The angle between two diagonals of a cube is

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(B) Let the side length of the cube be $a$. We place the cube in the coordinate system such that one vertex is at the origin $O(0, 0, 0)$. The coordin

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$\cos^{-1}(1/3)$

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(B) Let the side length of the cube be $a$. We place the cube in the coordinate system such that one vertex is at the origin $O(0, 0, 0)$. The coordinates of the vertices are $O(0, 0, 0)$,$A(a, 0, 0)$,$B(0, a, 0)$,$C(0, 0, a)$,$D(a, a, a)$,etc.Consider two diagonals of the cube,for example,the diagonal from $(0, 0, 0)$ to $(a, a, a)$ and the diagonal from $(a, 0, 0)$ to $(0, a, a)$.The direction vectors of these diagonals are $\vec{v_1} = (a, a, a)$ and $\vec{v_2} = (-a, a, a)$.The angle $	heta$ between these two vectors is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2$.$|\vec{v_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$ and $|\vec{v_2}| = \sqrt{(-a)^2 + a^2 + a^2} = a\sqrt{3}$.Therefore,$\cos 	heta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.Thus,$	heta = \cos^{-1}(1/3)$.

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