Maths - Plane, Surface and Area

From: mjb
Newsgroups: sci.math
Subject: How can we specify a plane in 'n' dimensions?
Date: Wed, 18 Apr 2007 10:30:14 +0100

How can we specify a plane (and rotations) in 'n' dimensions?

In 3D we can specify a plane by using two 3 dimensional base vectors (provided they are not parallel). In 3 dimensions we can use the vector cross product, however in higher dimensions the cross product does not apply and there in not a duality between vectors and bivectors.

Can we use the 'wedge' or exterior product instead?

A point in a 3D plane can be specified by a linear combination of these basis vectors. This requires 3 scalar values for each vector, so 6 scalar values in total. However we can reduce the number of degrees of freedom because the vectors can be unit length so:

b1x2 + b1y2 +b1z2 = 1
b2x2 + b2y2 +b2z2 = 1

where b1 and b2 are the two basis vectors.

We can also reduce the degrees of freedom by specifying the angle between the basis vectors, for instance, lets keep things simple by making the basis vectors mutually perpendicular:

b1x*b2x + b1y*b2y + b1z*b2z = 0

With these restrictions a plane in 3D has three degrees of freedom which is what we want.

So what about 4 dimensions? There are 6 degrees of freedom, represented by a linear combination of pairs of axes:

xy, yz, zx, wx, wy and wz.

So can we specify a plane by using two four dimensional base vectors?

This gives 8 scalar values, but again we can reduce this by 3 degrees of freedom by making the vectors unit length and mutually perpendicular. Which only leaves 5 degrees of freedom, so what have I got wrong?

This gets worse as the number of dimensions increases:

* In 5D there are 10 combinations of basis vectors.
* In 6D there are 15 combinations of basis vectors.

How can the product of two vectors have more degrees of freedom than both of the vectors being multiplied put together?

So is there a geometric interpretation of bivectors in 'n' dimensions? can any of the ways of defining a plane in 3D be extended? For instance as a linear combination of vectors, or by three points that it goes through, or by a linear equation.

Do these methods depend on 'accidents' or special features of 3D such as the duality of vectors and bivectors?

Martin


From: beeworks
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 18 Apr 2007 04:14:47 -0700

A plane in n-D space can be specified by n-2 independent linear equations. In 3-D we have

a1 * x1 + a2 * x2 + a3 * x3 = b

The plane is perpendicular to the vector a = (a1, a2, a3)^T and offset from the origin by b/|a| along the unit vector a/|a|. In 4-D we have

a11 * x1 + a12 * x2 + a13 * x3 = b1
a21 * x1 + a22 * x2 + a23 * x3 = b2

Each equation specifies a 3-D hyper-plane. The intersection of these two hyperplanes is a (2-D) plane.

In n-D the n-2 equations can be written more concisely as

A * X = B

where A = [aij] is a (n-2) by n matrix, X = (x1, x2, ..., xn)^T, and B
= (b1, b2, ..., bn)^T.

As for degrees of freedom, a plane in n-D space has (n+3)*(n-2)/2 d.o.f. A has n*(n-2) parameters and B has n-2 parameters for a total of (n+1)*(n-2) parameters. The rows of A can be constrained to be unit vectors and be mutually orthogonal, i.e.

A*A^T = I,

where I is the (n-2) by (n-2) identity matrix. This constraint on A reduces the free parameters by (n-1)*(n-2)/2, leaving

p = n*(n-2) + (n-2) - (n-1)*(n-2)/2 = (n+3)*(n-2)/2

free parameters. Thus, a plane in 3-D has 3 d.o.f., and in 4-D it has 7 d.o.f.

- MO


From: mjb
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: Thu, 19 Apr 2007 08:18:00 +0100

Thank you this is very interesting, effectively we have to add more constraints as the number of dimensions increases.

I thought for some reason (probably from reading about Geometric Algebra) that a plane (through the origin) in n dimensions could be represented by a bivector. Or at least a 'directed area' which would be an infinite plane plus a magnitude.

This now seems to tie up nicely with what you say except for a few discrepancys in the number of scalars required. As I understand it the dimension of the bivector is given by the second column Pascal's Triangle which is the binomial coefficient. = n! / (n-r)! r!

In this case: n! / (n-2)! 2!
so if n=3 then c=3*2/2 = 3
if n=4 then c=4*3*2/2*2=6
if n=5 then c=5*4*3*2/3*2*2=10

But this is for a 'directed area' so we can subtract 1

Number of scalars to represent infinite plane through origin
= (n! / (n-2)! 2!)-1

could the difference be removed if adjusted for an infinite plane through the origin?

Martin


From: beeworks
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 19 Apr 2007 05:59:26 -0700

Yes, I realized my error after posting. If one simply considers that a plane is specified by three points, and each point can be moved in the directions without changing what plane is specified, then one concludes there are 3*(n-2) parameters. I cannot at this time see what additional constrains must be applied to the n-2 simultaneous equations to reduce the number of free parameters to 3*(n-2).

- MO


From: Lee Rudolph
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 19 Apr 2007 09:06:43 -0400

beeworks writes:

...
>Yes, I realized my error after posting. If one simply considers that
>a plane is specified by three points, and each point can be moved in
>the directions without changing what plane is specified, then one
>concludes there are 3*(n-2) parameters. I cannot at this time see
>what additional constrains must be applied to the n-2 simultaneous
>equations to reduce the number of free parameters to 3*(n-2).

The Gram-Schmidt process, presumably (given that you're working over the reals, right?).

Lee Rudolph


From: Chip Eastham
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 18 Apr 2007 05:18:46 -0700

Yes, three (non-colinear) points determine a (two-dimensional) plane, whatever the dimension of the enclosing space. Rotate
the points, and this rotates the "generated" plane as well. The "plane" here is not necessarily a subspace; it is a subspace if and only if it happens to include the origin.

The points in the plane can be uniquely expressed in terms of the three generating points, say x,y,z, using (real) coefficients
whose sum is 1:

{ax + by + cz | a+b+c = 1}

The coefficients a,b,c in this definition can be be negative, just not all at the same time.

regards, chip


From: mjb
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: Thu, 19 Apr 2007 08:47:20 +0100

chip,

Thanks this is also very interesting.

Just to clarify, if the plane goes through the origin I guess that can be one of the points, so we only need two other points?

So this would need 2*n scalar values to specify which seems to conflict with the other part of this thread?

I like the idea that points are constraints on the possible position of the plane (that is each one subtracts some freedom) and yet we can still combine them additively. I'm still trying to think that through!

Martin


From: Chip Eastham
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 19 Apr 2007 06:36:50 -0700

Yes, if we are interested in planes that are subspaces (closed under scalar multiplication and vector addition), then the origin
is one point we can use, and we only need two nonzero points that are independent (not scalar multiples of each other) to specify the plane.

For clarity one often refers to planes not passing through the origin as affine planes, to emphasize that these are not required to be subspaces.

> So this would need 2*n scalar values to specify which seems to conflict with
> the other part of this thread?

The correspondance between (subspace) planes and 2*n scalar values is not one-to-one. We need on one hand the 2*n scalar
values to form two linear independent vectors, and on the other hand the same (2D subspace) plane can have many different bases.

Specifying a plane through the origin with a basis of two linearly independent vectors (points) has a dual formulation of "constraints" which amounts to a basis for the orthogonal complement of the plane through the origin.

Neither of these gives a "canonical" representation of the plane, by which I mean that one plane can have many representations
of these forms, e.g. two different bases for the same plane. As below we can impose a canonical representation by requiring
the basis to form a reduced row echelon matrix (when the vectors are taken to be rows of a 2xn matrix).

The number of vectors in a basis is invariably two for a plane, and the dimension of (number of vectors in a basis for) its
orthogonal complement (constraints) will invariably be n-2.

> I like the idea that points are constraints on the possible position of the
> plane (that is each one subtracts some freedom) and yet we can still
> combine them additively. I'm still trying to think that through!
>
> Martin

The points {u,v} in a basis give a parametric formulation of the
plane:

{ a*u + b*v | a,b any real scalars }

where the "constraints" basis gives a set of homogenous linear equations for the "solutions" that are points in the plane.

We may want to check whether two representations of a plane with bases of points are equivalent (correspond to the same
plane). Say we have {u1,v1} and {u2,v2} where matrix M1 with rows u1 and v1 is rank 2 (linear independence of {u1,v1} and similarly for matrix M2 with rows u2 and v2. These represent the same plane if and only if the matrix with rows u1,v1,u2,v2 is also rank 2, something that is easily found by row reduction aka Gauss elimination.

One can do a similar computation with two representations of the plane by "constraints" (bases for the respective orthogonal
complements). However for large n, this means reducing a matrix of 2(n-2) rows (rather than the four rows above), so the
constraint formulation is less efficient for this purpose if n is large. However for n = 3 the constraint formulation is more
efficient and compact, and for n = 4 the representations are tied.

regards, chip


From: beeworks
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 19 Apr 2007 09:13:54 -0700

The parameterization along the lines you suggest should be, I believe,

{a*u + b*v + c | u, v \m reals}

In your parameterization, where c = 0, the plane must pass through the origin.

In the parameterization that I give here, a, b, and c have 3*n parameters. However, we can constrain a and b to be unit vectors and a, b, and c to be mutually orthogonal. These constraints reduce the free parameters to 3*n - 5. Finally, we can replace a and b by

a' = cos(t) * a - sin(t) * b
b' = sin(t) * a + cos(t) * b

without changing what plane is specified. Then we can constrain t to a value that maximizes the number of zero elements in a. Thus the free parameters are reduced to

p = 3*n - 6

This is the same number of free parameters given by specifying the plane by three points, each of which has n-2 free parameters (movement orthogonal to the plane).

- MO


From: kunzmilan
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 20 Apr 2007 03:30:54 -0700

On Apr 18, 11:30 am, mjb wrote:
> How can we specify a plane (and rotations) in 'n' dimensions?
>
Points (3,0), (2,1), (1,2), (0,3) lie on a line in 2-dimensions. On this line, all poins lie which sum of coordinates is 3. The rotation axis is perpendicular to it.
Similarly we can construct triangles and tetrahedrons with constant sums of coordinates,and analogically n-dimensional simplices with constant sum of coordinates.
kunzmilan


From: beeworks
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 20 Apr 2007 04:02:17 -0700

On Apr 20, 6:30 am, kunzmilan wrote:
> On Apr 18, 11:30 am, mjb wrote:> How can we specify a plane (and rotations) in 'n' dimensions?
>
> Points (3,0), (2,1), (1,2), (0,3) lie on a line in 2-dimensions. On
> this line, all poins lie which sum of coordinates is 3. The rotation
> axis is perpendicular to it.
> Similarly we can construct triangles and tetrahedrons with constant
> sums of coordinates,and analogically n-dimensional simplices with
> constant sum of coordinates.
> kunzmilan

You need to allow weighted sums to specify a general simplex.


From: kunzmilan
Newsgroups: sci.math
Subject: Re: How can we specify a plane in 'n' dimensions?
Date: 26 Apr 2007 04:03:27 -0700

On Apr 20, 1:02 pm, beewo... wrote:
> On Apr 20, 6:30 am, kunzmilan wrote:
>
> > On Apr 18, 11:30 am, mjb wrote:> How can we specify a plane (and rotations) in 'n' dimensions?
>
> > Points (3,0), (2,1), (1,2), (0,3) lie on a line in 2-dimensions. On
> > this line, all poins lie which sum of coordinates is 3. The rotation
> > axis is perpendicular to it.
> > Similarly we can construct triangles and tetrahedrons with constant
> > sums of coordinates,and analogically n-dimensional simplices with
> > constant sum of coordinates.
> > kunzmilan
>
> You need to allow weighted sums to specify a general simplex.

N-dimensional simplices with sums 0, 1, 2, ... form simply plane
complexes.
Lines fill plane, and planes 3-dimensional space, etc..
>From 2 dimensional simplex we get a square by truncation points with
too high coordinates. A square with sides 0-3, includes the point
(3,3) but no points (6,0), (5,1), (4,2).
kunzmilan


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