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(B) Let the cube have side length $a$. Place the cube in a coordinate system such that its vertices are at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0

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$\operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}\right)$

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(B) Let the cube have side length $a$. Place the cube in a coordinate system such that its vertices are at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $(a,a,a)$.Consider the diagonal of the cube starting from the origin $(0,0,0)$ to $(a,a,a)$. The vector representing this diagonal is $\vec{d_1} = a\hat{i} + a\hat{j} + a\hat{k}$.Consider a face diagonal starting from the same origin $(0,0,0)$ to $(a,a,0)$. The vector representing this diagonal is $\vec{d_2} = a\hat{i} + a\hat{j} + 0\hat{k}$.The angle $	heta$ between these two vectors is given by $\cos 	heta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$.Calculating the dot product: $\vec{d_1} \cdot \vec{d_2} = (a)(a) + (a)(a) + (a)(0) = 2a^2$.Calculating the magnitudes: $|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$ and $|\vec{d_2}| = \sqrt{a^2 + a^2 + 0} = a\sqrt{2}$.Thus,$\cos 	heta = \frac{2a^2}{(a\sqrt{3})(a\sqrt{2})} = \frac{2}{\sqrt{6}} = \frac{\sqrt{4}}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.Therefore,$	heta = \operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}ight)$.

The angle between a diagonal of a cube and a diagonal of one of its faces,which are coterminus,is:

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