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