The angle between any two diagonals of a cube | vedclass

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(A) Let the side length of the cube be $a$. We consider a cube with vertices at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $

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

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(A) Let the side length of the cube be $a$. We consider a cube with vertices at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $(a,a,a)$.Two diagonals of the cube can be represented by vectors $\vec{d_1}$ and $\vec{d_2}$.Let $\vec{d_1}$ be the vector from $(0,0,0)$ to $(a,a,a)$,so $\vec{d_1} = a\hat{i} + a\hat{j} + a\hat{k}$.Let $\vec{d_2}$ be the vector from $(a,0,0)$ to $(0,a,a)$,so $\vec{d_2} = (0-a)\hat{i} + (a-0)\hat{j} + (a-0)\hat{k} = -a\hat{i} + a\hat{j} + a\hat{k}$.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: $\vec{d_1} \cdot \vec{d_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2$.Calculating the magnitudes: $|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$ and $|\vec{d_2}| = \sqrt{(-a)^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$.Thus,$\cos 	heta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.Therefore,$	heta = \cos ^{-1}\left(\frac{1}{3}ight)$.

The angle between any two diagonals of a cube is

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