Three points in $\mathbb R^5$

Question

How can I find the affine subspace $\pi$ that contains the following points?

$ P:(1,0,2,0,1)\\ Q:(0,1,0,1,0)\\ R:(2,0,1,0,2)$

Surely it is easy in $\mathbb R^3$, but I don't understand how can I determinate $\pi$ and the distance between the points in $\mathbb R^5$! (I need these information because I have to determine the area of the triangles that has P Q and R as vertices)

Answer 1

If what you’re ultimately after is the area of the triangle, you can simply use Heron’s formula $$\sqrt{s(s-a)(s-b)(s-c)},$$ where $a$, $b$ and $c$ are the side lengths and $s$ is the semi-perimeter $\frac12(a+b+c)$. The side lengths are just the distances between the points, which you can compute using the usual formula, e.g., $$a=PQ=\sqrt{(1-0)^2+(0-1)^2+(2-0)^2+(0-1)^2+(1-0)^2} = 2\sqrt2.$$

A parametric equation for the plane isn’t really any different than it would be for points in $\mathbb R^3$. Pick one as your “base” point and compute the two vectors to the other points, e.g., $P+s(Q-P)+t(R-P)$. Alternatively, you could use three parameters: $${m_1P+m_2Q+m_3R\over m_1+m_2+m_3}.$$ (This should look familiar to you if you’re encountered barycentric coordinates.)

Since you’re working in $\mathbb R^5$, you will need three implicit Cartesian equations of the form $a_0+a_1x_1+a_2x_2+a_3x_3+a_4x_4+a_5x_5=0$. Plugging in the coordinates of the three points will generate a system of three linear equations in the coefficients $a_i$. A basis of the three-dimensional solution space will give you three independent equations. In other words, compute a basis for the null space of the matrix $$\begin{bmatrix}1&1&0&2&0&1\\1&0&1&0&1&0\\1&2&0&1&0&2\end{bmatrix}.$$ This, too, is no different from the way one might go about doing this in $\mathbb R^3$, but in that case the null space will only be one-dimensional.

Answer 2

I'm not sure exactly what form you hope for for a presentation of $\pi$, but let me address the part your question about how to determine the area of the triangle with those vertices.

You can find the area of a triangle knowing an angle along with the lengths of the two incident sides. In particular, you need only the sine of that angle. Consider angle $\angle PQR$. The length of $PQ$ is $\sqrt{(1-0)^2+(0-1)^2+(2-0)^2+(0-1)^2+(1-0)^2}=2\sqrt{2}$, and the length of $QR$ is $\sqrt{(0-2)^2+(1-0)^2+(0-1)^2+(1-0)^2+(0-2)^2}=\sqrt{11}$. We can get the cosine using dot products $$(P-Q)\cdot (R-Q)=\lVert PQ\rVert \lVert QR\rVert\cos(\angle PQR),$$ which evaluates to $8=2\sqrt{22}\cos(\angle PQR)$, hence $\cos(\angle PQR)=\sqrt{8/11}$. Using the identity $\cos^2\theta+\sin^2\theta=1$, we see $\lvert\sin(\angle PQR)\rvert = \sqrt{3/11}$. Thus, the angle of the triangle is $\frac{1}{2}\lVert PQ\rVert \lVert QR\rVert\lvert\sin(\angle PQR)\rvert=\sqrt{6}$.