The angle between any two diagonals of a cube is

Let ‘a’ is the length of each side of cube. Let one of the vertex of the cube is the origin $\text{'}o\text{'}$ and the coordinate axes along the 3 edges $\overline{OA},\overline{OB}$ and $\overline{OC}$ are passing through origin The four diagonals are $\overline{OG},\overline{AF},\overline{CE},\overline{BD}$

$\begin{array}{l}O=(0,0,0)\\ C=(0,0,a)\\ F=(0,a,a)\end{array} \begin{array}{l}A=(a,0,0)\\ G=(a,a,a)\\ D=(a,0,a)\end{array} \begin{array}{l}B=(0,a,0)\\ E=(a,a,0)\end{array}$

dr’s of   $\overline{OG} =(x_2-x_1,y_2-y_1,z_2-z_1)$
$$\begin{gathered} =(a-0,a-0,a-0) \\ =(a,a,a) \end{gathered}$$                    
                    
dc’s of  $\overline{OG}=(\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}})$

$$\begin{gathered} =(\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}}) \\ =(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}) \end{gathered}$$

dr’s of   $=(0-a,a-0,a-0)=(-a,a,a)$
dc’s of  $\overline{AF} =(\frac{-a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}})$

$$\begin{gathered} =(\frac{-a}{a\sqrt{3}},\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}}) =(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}) \\ Let\theta istheanglebetweenthediagonals\overline{OG}\&\overline{AF} \\ \cos \theta =|l_1{l}_2+m_1{m}_2+n_1{n}_2| \\ =|\frac{1}{\sqrt{3}}(-\frac{1}{\sqrt{3}})+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)| \\ =|-\frac{1}{3}+\frac{1}{3}+\frac{1}{3}| \\ \cos \theta =\frac{1}{3}\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{3}\right) \end{gathered}$$