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

Consider a cube of side length $a$ with one of its vertices at the origin $(0,0,0)$ and edges along the coordinate axes.

The vector representing an edge along the x-axis is $\vec{e} = a\hat{i}$.

The vector representing the diagonal of the cube from the origin is $\vec{d} = a\hat{i} + a\hat{j} + a\hat{k}$.

The angle $\theta$ between the diagonal and the edge is given by $\cos\theta = \dfrac{\vec{d} \cdot \vec{e}}{|\vec{d}| |\vec{e}|}$.

$\vec{d} \cdot \vec{e} = (a)(a) + (a)(0) + (a)(0) = a^2$

$|\vec{d}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$

$|\vec{e}| = a$

$\cos\theta = \dfrac{a^2}{a\sqrt{3} \times a} = \dfrac{1}{\sqrt{3}}$

Using trigonometric identities, if $\cos\theta = \dfrac{1}{\sqrt{3}}$, then the adjacent side is $1$ and the hypotenuse is $\sqrt{3}$.

The opposite side is $\sqrt{(\sqrt{3})^2 - 1^2} = \sqrt{2}$.

Therefore, $\tan\theta = \dfrac{\sqrt{2}}{1} = \sqrt{2}$.

$\theta = \tan^{-1}(\sqrt{2})$

Answer: $\tan^{-1}\left(\sqrt{2}\right)$