Rotation in multidimensional space

Question

I have a regular pentagon, with coordinates

\begin{align*} x(1) &= (2, 0, 1, 1, 0) \\ x(2) &= (1, 1, 0, 2, 0) \\ x(3) &= (0, 2, 0, 1, 1) \\ x(4) &= (0, 1, 1, 0, 2) \\ x(5) &= (1, 0, 2, 0, 1) \end{align*}

the side of pentagon is 2. I would like to know if there is way to find hyperplane, so that this pentagon in respect to the hyperplane be 2 dimensional?

Remark: the central point can be found by directive ways

$$ \left(\frac{4}{5}+\frac{1}{5} \sqrt{5 \left(\frac{1}{\sin \left(\frac{\pi }{5}\right)}\right)^2-14}\right) (1,1,1,1,1)$$

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\vx}{\mathbf{x}}\newcommand{\vv}{\mathbf{v}}$Generally, if $\vx_{0}$, ..., $\vx_{k}$ are coplanar points of $\Reals^{n}$ such that $\vx_{0}$, $\vx_{1}$, and $\vx_{2}$ are non-collinear, then the displacement vectors $\vv_{1} = \vx_{1} - \vx_{0}$ and $\vv_{2} = \vx_{2} - \vx_{0}$ are linearly independent, and parallel to the plane containing all the points; that is, each $\vx_{j}$ is uniquely of the form $$ \vx_{0} + s \vv_{1} + t \vv_{2} \tag{1} $$ for some scalars $s$, $t$.

The set of points of the form (1) is a parametric description of the plane you seek. (Since the plane lies in $\Reals^{5}$, you would need three equations to describe this plane as the solution set of a linear system.)

Presumably you'll take $\vx_{0}$ to be the center of the pentagon, and $\vx_{1}$, ..., $\vx_{5}$ to be the vertices. In this case, if you perform the Gram-Schmidt algorithm on the ordered basis $(\vv_{1}, \vv_{2})$, obtaining an orthonormal basis $(\Basis_{1}, \Basis_{2})$ with $\vv_{1} = r\Basis_{1}$, then for each $j = 1$, ..., $5$, $$ \vx_{j} = \vx_{0} + (r\cos \theta_{j}) \Basis_{1} + (r\sin \theta_{j}) \Basis_{2},\qquad \theta_{j} = \tfrac{2(j-1)\pi}{5}. $$

Answer 2

The figure you have described is not a regular pentagon. A regular pentagon with side $2$ would have diagonal $1 + \sqrt 5$, but the corresponding "diagonal" of this figure (for example, the segment from $x(1)$ to $x(3)$) has length $\sqrt{10}$.

We can take five vectors parallel to the five sides of the "pentagon" by taking the differences of coordinates of successive vertices, $x(k+1) - x(k)$. If the five vertices all lay in one two-dimensional plane then these five vectors would all lie in a two-dimensional subspace parallel to that plane. But in fact we can take linear combinations of these five vectors to form a set of four vectors such as the following:

$$ v_1 = \pmatrix{1 \\ -1 \\ 0 \\ 0 \\ 0}, \quad v_2 = \pmatrix{0 \\ 1 \\ -1 \\ 0 \\ 0}, \quad v_3 = \pmatrix{0 \\ 0 \\ 1 \\ -1 \\ 0}, \quad v_4 = \pmatrix{0 \\ 0 \\ 0 \\ 1 \\ -1}, \quad $$

These vectors span a four-dimensional subspace, so the five vertices do not lie in a single plane.

So the five points do not form any object we would normally call a pentagon, and the object they do form is four-dimensional.

You may want to clarify what you mean by "two-dimensional" in the question.