The eight vertices of a cube centered at $(0, 0, 0)$ of side length $2$ are at $(±1, ±1, ±1)$. Find the four vertices of the cube, starting with $(1, 1, 1)$, that form a regular tetrahedron. Confirm your answer by finding the length of an edge and explaining why all edges have the same length.
I was trying to choose any three other vertices from the cube, ensuring that they are equidistant from the starting vertex $(1, 1, 1)$. To form an equilateral triangle, those are:
$$(1, -1, -1),(-1, -1, 1),(-1, 1, -1)$$
Is this correct?