If alpha is the angle between any two diagona | vedclass

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(C) Let the vertices of the cube be $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)$.Consider two diagonals of the cube, for e

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(C) Let the vertices of the cube be $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)$.Consider two diagonals of the cube, for example, the vectors $\vec{d_1} = (a,a,a)$ and $\vec{d_2} = (-a,a,a)$.The angle $\alpha$ between them is given by $\cos \alpha = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{-a^2+a^2+a^2}{3a^2} = \frac{1}{3}$.Now, consider a diagonal of the cube $\vec{d_1} = (a,a,a)$ and a diagonal of a face that intersects it, for example, $\vec{f} = (a,a,0)$.The angle $\beta$ between them is given by $\cos \beta = \frac{\vec{d_1} \cdot \vec{f}}{|\vec{d_1}| |\vec{f}|} = \frac{a^2+a^2+0}{\sqrt{3a^2} \sqrt{2a^2}} = \frac{2a^2}{a^2\sqrt{6}} = \sqrt{\frac{2}{3}}$.Thus, $\cos^2 \beta = \frac{2}{3}$.Finally, $\cos \alpha + \cos^2 \beta = \frac{1}{3} + \frac{2}{3} = 1$.

If $\alpha$ is the angle between any two diagonals of a cube and $\beta$ is the angle between a diagonal of a cube and a diagonal of its face, which intersects this diagonal of the cube, then $\cos \alpha + \cos^2 \beta =$

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