If θ is the acute angle between the diagonals of a cube, then which one of the following is correct?

[d]Let there be cube of side ?a? Co-ordinates of its vertices O, A, B, C, D, E, F have be marked in the figure. Diagonals are OE, FC, GB and AD. Direction ratios (dr3) of these diagonals are: OE ⟨(a−0),(a−0),(a−0)⟩=(a,a,a)$FC⟨\left(-a,a,-a\right)⟩;GB⟨\left(-a,a,a\right)⟩\text{and}$AD⟨(a,a,−a)⟩$Theirdcsare:$OE,⟨aa2+a2+a2−−−−−−−−−−√,aa2+a2+a2−−−−−−−−−−√,aa2+a2+a2−−−−−−−−−−√⟩$=⟨13\surd ,13\surd ,13\surd ⟩$AD,⟨aΣa2−−−√,aΣa2−−−√,−aΣa2−−−√⟩=⟨13√,13√,13√⟩$Ang\le ,\theta ,betweenAD\text{and}OEisgivenby$cosθ=±13√×13√+13√×13√−13√×13√{(13√)2+(13√3)+(13√)2}−−−−−−−−−−−−−−−−−−−−−−−√{(13√)2+(13√)2+(13√)2}$=131\times 1=\pm 13$Since the cube is in positive octant, we take +13.$So,\cos \theta =13\Rightarrow \theta >60\prime$[Since value of cosθ decreases as θ increases in 0 to 90∘.cosθ=1 when θ=0∘ and cosθ=0 whenθ=90∘]