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

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

Given $|\mathrm{A}|=2$ and $|\mathrm{B}|=4$

$(a)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=0$

$\therefore \quad 2 \times 4 \cos \theta=0$

$\therefore \quad \cos \theta=0=\cos 90^{\circ}$

$\therefore \quad \theta=90^{\circ}$

$\therefore$ Option $(a)$ matches with option $(ii)$.

$(b)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=8$

$\therefore 2 \times 4 \cos \theta=8$

$\therefore \quad \cos \theta=1=\cos 0^{\circ}$

$\therefore \quad \theta=0^{\circ}$

$\therefore$ Option $(b)$ matches with option $(i)$.

$(c)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=4$

$\therefore \quad 2 \times 4 \cos \theta=4$ $\therefore \quad \cos \theta=\frac{1}{2}=\cos 60^{\circ}$ $\therefore \quad \theta=60^{\circ}$

$\therefore$ Option $(c)$ matches with option $(iv)$.

$(d)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=-8$

$\therefore \quad 2 \times 4 \cos \theta=-8$

$\therefore \quad \cos \theta=-1=\cos 180^{\circ}$

$\therefore \quad \cos \theta=180^{\circ}$

Option $(d)$ matches with option $(iii)$.