Find the angle between two diagonals of a cube.

Let ‘a’ be the length of each side of cube. Let one of the vertex of the cube is the origin ‘O’ and the coordinate axes along the 3 edges $\overline{\mathrm{OA}},\overline{\mathrm{OB}}\text{ and }\overline{\mathrm{OC}}\text{ are passing through origin }$
$\text{ The four diagonals are }\overline{\mathrm{OG}},\overline{\mathrm{AF}},\overline{\mathrm{CE}},\overline{\mathrm{BD}}$
$\begin{array}{l}\mathrm{O}=(0,0,0)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{A}=(\mathrm{a},0,0)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{B}=(0,\mathrm{a},0)\\ \mathrm{C}=(0,0,\mathrm{a})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{G}=(\mathrm{a},\mathrm{a},\mathrm{a})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{E}=(\mathrm{a},\mathrm{a},0)\\ \mathrm{F}=(0,\mathrm{a},\mathrm{a})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{D}=(\mathrm{a},0,\mathrm{a})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

$\text{ dr's of }\overline{\mathrm{OG}}=\left({\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{y}}_2-{\mathrm{y}}_1,{\mathrm{z}}_2-{\mathrm{z}}_1\right)$
= (a – 0, a – 0, a – 0) = (a, a, a)
$\text{ dc's of }\overline{\mathrm{OG}}$
$\begin{array}{l}=\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}}\right)\\ =\left(\frac{\mathrm{a}}{\mathrm{a}\sqrt{3}},\frac{\mathrm{a}}{\mathrm{a}\sqrt{3}},\frac{\mathrm{a}}{\mathrm{a}\sqrt{3}}\right)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\end{array}$
d.r’s of $\overline{\mathrm{AF}}=(0-\mathrm{a},\mathrm{a}-0,\mathrm{a}-0)=(-\mathrm{a},\mathrm{a},\mathrm{a})$
dc’s of AF
$\begin{array}{l}=\left(\frac{-\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}}\right)\\ =\left(\frac{-\mathrm{a}}{\mathrm{a}\sqrt{3}},\frac{\mathrm{a}}{\mathrm{a}\sqrt{3}},\frac{\mathrm{a}}{\mathrm{a}\sqrt{3}}\right)=\left(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\end{array}$
Let $\mathrm{\theta }$ be the angle between the diagonals $\overline{OG} and \overline{AF}$ then
$\begin{array}{l} \\ \cos \mathrm{\theta }=\left|{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2\right|\\ =\left|\frac{1}{\sqrt{3}}\left(-\frac{1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)\right|\\ =\left|-\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right|\\ \cos \mathrm{\theta }=\frac{1}{3}\Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{1}{3}\right)\end{array}$