Equation of a plane determined by three or four points in R4

[Rorschach]

How do you find the equation of a plane in determined by say three or four points in $\mathbb{R}^4$?

In $\mathbb{R}^{3}$ we would use the cross product to find normal vector, but that doesn't apply in $\mathbb{R}^4$.

 

[HOI]

Exactly what do you mean by a "plane" in four dimensions? A two dimensional, flat, surface or a three dimensional, flat, subspace?

If you mean a two dimensional surface then it can be determined by three points. Calling the extra dimension "t", we can write the plane as x= Au+ Bv+ C, y= Du+ Ev+ F, z= Gu+ Hv+ I, t= Ju+ Kv+ L. That is, since it is two dimensional it can be written in terms of two parameters and since it is a plane those functions can be taken to be linear. Given three points, \(\displaystyle (x_1, y_1, z_1, t_1)\), \(\displaystyle (x_2, y_2, z_2, t_2)\) and \(\displaystyle (x_3, y_3, z_3, t_3)\), and taking parameters u and v such that the first point corresponds to u= 0, v= 0, the second to u= 1, v= 0, and the third to u= 0. v= 1 we have 12 linear equations
\(\displaystyle x_1= C\)
\(\displaystyle y_1= F\)
\(\displaystyle z_1= I\)
\(\displaystyle t_1= L\)
\(\displaystyle x_2= A+ C\)
\(\displaystyle y_2= D+ F\)
\(\displaystyle z_2= G+ I\)
\(\displaystyle t_2= J+ L\)
\(\displaystyle x_3= B+ C\)
\(\displaystyle y_3= E+ F\)
\(\displaystyle z_3= H+ I\)
\(\displaystyle t_3= K+ L\)
to solve for the 12 coefficients.

If you mean a three dimensional subspace of the four dimensional space you will need four points. A three dimensional subspace will require three parameters. Now we would write, in parametric equations
\(\displaystyle x= Au+ Bv+ Cw+ D\)
\(\displaystyle y= Eu+ Fv+ Gw+ H\)
\(\displaystyle z= Iu+ Jv+ Kw+ L\)
\(\displaystyle t= Mu+ Nv+ Ow+ P\).
Taking the four points to correspond to parameters u= v= w= 0; u= 1, v= w= 0; v= 1, u= w= 0; and w= 1, u= v= 0, we have a total of 16 linear equations to solve for the 16 coefficients.

 

[I like Serena]

How do you find the equation of a plane in determined by say three or four points in $\mathbb{R}^4$?

In $\mathbb{R}^{3}$ we would use the cross product to find normal vector, but that doesn't apply in $\mathbb{R}^4$.

Hi Rorschach,

There is a generalized cross product.
For 4 dimensions it is:
$$\mathbf n = \begin{vmatrix}a_1&a_2&a_3&a_4 \\ b_1&b_2&b_3&b_4 \\ c_1&c_2&c_3&c_4 \\ \mathbf e_1&\mathbf e_2&\mathbf e_3&\mathbf e_4\end{vmatrix} \tag 1$$
where $\mathbf e_i$ are the standard unit vectors, and where $|\cdot|$ denotes the determinant.
The result $\mathbf n$ is a vector that is perpendicular to $\mathbf a, \mathbf b$, and $\mathbf c$.

We can use it to find the equation of a hyperplane, which is a 3D plane in 4 dimensions.More generally, suppose $\mathbf v_1, ..., \mathbf v_k$ are the vectors that span a plane.
Then we can find the normal vectors by solving $\mathbf v_1 \cdot \mathbf n = 0, ..., \mathbf v_k \cdot \mathbf n = 0$.
Or in matrix notation:
$$\begin{bmatrix}\mathbf v_1 & ... & \mathbf v_k\end{bmatrix}^T\mathbf n = 0 \tag 2$$
We can solve it for instance with Gaussian elimination.