Thread Subject:

Project points using null()

Hello everyone,

I'm trying to use a technique to project 3d points onto a plane of which I have the normal vector and a point.

By reading through this newsgroup (this post in particular: http://www.mathworks.com/matlabcentral/newsreader/view_thread/169417) I understood that I get a projection by multiplying my matrix containing the point coordinates with the null function of the normal vector of the plane:

[projection]=[Points]*null([normal vector])

It works beautifully, but I can't wrap my head around how reason it actually works ;)..In what kind of coordinate system/frame of reference do I have to see these resulting coordinates? I'd like to transform them back into my global Cartesian system..
Thanks for any help on this..

Stan

"Stan " <zufallsacc@yahoo.com> wrote in message <i8i2vk$77u$1@fred.mathworks.com>...
> Hello everyone,
>
> I'm trying to use a technique to project 3d points onto a plane of which I have the normal vector and a point.
>
> By reading through this newsgroup (this post in particular: http://www.mathworks.com/matlabcentral/newsreader/view_thread/169417) I understood that I get a projection by multiplying my matrix containing the point coordinates with the null function of the normal vector of the plane:
>
> [projection]=[Points]*null([normal vector])
>
> It works beautifully, but I can't wrap my head around how reason it actually works ;)..In what kind of coordinate system/frame of reference do I have to see these resulting coordinates? I'd like to transform them back into my global Cartesian system..
> Thanks for any help on this..
=========

The above is correct only for planes passing through the origin. More generally, you would do the following, which would also produce output in your original coordinate system.

basisPlane=null(normalVector); %basis for the plane

basisCoefficients= bsxfun(@minus,Points,pointInplane)*basisPlane ;

theResult=bsxfun(@plus,basisCoefficients*basisPlane.', pointInplane),

"Stan " <zufallsacc@yahoo.com> wrote in message <i8i2vk$77u$1@fred.mathworks.com>...
> Hello everyone,
>
> I'm trying to use a technique to project 3d points onto a plane of which I have the normal vector and a point.
>
> By reading through this newsgroup (this post in particular: http://www.mathworks.com/matlabcentral/newsreader/view_thread/169417) I understood that I get a projection by multiplying my matrix containing the point coordinates with the null function of the normal vector of the plane:
>
> [projection]=[Points]*null([normal vector])
>
> It works beautifully, but I can't wrap my head around how reason it actually works ;)..In what kind of coordinate system/frame of reference do I have to see these resulting coordinates? I'd like to transform them back into my global Cartesian system..
> Thanks for any help on this..
>
> Stan
- - - - - - - - - - -
  Let P be the m x 3 array of the 3D points to be projected, let Q be the 1 x 3 vector of the given point on the plane, let N be the 1 x 3 vector of the normal direction to the plane, and let P0 be the m x 3 array of points orthogonally projected from P onto the plane. Then do this:

 N = N/norm(N); % <-- do this if N is not normalized
 N2 = N.'*N;
 P0 = P*(eye(3)-N2)+repmat(Q*N2,m,1);

(You can also do that last line using bsxfun.)

  This can be derived from the single-point vector equation

 p0 = p - dot(p-q),n)*n

where p is a vector to a point to be projected, q is a vector to the point on the plane, n is the unit normal vector to the plane, and p0 is the orthogonal projection of p onto the plane.

Roger Stafford

Thank you both for your help, this works beautifully!

Kind regards,
Stan

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <i8ij1r$39r$1@fred.mathworks.com>...
> "Stan " <zufallsacc@yahoo.com> wrote in message <i8i2vk$77u$1@fred.mathworks.com>...
> > Hello everyone,
> >
> > I'm trying to use a technique to project 3d points onto a plane of which I have the normal vector and a point.
> >
> > By reading through this newsgroup (this post in particular: http://www.mathworks.com/matlabcentral/newsreader/view_thread/169417) I understood that I get a projection by multiplying my matrix containing the point coordinates with the null function of the normal vector of the plane:
> >
> > [projection]=[Points]*null([normal vector])
> >
> > It works beautifully, but I can't wrap my head around how reason it actually works ;)..In what kind of coordinate system/frame of reference do I have to see these resulting coordinates? I'd like to transform them back into my global Cartesian system..
> > Thanks for any help on this..
> >
> > Stan
> - - - - - - - - - - -
> Let P be the m x 3 array of the 3D points to be projected, let Q be the 1 x 3 vector of the given point on the plane, let N be the 1 x 3 vector of the normal direction to the plane, and let P0 be the m x 3 array of points orthogonally projected from P onto the plane. Then do this:
>
> N = N/norm(N); % <-- do this if N is not normalized
> N2 = N.'*N;
> P0 = P*(eye(3)-N2)+repmat(Q*N2,m,1);
>
> (You can also do that last line using bsxfun.)
>
> This can be derived from the single-point vector equation
>
> p0 = p - dot(p-q),n)*n
>
> where p is a vector to a point to be projected, q is a vector to the point on the plane, n is the unit normal vector to the plane, and p0 is the orthogonal projection of p onto the plane.
>
> Roger Stafford

Stan,

Which of the two works beautifully? Have you implemented Roger's method? In his formulation N2 is 1 since N was normalized. Thus, when you check the equation for P0, you see that P0 is independent of N. Therefore, the equation is wrong. Agree?
"Stan " <zufallsacc@yahoo.com> wrote in message <i8mj09$i65$1@fred.mathworks.com>...
> Thank you both for your help, this works beautifully!
>
> Kind regards,
> Stan
>
> "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <i8ij1r$39r$1@fred.mathworks.com>...
> > "Stan " <zufallsacc@yahoo.com> wrote in message <i8i2vk$77u$1@fred.mathworks.com>...
> > > Hello everyone,
> > >
> > > I'm trying to use a technique to project 3d points onto a plane of which I have the normal vector and a point.
> > >
> > > By reading through this newsgroup (this post in particular: http://www.mathworks.com/matlabcentral/newsreader/view_thread/169417) I understood that I get a projection by multiplying my matrix containing the point coordinates with the null function of the normal vector of the plane:
> > >
> > > [projection]=[Points]*null([normal vector])
> > >
> > > It works beautifully, but I can't wrap my head around how reason it actually works ;)..In what kind of coordinate system/frame of reference do I have to see these resulting coordinates? I'd like to transform them back into my global Cartesian system..
> > > Thanks for any help on this..
> > >
> > > Stan
> > - - - - - - - - - - -
> > Let P be the m x 3 array of the 3D points to be projected, let Q be the 1 x 3 vector of the given point on the plane, let N be the 1 x 3 vector of the normal direction to the plane, and let P0 be the m x 3 array of points orthogonally projected from P onto the plane. Then do this:
> >
> > N = N/norm(N); % <-- do this if N is not normalized
> > N2 = N.'*N;
> > P0 = P*(eye(3)-N2)+repmat(Q*N2,m,1);
> >
> > (You can also do that last line using bsxfun.)
> >
> > This can be derived from the single-point vector equation
> >
> > p0 = p - dot(p-q),n)*n
> >
> > where p is a vector to a point to be projected, q is a vector to the point on the plane, n is the unit normal vector to the plane, and p0 is the orthogonal projection of p onto the plane.
> >
> > Roger Stafford

"Kadir" wrote in message <j9api5$eml$1@newscl01ah.mathworks.com>...
> Stan,
>
> Which of the two works beautifully? Have you implemented Roger's method? In his formulation N2 is 1 since N was normalized. Thus, when you check the equation for P0, you see that P0 is independent of N. Therefore, the equation is wrong. Agree?
> "Stan " <zufallsacc@yahoo.com> wrote in message <i8mj09$i65$1@fred.mathworks.com>...
> > Thank you both for your help, this works beautifully!
- - - - - - - -
  Kadir, what makes you think N2 equals 1? It is the matrix product of the 3-by-1 column vector, N.', by the 1-by-3 row vector, N, producing a 3-by-3 matrix which is neither the scalar 1 nor eye(3). P0, in fact, very much depends on N (unless P is coincident with Q.)

Roger Stafford

You're right. Sorry, because I generally think about points as column vectors.

"Roger Stafford" wrote in message <j9bno5$mnj$1@newscl01ah.mathworks.com>...
> "Kadir" wrote in message <j9api5$eml$1@newscl01ah.mathworks.com>...
> > Stan,
> >
> > Which of the two works beautifully? Have you implemented Roger's method? In his formulation N2 is 1 since N was normalized. Thus, when you check the equation for P0, you see that P0 is independent of N. Therefore, the equation is wrong. Agree?
> > "Stan " <zufallsacc@yahoo.com> wrote in message <i8mj09$i65$1@fred.mathworks.com>...
> > > Thank you both for your help, this works beautifully!
> - - - - - - - -
> Kadir, what makes you think N2 equals 1? It is the matrix product of the 3-by-1 column vector, N.', by the 1-by-3 row vector, N, producing a 3-by-3 matrix which is neither the scalar 1 nor eye(3). P0, in fact, very much depends on N (unless P is coincident with Q.)
>
> Roger Stafford

Roger,
Given a plane defined in constants XCoeff, YCoeff, CCoeff (see http://www.mathworks.co.uk/support/solutions/en/data/1-1AVW5/index.html?solution=1-1AVW5)
        z = XCoeff * x + YCoeff * y + CCoeff
how would you calculate Q and N for your method?

I tried
Q = [1, 1, (XCoeff+YCoeff+CCoeff)];
N = [XCoeff, YCoeff, -1];

and while the projected points do all lie on the plane they are not orthogonally projected onto the plane?

Best wishes

Steven

> Let P be the m x 3 array of the 3D points to be projected, let Q be the 1 x 3 vector of the given point on the plane, let N be the 1 x 3 vector of the normal direction to the plane, and let P0 be the m x 3 array of points orthogonally projected from P onto the plane. Then do this:
>
> N = N/norm(N); % <-- do this if N is not normalized
> N2 = N.'*N;
> P0 = P*(eye(3)-N2)+repmat(Q*N2,m,1);
>
> (You can also do that last line using bsxfun.)
>
> This can be derived from the single-point vector equation
>
> p0 = p - dot(p-q),n)*n
>
> where p is a vector to a point to be projected, q is a vector to the point on the plane, n is the unit normal vector to the plane, and p0 is the orthogonal projection of p onto the plane.
>
> Roger Stafford

Thanks to this post was also trying to do this and now problem resolved after long

makeyourassignment.com

"Steven " <steven_ew@hotmail.com> wrote in message <jsmja3$a79$1@newscl01ah.mathworks.com>...
> Roger,
> Given a plane defined in constants XCoeff, YCoeff, CCoeff (see http://www.mathworks.co.uk/support/solutions/en/data/1-1AVW5/index.html?solution=1-1AVW5)
> z = XCoeff * x + YCoeff * y + CCoeff
> how would you calculate Q and N for your method?

 z = XCoeff * x + YCoeff * y + CCoeff

is equivalent to

< (x,y,z) . (XCoeff, YCoeff , -1) > = -CCoeff

<.,.> is dot product. Thus the normal vector is N = (XCoeff, YCoeff , -1) .

% Bruno