Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Point on Line Segment in 3D ? Fast Algorithm
Date: Wed, 30 Jun 2010 20:04:04 +0000 (UTC)
Organization: Xoran Technologies
Lines: 9
Message-ID: <i0g7vk$jq2$1@fred.mathworks.com>
References: <6a9sqqF35gdblU1@mid.uni-berlin.de> <i0g164$m97$1@fred.mathworks.com> <i0g60c$afg$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 1277928244 20290 172.30.248.38 (30 Jun 2010 20:04:04 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Wed, 30 Jun 2010 20:04:04 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 1440443
Xref: news.mathworks.com comp.soft-sys.matlab:649423

"Kevin " <kablack@umich.edu> wrote in message <i0g60c$afg$1@fred.mathworks.com>...
> How could this method be used for finding if a point was located beneath a surface.  Such as a matrix of points?
=============

I doubt it could. 

In the case of a surface, you would need to first find an equation that describes the surface and fits the points. In the case of the line segment, that's trivial, because it is straightforward how to fit a line segment to 2 points.

Once you have your surface equation c(x)=0, determining whether a point P lies in, under, or above the surface is a matter of looking at the sign of c(P).