From 2-D to Higher Dot Products

Consider a regular tetrahedron with vertices $A, B, C, D$ and center $O.$ Find $\cos\angle AOB.$

Enter your answer in simpest form and without any extra spaces or LaTeX formatting.

Hint: Use the following: $$\cos\angle AOB = \frac{\vec{OA}\cdot\vec{OB}}{\left|\vec{OA}\right|\left|\vec{OB}\right|}.$$ Now, you need to figure out: in what coordinate system can you compute all of these in the most convenient way?