POV-Ray : Newsgroups : povray.general : Plane passing through 3 points : Re: Plane passing through 3 points Server Time
25 May 2022 06:33:57 EDT (-0400)
  Re: Plane passing through 3 points  
From: Tor Olav Kristensen
Date: 1 May 2022 16:45:00
Message: <web.626ef069e0a3ca27892957989db30a9@news.povray.org>
"Tor Olav Kristensen" <tor### [at] TOBEREMOVEDgmailcom> wrote:
> kurtz le pirate <kur### [at] gmailcom> wrote:
> > Hello,
> >
> > I have 3 points A, B and C
> > I would like the "plane" object to pass through these three points.
> >
> > It is easy to determine the equation of a plane that passes through
> > these three points. We obtain this equation : ax + by + cz + d = 0
> >
> > I try to do : plane { <a, b, c>, d }
> >
> > but the plane does not pass through the points.
> >
> >
> > Documentation at <http://www.povray.org/documentation/view/3.7.1/297/>
> > indicates that the object "plane { <A, B, C>, D }" represent the plane
> > defined by : A*x + B*y + C*z - D*sqrt(A^2 + B^2 + C^2) = 0
> >
> > ... I must be dumb but I can't make the link with my plan equation.
> >
> >
> > A little help ?
>
> I would prefer to rewrite this equation:
>
> A*x + B*y + C*z - D*sqrt(A^2 + B^2 + C^2) = 0
>
> - like this:
>
> A*X + B*Y + C*Z = D*sqrt(A^2 + B^2 + C^2)

Since this has the same effect as normalizing the normal vector,
the statement "that is only true if the normal vector has unit length",
in the documentation page that you linked to, seems false to me.

I.e.: The normal vector in the plane statement does not need to be
normalized.


> If we let L = sqrt(A^2 + B^2 + C^2)
>
> - and we then divide by L on both sides, we get this equation:
>
> A/L*X + B/L*Y + C/L*Z = D
>
> The left side of the equation equals the dot product between
> two vectors; vN and vP, where vN = <A/L, B/L, C/L> and
> vP = <X, Y, Z>. I.e.:
>
> vdot(vN, vP) = D
>
> Since L = vlength(<A, B, C>), vN is a normalized vector;
>
> vN = <A, B, C>/L = vnormalize(<A, B, C>)
>
> Now if vN is the normalized normal vector to a plane - and if
> vP is the position vector to any point in that plane, then the
> plane is defined by all points in 3D space, where the dot
> product between these two vectors equals the distance; D from
> the origin to the plane in the direction of the plane's normal
> vector.

Hmmm.. not so precise. I try again:

Now if vN is the normalized normal vector to a plane - and if
vP is a position vector to any point in 3D space, then the
plane is defined by (or contains ?) all possible points vP
whose the dot product with vN equals D; the distance from the
origin to the plane in the direction of vN.

--
Tor Olav
http://subcube.com
https://github.com/t-o-k


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