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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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The scalar product of two vector $\overrightarrow{A}$ and $\overrightarrow{B}$ is $5 \sqrt 3$ and the magnitude of their vector product is 5. The angle between the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is

$\begin{array}{1 1}(A)\;30^{\circ} \\(B)\;45^{\circ} \\(C)\;60^{\circ} \\(D)\;90^{\circ} \end{array} $

Can you answer this question?
 
 

1 Answer

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$|\overrightarrow{A} .\overrightarrow{B}|=AB \cos \theta$
$| \overrightarrow{A} \times \overrightarrow{B}|=AB \sin \theta$
$ \tan \theta = \large\frac{|\overrightarrow{A} \times \overrightarrow{B}|}{|\overrightarrow{A}.\overrightarrow{B}|}$
$ \qquad= \large\frac{5}{5 \sqrt 3 }$
$\qquad= \large\frac{1}{\sqrt 3}$
Hence A is the correct answer.
answered Jul 16, 2014 by meena.p
 

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