The angle between two diagonals of a cube is:

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(C) Consider a cube with side length $1$. Let the vertices be represented in a $3D$ coordinate system as shown in the figure.Let the two diagonals be

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$\cos^{-1}\left(\frac{1}{3}\right)$

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(C) Consider a cube with side length $1$. Let the vertices be represented in a $3D$ coordinate system as shown in the figure.Let the two diagonals be $\vec{OA}$ and $\vec{BC}$.The coordinates are $O(0,0,0)$,$A(1,1,1)$,$B(1,0,0)$,and $C(0,1,1)$.The vector $\vec{OA} = (1-0, 1-0, 1-0) = (1, 1, 1)$.The vector $\vec{BC} = (0-1, 1-0, 1-0) = (-1, 1, 1)$.The angle $	heta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.$\vec{OA} \cdot \vec{BC} = (1)(-1) + (1)(1) + (1)(1) = -1 + 1 + 1 = 1$.$|\vec{OA}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.$|\vec{BC}| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$.Therefore,$\cos 	heta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$.Thus,$	heta = \cos^{-1}\left(\frac{1}{3}ight)$.

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