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 $-II$
$(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$	$(i)$ $\theta = \,{30^o}$
$(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$	$(ii)$ $\theta = \,{45^o}$
$(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$	$(iii)$ $\theta = \,{90^o}$
$(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$	$(iv)$ $\theta = \,{0^o}$

$(a)$ $|\overrightarrow{ A } \times \overrightarrow{ B }|= AB \sin \theta$

$\therefore 0=2 \times 4 \times \sin \theta$

$\therefore 0=\sin \theta$

$\therefore \theta=0^{\circ}$ તેથી $(a-iv)$

$(b)$ $|\vec{A} \times \vec{B}|=A B \sin \theta$

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

$\therefore 1=\sin \theta$

$\therefore \theta=90^{\circ}$ તેથી (b - iii)

$(c)$ $|\overrightarrow{ A } \times \overrightarrow{ B }|= AB \sin \theta$

$\therefore 4=2 \times 4 \times \sin \theta$

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

$\therefore \theta=30^{\circ}$ તેથી $(c-i)$

$(d)$ $|\vec{A} \times \vec{B}|=A B \sin \theta$

$\therefore 4 \sqrt{2}=2 \times 4 \times \sin \theta$

$\therefore \frac{1}{\sqrt{2}}=\sin \theta$

$\therefore \theta=45^{\circ}$ તેથી $(d-ii)$