Let edge of a cube be 1 unit.
The diagonals of a cube are $O A$ and $B C$.
So, DR's of diagonals $O A$ are $(1,1,1)$ and $B C$ are $(0-1,1,1)$, i.e., $(-1,1,1)$
Now, angle between diagonals,
$\cos \theta =\frac{1(-1)+1(1)+1(1)}{\sqrt{1^{2}+1^{2}+1^{2}} \sqrt{(-1)^{2}+1^{2}+1^{2}}}$
$=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}} \sqrt{1^{2}+1^{2}+1^{2}}}$
$=\frac{1}{\sqrt{3} \sqrt{3}}=\frac{1}{3}$
$\Rightarrow \theta =\cos ^{-1}(1 / 3)$
