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Consider a unit cube with one one corner at the origin what is the angle between any two diagonals?

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Answer:

1.231 rad

Explanation:

Suppose that the vertex opposite the origin is (a, a, a), therefore, the geometric vector A = (a, a, a) is a diagonal of the cube. There are two opposite vertices (0, a, 0) and (a, 0, a), with which B = (a, 0, a) - (a, 0, a) = (a, -a, a) is the diagonal that intersects the first diagonal.

Now,
\theta = arccos((A\bullet B)/(\left\| A \right\| \left\| B \right\|)) = arccos((a^2)/(3a^2)) = arccos((1)/(3)) = 1.231 rad

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