The angle between two diagonals of a cube is:

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(A) Let the vertices of the cube be $O(0,0,0)$, $A(a,0,0)$, $B(a,a,0)$, $C(a,a,a)$, $D(0,a,a)$, $E(0,0,a)$, $F(a,0,a)$, and $G(0,a,0)$.Consider two di

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$\cos ^{-1}\left(\frac{1}{3}\right)$

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(A) Let the vertices of the cube be $O(0,0,0)$, $A(a,0,0)$, $B(a,a,0)$, $C(a,a,a)$, $D(0,a,a)$, $E(0,0,a)$, $F(a,0,a)$, and $G(0,a,0)$.Consider two diagonals of the cube, for example, the diagonal connecting $(0,0,0)$ to $(a,a,a)$ and the diagonal connecting $(a,0,0)$ to $(0,a,a)$.The vector along the first diagonal is $\vec{v_1} = a\hat{i} + a\hat{j} + a\hat{k}$.The vector along the second diagonal is $\vec{v_2} = -a\hat{i} + a\hat{j} + a\hat{k}$.The angle $	heta$ between these two vectors is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2$.$|\vec{v_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$.$|\vec{v_2}| = \sqrt{(-a)^2 + a^2 + a^2} = a\sqrt{3}$.$\cos 	heta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.Therefore, $	heta = \cos ^{-1}\left(\frac{1}{3}ight)$.

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