Question

Find the angle between a diagonal of a cube and one of its edges

Answer 1

Let $k$ be the length of an edge and introduce a coordinate system. If we let $\mathbf{u}_{1}=(k, 0,0), \mathbf{u}_{2}=(0, k, 0)$, and $\mathbf{u}_{3}=(0,0, k)$, then the vector $$ \mathbf{d}=(k, k, k)=\mathbf{u}_{1}+\mathbf{u}_{2}+\mathbf{u}_{3} $$ is a diagonal of the cube. It follows that the angle $\theta$ between $\mathbf{d}$ and the edge $\mathbf{u}_{1}$ satisfies $$ \cos \theta=\frac{\mathbf{u}_{1} \cdot \mathbf{d}}{\left\|\mathbf{u}_{1}\right\|\|\mathbf{d}\|}=\frac{k^{2}}{(k)\left(\sqrt{3 k^{2}}\right)}=\frac{1}{\sqrt{3}} $$ With the help of a calculator we obtain $$ \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.74^{\circ} $$