Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: plane equation with a radius Date: Mon, 15 Mar 2010 18:38:05 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 27 Message-ID: <hnluqd$bb4$1@fred.mathworks.com> References: <hnl3gl$2lm$1@fred.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-03-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1268678285 11620 172.30.248.38 (15 Mar 2010 18:38:05 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Mon, 15 Mar 2010 18:38:05 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:616995 "Junghyun " <wkhdntnkpst@yahoo.com> wrote in message <hnl3gl$2lm$1@fred.mathworks.com>... > Hello, > > I need to find the plane equation if 2 eigenvectors at a given point in a 3D space and a radius are given. > How can I formulate it in Matlab? > Thanks, > > Junghyun Your question is not at all clear to me Junghyun and I can only guess at what you mean. My speculation is that you wish to define the circle which lies in a plane parallel to two given vectors a and b, and contains a given point c, with the circle's center positioned at this c and with its radius equal to a given value r. If this is correct, do the following. If a and b are unit vectors and orthogonal, this circle can be expressed as follows: p = c + r*(cos(t)*a + sin(t)*b); % Point p traces out the circle in 3D More generally if a and b are any two non-parallel vectors, the circle can be expressed as: u = a/norm(a); % u is normalization of vector a v = cross(cross(a,b),a); % v is parallel to the plane and orthog. to u v = v/norm(v); % Normalize vector v p = c + r*(cos(t)*u + sin(t)*v); % Point p traces out the circle in 3D In either case, as parameter t varies from 0 to 2*pi, p starts at a, rotates past b, and on around again in a full circle to a at t = 2*pi. The assumption is made that a and b are not parallel. Otherwise the circle is not uniquely determined. Roger Stafford