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# /sci/ - Science & Math

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Anonymous 06/28/14(Sat)17:12 UTC+1 No.6616408 Report

Any 2 points in space are colinear, and a line is a 1-dimensional object.
Any 3 points in space are coplanar, and a plane is a 2-dimensional object.

So what 3-dimensional object do any 4 points in space co-exist on? Can we use this to go even further, say a 4-dimensional object with 5 correlated points? Or does the basic fact that space is 3-dimensional only limit this to the 3rd dimension?
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Anonymous 06/28/14(Sat)17:16 UTC+1 No.6616411 Report

>>6616408
>So what 3-dimensional object do any 4 points in space co-exist on?
it should be fucking obvious that its a volume.
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Your ten last books 06/28/14(Sat)17:16 UTC+1 No.6616412 Report

>>6616408
Here you are:
http://en.wikipedia.org/wiki/Hyperplane
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Anonymous 06/28/14(Sat)17:17 UTC+1 No.6616414 Report

http://en.wikipedia.org/wiki/Hyperplane
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Anonymous 06/28/14(Sat)17:18 UTC+1 No.6616417 Report

>>6616412
Damn you beat me :(
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Anonymous 06/28/14(Sat)17:24 UTC+1 No.6616431 Report

>>6616417
froggy wins
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Anonymous 06/28/14(Sat)17:45 UTC+1 No.6616458 Report

>>6616408
Yes. Fist let me introduce a more formal definition of "colinear" , "coplanar", ...ect.

Introducing the notion of a set: \mathbb{R}^{n}.

Two points can form a one dimensional subset in \mathbb{R}^{n} by this process:

let \mathbf{v}_1 and \mathbf{v}_2 be two points in \mathbb{R}^{n}. i.e. \mathbf{v}_1 = (x_1,x_2,..., x_n). and \mathbf{v}_2 = (y_1,y_2,..., y_n).
You can take the difference of the two vectors to get a new vector \mathbf{v}_3 :
\mathbf{v}_3 = \mathbf{v}_2 - \mathbf{v}_1.
The subset \mathbf{S} is defined by:
\mathbf{S} := \left \{ \mathbf{v_1} + a \mathbf{v}_3 : a \in \mathbb{R} \right \}

The two points are said to be coplanar if they are in the same subset \mathbf{S} .

For higher dimensions, consider a collection of k vectors in \mathbb{R}^n .
\mathbf{v}_1, \mathbf{v}_2, ... , \mathbf{v}_k \in \mathbb{R}^n .

and consider the k-1 vectors:
\mathbf{v}_{12} = \mathbf{v}_2 - \mathbf{v}_1
\mathbf{v}_{13} = \mathbf{v}_3 - \mathbf{v}_1
....
\mathbf{v}_{1k} = \mathbf{v}_k - \mathbf{v}_1 .

These can form a subset of dimension k-1 s.t.
\mathbf{S}_{k-1} := \left \{ \mathbf{v}_1 + c_1\mathbf{v}_{12} + c_2\mathbf{v}_{13} + ... + c_{k-1}\mathbf{v}_{1k}: c_i \in \mathbb{R} \right \}
Points are said to be co-hyperplanar to these points if they are found in the same subset \mathbf{S}_{k-1} .
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Anonymous 06/28/14(Sat)19:20 UTC+1 No.6616627 Report

>>6616458
>rewriting wikpedia
>how courageous you are
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Anonymous 06/29/14(Sun)05:01 UTC+1 No.6617395 Report

>>6616627
I did not even read the Wikipedia article. I started with the basic definitions of planes and work from there.
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Anonymous 06/29/14(Sun)07:44 UTC+1 No.6617648 Report

>>6617395
then you should participate in wikipedia.
Wiki pages aren't pruned automatically.

It'll be kinda rewarding for you
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Anonymous 06/29/14(Sun)23:34 UTC+1 No.6618678 Report

>>6617648
I actually have contributed to a few articles.

Here's one of my contributions to the Area of an ellipse:

It's the one involving the coordinate transformation \mathbf{T}(r,\theta) (last formula). It's been there for a year. so I guess people like it. But I contribute little, because I don't feel confident all the time that my way is the right way. It just so happened that I knew a clean way to prove something.
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Anonymous 06/29/14(Sun)23:36 UTC+1 No.6618682 Report

>>6618678