Equations of a Plane/Hyperplane

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To find the equation of a plane in 3-dimensional space given three points, substitute the coordinates of the points into the plane equation and solve the resulting system of equations. The discussion highlights that the normal vector can be determined using the cross product of vectors formed by the points. For generalizing to n-dimensional space, the concept of using determinants and vectors is suggested, with hints towards matrix methods for a more elegant solution. The conversation emphasizes the importance of understanding the normal vector in both 3D and n-dimensional cases. Overall, the thread provides insights into solving for planes and hyperplanes in various dimensions.
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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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