Let there be cube of side 'a'. Co-ordinates of its vertices $\mathrm{O}, \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}$ have bemarked in the figure. Diagonals
are $\mathrm{OE}, \mathrm{FC}, \mathrm{GB}$ and $\mathrm{AD}$. Direction ratios $\left(\mathrm{dr}_{3}\right.$ ) of these diagonals are : $\mathrm{OE}\langle(\mathrm{a}-0),(\mathrm{a}-0),(\mathrm{a}-0)\rangle=(\mathrm{a}, \mathrm{a}, \mathrm{a})$
$F C\langle(-a, a,-a)\rangle ; G B\langle(-a, a, a)\rangle$ and $A D\langle(a, a,-a)\rangle$
Their des are :
OE, $\left\langle\frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}\right\rangle$
$=\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$
$\mathrm{AD},\left\langle\frac{\mathrm{a}}{\sqrt{\Sigma \mathrm{a}^{2}}}, \frac{\mathrm{a}}{\sqrt{\Sigma \mathrm{a}^{2}}}, \frac{-\mathrm{a}}{\sqrt{\Sigma \mathrm{a}^{2}}}\right\rangle=\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right\rangle$
Angle, $\theta$, between $\mathrm{AD}$ and $\mathrm{OE}$ is given by
$\cos \theta=\pm \frac{\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}}{\sqrt{\left\{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}\right\}\left\{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(-\frac{1}{\sqrt{3}}\right)^{2}\right\}}}$
$=\frac{\frac{1}{3}}{1 \times 1}=\pm \frac{1}{3}$
Since the cube is in positive octant, we take $+\frac{1}{3}$.
So, $\cos \theta=\frac{1}{3} \Rightarrow \theta>60$ '
[Since value of $\cos \theta$ decreases as $\theta$ increases in 0 to $90^{\circ} \cdot \cos \theta=1$ when $\theta=0^{\circ}$ and $\cos \theta=0$ when $\theta=90^{\circ}$ ]