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3-1.Vectors
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What is the angle between $\vec A\,\,$ and $\vec B\,\,$ if $\vec A\,\,$ and $\vec B\,\,$ are the adjacent sides of a parallelogran drawn from a common point and the area of the parallelogram is $\frac {AB}{2}$
A
$\frac {\pi }{3}$
B
$\frac {\pi }{6}$
C
$\frac {\pi }{2}$
D
$\pi $
Solution
$|\vec{A} \times \vec{B}|=\frac{A B}{2}$
$\Rightarrow \mathrm{AB} \sin \theta=\frac{\mathrm{AB}}{2} \Rightarrow \sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}=\frac{\pi}{6}$
Standard 11
Physics
