in 3D,we can test |(p2-p1)*(p3-p1)|==0,where p1,p2,p3 are 3D-vectors that represent the three points. in n-dimension,i think we can let A=(a1,a2,...,an)=p2-p1, and B=(b1,b2,...,bn)=p3-p1, and test every elements of the matrix (ATB- BTA). That is ai*bj-aj*bi.
On 5月27日, 上午7时22分, Feng <[EMAIL PROTECTED]> wrote: > Hi all! > > Given 3 points in 3D, what is the fast and numerically stable way to > test if they form a triangle? > > I am thinking computing the determinant of the square matrix formed by > the 3 points and testing if the determinant is nonzero. But I am not > sure. > > What about the case for high dimensions, i.e. 4D, 5D ... > > Thanks! --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
