If $|\vec A| = 2$ and $|\vec B| = 4$,then match the relation in Column $-I$ with the angle $\theta$ between $\vec A$ and $\vec B$ in Column $-II$.

Column $-I$	Column $-II$
$(a) \vec A \cdot \vec B = 0$	$(i) \theta = 0^\circ$
$(b) \vec A \cdot \vec B = +8$	$(ii) \theta = 90^\circ$
$(c) \vec A \cdot \vec B = 4$	$(iii) \theta = 180^\circ$
$(d) \vec A \cdot \vec B = -8$	$(iv) \theta = 60^\circ$

(A) Given $|A| = 2$ and $|B| = 4$. The dot product is defined as $\vec A \cdot \vec B = |A||B| \cos \theta = 8 \cos \theta$.
$(a) \vec A \cdot \vec B = 0 \implies 8 \cos \theta = 0 \implies \cos \theta = 0 \implies \theta = 90^\circ$. Matches $(ii)$.
$(b) \vec A \cdot \vec B = 8 \implies 8 \cos \theta = 8 \implies \cos \theta = 1 \implies \theta = 0^\circ$. Matches $(i)$.
$(c) \vec A \cdot \vec B = 4 \implies 8 \cos \theta = 4 \implies \cos \theta = 1/2 \implies \theta = 60^\circ$. Matches $(iv)$.
$(d) \vec A \cdot \vec B = -8 \implies 8 \cos \theta = -8 \implies \cos \theta = -1 \implies \theta = 180^\circ$. Matches $(iii)$.