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

• MHB
• Rorschach
In summary, to find the equation of a plane in $\mathbb{R}^4$ determined by three or four points, we can use either a two-dimensional or three-dimensional approach. For a two-dimensional surface, we can use linear equations with two parameters to find the coefficients. For a three-dimensional subspace, four points and three parameters are needed to solve for the coefficients. In both cases, we can use the generalized cross product or solve a system of equations to find the normal vector, which is perpendicular to the plane.
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 $\mathbb{R}^4$.

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.

Rorschach said:
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.

## 1. What is the equation of a plane in R4?

The equation of a plane in R4 is represented as Ax + By + Cz + Dw = E, where A, B, C, D, and E are constants and x, y, z, and w are variables.

## 2. How many points are needed to determine a plane in R4?

A plane in R4 can be determined by three or four points. Three non-collinear points are needed for a unique solution, while four points can be used to check for consistency.

## 3. Can the equation of a plane in R4 be written in different forms?

Yes, the equation of a plane in R4 can also be written in vector form as p = p0 + t1v1 + t2v2, where p is a point on the plane, p0 is a known point on the plane, and v1 and v2 are two non-parallel vectors in the plane.

## 4. How do you determine if a point lies on a plane in R4?

If a point (x,y,z,w) satisfies the equation Ax + By + Cz + Dw = E, then it lies on the plane. Alternatively, if the vector from a known point on the plane to the given point is orthogonal to the plane's normal vector, then the point lies on the plane.

## 5. Can the equation of a plane in R4 be used to find the distance between the plane and a point?

Yes, the distance d from a plane Ax + By + Cz + Dw = E to a point (x0,y0,z0,w0) can be found using the formula d = |Ax0 + By0 + Cz0 + Dw0 - E| / sqrt(A^2 + B^2 + C^2 + D^2).

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