A cube has 8 vertices (±1, ±1, ±1). We draw vectors from the center (0, 0, 0) to three top vertices. What is the volume of the parallelepiped formed?

Question

for clarification: the top vertices are those with z = 1.

I figured I could simply use $P = (1,1,1) Q = (-1,1,1) R = (1,-1,1) S = (-1,-1,1)$

Which would give $ PQ = <-2,0,0> PR = <0,-2,0> PS = <-2,-2,0> $

And then use the formula for the volume of a parellelepiped $PQ . (PR X PS)$ however this gave me a result of $0$. I do not think I made a mistake in calculating, so I'm assuming I am not using the correct method.

Any help would be greatly appreciated! thank you.

Answer 1

Choose 3 of $P,Q,R, S$ as above. It doesn't matter which 3.

The volume of the parallelepiped, formed with theses points and the origin equals the absolute value of determinant of the matrix formed from those points.

$\left| \begin{array}{} 1&1&1\\-1&1&1\\1&-1&1 \end{array}\right| = 4$