The angle between a diagonal of a cube and one of its edges is

Let $\mathbf{a}=a_1\mathbf{i},\mathbf{b}=a_1\mathbf{j}\text{ and }\mathbf{c}=a_1\mathbf{k}$ Then the  vector $\mathbf{d}=a_1(\mathbf{i}+\mathbf{j}+\mathbf{k})$ is a diagonal of the cube. The angle $\theta$ between $\mathit{d}$ and $\mathit{a}$ is given by

$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{d}}{\left|\mathbf{a}\right|\left|\mathbf{d}\right|}=\frac{a_1^2}{a_1\left(\sqrt{3a_1^2}\right)}=\frac{1}{\sqrt{3}}$.