The angle between a diagonal of a cube and on | vedclass

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(A) Let the side length of the cube be $a$. Place the cube in a coordinate system such that one vertex is at the origin $(0,0,0)$ and the edges lie al

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

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(A) Let the side length of the cube be $a$. Place the cube in a coordinate system such that one vertex is at the origin $(0,0,0)$ and the edges lie along the $x, y,$ and $z$ axes.The vertices are $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $(a,a,a)$.Consider the edge along the $x$-axis, represented by the vector $\vec{u} = (a, 0, 0)$.Consider the diagonal of the cube from the origin to $(a, a, a)$, represented by the vector $\vec{v} = (a, a, a)$.The angle $	heta$ between these two vectors is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.$\vec{u} \cdot \vec{v} = (a)(a) + (0)(a) + (0)(a) = a^2$.$|\vec{u}| = \sqrt{a^2 + 0^2 + 0^2} = a$.$|\vec{v}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$.$\cos 	heta = \frac{a^2}{a \cdot a\sqrt{3}} = \frac{1}{\sqrt{3}}$.Therefore, $	heta = \cos^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

The angle between a diagonal of a cube and one of its edges is ...................

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