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Find the angle between a diagonal of a cube and one of its edges
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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 $$\theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.74^{\circ}$$
by Platinum (130,566 points)

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