829/1204817782265479280
(they add normal vectors for vertices (i, i+1, i+2) and normalize, but 
for 2nd example the sum is the null vector).

Regards,

Hermann.