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Re: Are points co-planar in (numDimensions-1)?

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  • Subject: [mg43327] Re: Are points co-planar in (numDimensions-1)?
  • From: "AngleWyrm" <no_spam_anglewyrm at hotmail.com>
  • Date: Sun, 24 Aug 2003 04:55:41 -0400 (EDT)
  • References: <bi195e$akp$1@smc.vnet.net> <bi7nu3$pc8$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Olaf Rogalsky" <Olaf.Rogalsky at physik.uni-erlangen.de> wrote in message
news:bi7nu3$pc8$1 at smc.vnet.net...
> AngleWyrm wrote:
> > Given some n-dimensional vectors, are they coplanar in n-1? Let a1, a2, ..., an be vectors. If
they
> > are coplanar, then there exists a set of coefficients {k1, k2, ..., kn}, not all zero, which
satisfy
> > the equation: k1 a1 + k2 a2 + ... + kn an = 0.
> I don't agree. This is the definition of linear independence of n vectors, not of coplanarity.
> A set of k vectors {a_i in Vn | i=1..k} in a n-dimensional vector space Vn are said to be
coplanar,
> if all k vectors {a_i in Vn | i=1..k} are elements of a (n-1)-dimensional affine subspace U of Vn.

If a and b are non-parallel vectors (and not 0), and c is coplanar with a and b, then it is possible
to transform to c from a and b like so:
c = k1a + k2b           // where k1 and k2 are some factors.
k1a + k2b - c = 0     // rearranging the equation by subracting c

Otherwise, if a and b are parallel, there is some factor so that say, b=k3a. Then
k3a - b = 0    // also by subtraction

Thus, to summarize, if a,b, and c are coplanar (and not 0), then there exists some relation of the
form:
k1 a + k2 b + k3 c = 0


> Solve[ Sum[ k\_i sample[[i]], {i, numDimensions} ] ==0, Table[k\_i,{i,1,numDimensions}] ]
Thank you, this helped.


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