Proving Uniqueness of Affine Plane Containing S & Weak-Parallel to T

  • Thread starter Thread starter A_B
  • Start date Start date
  • Tags Tags
    Plane Uniqueness
A_B
Messages
87
Reaction score
1

Homework Statement


S and T are two affine lines in \mathbb{A}^3 that are not parallel and S\cap T=\emptyset.

Show there is a unique affine plane R that contains S and is weak parallel with T.

The Attempt at a Solution


Existence is easy, if S=p+V and T=q+W then R=p+(V+W) satisfies the conditions.

To prove uniqueness I assume planes R and Q both satisfy all conditions. They both contain S so they can be written as p+(V+vectorspace). That vectorspace must be W since the planes must be weak parallel with Tso both R and Q are equal to p+(V+W).

Is this good?
If it is, it still seems very ugly to me, is there a better way to do it?

Thanks
Alex
 
Physics news on Phys.org
What is ugly about the solution?? It seems nice...

There is a little detail missing though. For uniqueness, you must use somewhere that S and T are not parallel.
 
Thanks micromass!

Ok,The direction of R and Q must contain W, since V does not contain W, and is not a subspace of W (S and T are not parrallel and have dimensions 1) , "vectorspace" must be W
good now?

I feel it's ugly because it largely repeats the construction for existence.

thanks again.
 
A_B said:
Thanks micromass!

Ok,The direction of R and Q must contain W, since V does not contain W, and is not a subspace of W (S and T are not parrallel and have dimensions 1) , "vectorspace" must be W
good now?

That's better.

I feel it's ugly because it largely repeats the construction for existence.

That's exactly why I think it's a pretty argument :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top