Equations of a Plane/Hyperplane

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Homework Help Overview

The discussion revolves around finding the equation of a plane given three points in three-dimensional space and generalizing this concept to n-dimensional space. The subject area includes geometry and linear algebra, particularly focusing on planes and hyperplanes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to substitute points into the plane equation and expresses uncertainty about generalizing to n dimensions. Some participants suggest finding the normal vector and using determinants, while others question the method of using coordinates versus vectors.

Discussion Status

Participants are exploring different approaches to the problem, including hints about using normal vectors and determinants. There is an ongoing exchange of ideas regarding the transition from three-dimensional to n-dimensional cases, with no explicit consensus reached yet.

Contextual Notes

There is mention of potential constraints related to the methods used, such as whether to employ coordinate systems or vector representations. The original poster expresses difficulty with the algebra involved in the n-dimensional case.

AngelofMusic
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Homework Statement



Given 3 points in 3-dimensional space, Find the equation of a plane containing those 3 points. How would you generalize this to n points in n-dimensional space?

Homework Equations



Equation of a Plane: ax + by + cz + d = 0 (1)
Equation of a Hyperplane: a_1 x_1 + ... + a_n x_n + d=0 (2)

The Attempt at a Solution



For the 3-D case, I simply substituted the 3 points into the equation. For example, p1 = (x1, y1, z1), p2 = (x2, y2, z2), p3=(x3, y3, z3).

ax1 + by1 + cz1 = -d
ax2 + by2 + cz2 = -d
ax3 + by3 + cz3 = -d

ax1 + by1 + cz1 = ax2 + by2 + cz2
=> a(x1-x2) + b(y1-y2) + c(z1-z2) = 0

Similarly,

a(x1-x3) + b(y1-y3) + c(z1-z3) = 0

and

a(x2-x3) + b(y2-y3) + c(z2-z3) = 0

Which is 3 equations with 3 unknowns that can be solved.

Where I'm stumped is how I would generalize this to the n-dimensional case. I have a feeling that maybe I'm doing this the 'dumb' way and there's a far more elegant solution (perhaps involving matrices?) for solving the 3-D case that will extend more easily to the n-dimensional case.

Any help would be appreciated!
 
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AngelofMusic said:
Given 3 points in 3-dimensional space, Find the equation of a plane containing those 3 points. How would you generalize this to n points in n-dimensional space?

Hi AngelofMusic! :smile:

Hint: find the normal (and don't use coordinates, use whole vectors) …

try the 3D case first: for three vectors a b and c, what can you say about the normal? :wink:
 
The plane is the determinant:

\begin{vmatrix}<br /> x-x_1 &amp; y-y_1 &amp; z-z_1\\ <br /> x_2-x_1 &amp; y_2-y_1 &amp; z_2-z_1 \\ <br /> x_3-x_1 &amp; y_3-x_1 &amp; z_3-z_1 <br /> \end{vmatrix}=0

Where M_1(x_1,y_1), M_2(x_2,y_2), M_3(x_3,y_3).
 
tiny-tim said:
Hi AngelofMusic! :smile:

Hint: find the normal (and don't use coordinates, use whole vectors) …

try the 3D case first: for three vectors a b and c, what can you say about the normal? :wink:

One such normal would be n = (a-b)x(b-c), right? So for the n-dimensional case, would I just repeatedly take cross products of the vectors? The wikipedia page on surface normals has a neat solution where n=(AA^T + bb^T)^{-1} b, but they don't show how that is derived.

Thanks a lot for the help so far! I can deal with 3-D cases relatively well, but my mind just isn't wrapping around the n-dimensional algebra very well at the moment.
 

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