The acute angle between the diagonals OG and AD of the cuboid (shown in the figure) is

cos^-1\((\frac{1}{9})\)

$\vec{OG}=(2\hat i + 2 \hat j + \hat k)$

$\vec{AD}=\vec{OD}-\vec{OA}=(2\hat i+\hat k)-(2\hat j)$

$= 2\hat i - 2 \hat j + \hat k$

Let θ be the angle between $\vec{OG} \And \vec{AD}$

$\cos θ =\frac{4-4+1}{3×3}⇒\cos θ =\frac{1}{9}$

$∴θ =\cos^{-1}(\frac{1}{9})$