Given $n$ points in $\mathbb{R}^n$, there exists exactly one convex polytope surrounded by $n$ points on a hyperplane of $\mathbb{R}^n$. When $n$ is 2, it is a line on a plane, and when $n$ is 3, it is a triangle in a space.
$\mathbf{v}_{n, j}$ denotes the $j$-th vector in $\mathbb{R}^n$, and then
$(a).\ (n=2)$ the length of the line is $| \mathbf{v}_{2, 2} - \mathbf{v}_{2, 1} |$,
$(b).\ (n=3)$ the area of the surface is $1/2 \cdot |(\mathbf{v}_{3, 3}-\mathbf{v}_{3, 1})\times(\mathbf{v}_{3, 2}-\mathbf{v}_{3, 1})|$.
How to calculate the volume closed by 4 vectors in $\mathbb{R}^4$? What is the general form when $n>3$?
This shape is known as a simplex. If we rotate and translate the simplex so that one of its vertices is $(0,\ldots,0)$ and all vertices lie in the plane $x_n=0$ then its volume is $$\pm\frac1{(n-1)!}\det\pmatrix{a_{1,1}&a_{1,2}&\cdots&a_{1,n-1}\\ a_{2,1}&a_{2,2}&\cdots&a_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n-1,1}&a_{n-1,2}&\cdots&a_{n-1,n-1}\\}$$ where the other vectors are $(a_{k,1},a_{k,2},\ldots,a_{k,n-1},0)$.