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.

the vertices (1,1,1), (1,0,0), (0,1,0) and (0,0,1) form a tetrahedron. The length of each side is √2

Step-by-step explanation:

The cube has 8 vertices: (0,0,0), (1,1,0), (0,1,0), (1,0,0), (0,0,1), (0,1,1), (1,0,1), (1,1,0) and (1,1,1). The first four of them are the vertices of the bottom square and the last four are the vertices of the upper square of the cube.

We will take two non-consecutive vertices from each square. For the upper one we take (1,1,1) as the problem suggests, and (0,0,1), which is not consecutive from (1,1,1) and its distance is √2. The non consecutive vertices from the bottom square respect to the vertex (1,1,1) are (0,0,0), (0,1,0) and (1,0,0).

We take (0,1,0) and (1,0,0) because (0,0,0) is consecutive from (0,0,1) hence its distance from it is not √2, but 1.

Note that we take (1,1,1), (0,0,1), (0,1,0) and (1,0,0). If we take any two vertices and compare them toguether we will notice that both of those vertices differ in two places and are equal in the other. In the places where they differ one has the value 1 and the other 0, so the distance between those vertices is √(1²+1²) = √2.

Thus, the vertices (1,1,1), (1,0,0), (0,1,0) and (0,0,1) form a tetrahedron.

idk

step-by-step explanation:

your victory is right around the corner. never give up.

- niki minaj

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: