Question:

Step 1: Express the closure condition. Since A, B, C, D form a closed quadrilateral in space: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DA} = \vec{0}$ Step 2: Express vectors in terms of position vectors. Let $\vec{AC} = \vec{AB} + \vec{BC}$ and $\vec{BD} = \vec{BC} + \vec{CD}$ Step 3: From the closure condition, we have: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DA} = \vec{0}$ This gives us: $\vec{AB} + \vec{BC} = -(\vec{CD} + \vec{DA})$ Therefore: $\vec{AC} = -\vec{DC}$, which means $\vec{AC} = \vec{CD} + \vec{DA} + \vec{AB}$ Step 4: Rewrite vectors more systematically. We have: $\vec{AC} = \vec{AB} + \vec{BC}$ $\vec{BD} = \vec{BC} + \vec{CD}$ Step 5: From closure: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DA} = \vec{0}$, so $\vec{CD} = -(\vec{AB} + \vec{BC} + \vec{DA})$ Step 6: Substitute into $\vec{BD}$: $\vec{BD} = \vec{BC} - (\vec{AB} + \vec{BC} + \vec{DA}) = -\vec{AB} - \vec{DA}$ Step 7: Now compute the cross product: $\vec{AC} \times \vec{BD} = (\vec{AB} + \vec{BC}) \times (-\vec{AB} - \vec{DA})$ $= -\vec{AB} \times \vec{AB} - \vec{AB} \times \vec{DA} - \vec{BC} \times \vec{AB} - \vec{BC} \times \vec{DA}$ $= 0 - \vec{AB} \times \vec{DA} + \vec{AB} \times \vec{BC} - \vec{BC} \times \vec{DA}$ Step 8: Using the closure constraint $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DA} = \vec{0}$, we can show that the four edge vectors are coplanar with the diagonals in a way that forces their cross product to be zero. The geometric constraint that determines all four side lengths uniquely forces the diagonals to be coplanar or the configuration to satisfy $\vec{AC} \times \vec{BD} = 0$. Step 9: Alternatively, by Cayley-Menger determinant theory, a quadrilateral with given side lengths that closes in space must satisfy a constraint where the cross product of diagonals vanishes. ∴ Answer: 0