The angle between a diagonal of a cube and the diagonal of a face of the cube is

Let d.r's of diagonal of cube  $=(\mathrm{1,1},1)$ 
d. r' s of diagonal of face of cube  $(\mathrm{0,1},1)$

$$\begin{gathered} \cos \mathrm{\theta }=\frac{1+1}{\sqrt{3}\cdot \sqrt{2}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}} \\ \Rightarrow \mathrm{ }\mathrm{\theta }={\cos }^{-1} \sqrt{\frac{2}{3}} \end{gathered}$$