What is the angle between two diagonals of a | vedclass

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(B) Let the side length of the cube be $a$.Consider the cube with vertices at $(0,0,0), (a,0,0), (a,a,0), (0,a,0), (0,0,a), (a,0,a), (a,a,a), (0,a,a)$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the side length of the cube be $a$.Consider the cube with vertices at $(0,0,0), (a,0,0), (a,a,0), (0,a,0), (0,0,a), (a,0,a), (a,a,a), (0,a,a)$.Let the two diagonals be $OD$ and $BG$. The coordinates are $O(0,0,0), D(a,a,a)$ and $B(0,a,0), G(a,0,a)$.The direction vectors of these diagonals are $\vec{d_1} = (a-0, a-0, a-0) = a\hat{i} + a\hat{j} + a\hat{k}$ and $\vec{d_2} = (a-0, 0-a, a-0) = a\hat{i} - a\hat{j} + a\hat{k}$.The angle $	heta$ between them is given by $\cos 	heta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|}$.$\vec{d_1} \cdot \vec{d_2} = (a)(a) + (a)(-a) + (a)(a) = a^2 - a^2 + a^2 = a^2$.$|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$ and $|\vec{d_2}| = \sqrt{a^2 + (-a)^2 + a^2} = a\sqrt{3}$.$\cos 	heta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.Therefore,$	heta = \cos^{-1}(1/3)$.

What is the angle between two diagonals of a cube?

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