જો $\left| {\vec  A } \right|\, = \,2$ અને $\left| {\vec  B } \right|\, = \,4$ હોય, તો કોલમ $-II$માં આપેલા ખૂણાને અનુરૂપ કોલમ $-I$માં આપેલા યોગ્ય સંબંધ સાથે જોડો.

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

$(a-i i),(b-i),(c-i v),(d-i i i)$

$(a)$ $\overrightarrow{ A } \cdot \overrightarrow{ B }= ABcos \theta$

$0=A B \cos \theta$

$\therefore \quad \cos \theta=0 \quad(\because A \neq 0, B \neq 0)$

$\therefore \quad \theta=90^{\circ}$ તેથી $(a-ii)$

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

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

$1$$=\cos \theta$

$=0^{\circ}$ તેથી $( b - i )$

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

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

$\therefore \frac{1}{2}=\cos \theta$

$\therefore \theta=60^{\circ}$ તેથી $(c - iv)$

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

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

$\therefore \quad-1=\cos \theta$

$\therefore \quad \theta=180^{\circ}$ તેથી $(d-iii)$