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Topic: Project points using null()
Replies: 9   Last Post: Jul 1, 2012 5:04 AM

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Steven

Posts: 4
Registered: 6/30/12
Re: Project points using null()
Posted: Jun 30, 2012 6:10 AM
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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




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