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Home  >>  TN XII Math  >>  Vector Algebra
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Find the angle between two vectors $\overrightarrow{a} $ and $\overrightarrow{b} $ if $|\overrightarrow{a} \times \overrightarrow{b}|=\overrightarrow{a}.\overrightarrow{b}$

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  • $ \overrightarrow a.\overrightarrow b=| \overrightarrow a|| \overrightarrow b| \cos \theta$ $ \therefore \cos \theta = \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|} \Rightarrow \theta = \cos^{-1} \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|}$
  • For two vectors $ \overrightarrow a \: and \: \overrightarrow b$, the vector product $ \overrightarrow a$ x $ \overrightarrow b=|\overrightarrow a||\overrightarrow b| \sin \theta \hat n$ with $ \hat n \perp $ to $ \overrightarrow a \: and \: \overrightarrow b\: and \: \overrightarrow a, \overrightarrow b, \hat n$ forming a right handed system.
$ | \overrightarrow a \times \overrightarrow b| = |\overrightarrow a||\overrightarrow b| \sin \theta$
$ \overrightarrow a.\overrightarrow b = |\overrightarrow a||\overrightarrow b| \cos \theta$
$ |\overrightarrow a||\overrightarrow b| \sin \theta = |\overrightarrow a|.|\overrightarrow b| \cos \theta$
$ \Rightarrow \tan \theta = 1 \:\: \: \: \: 0 \leq \theta \leq \pi$
$ \therefore \theta = \tan^{-1}1 = \large\frac{\pi}{4}$
The angle between the vectors is $ \large\frac{\pi}{4}$
answered Jun 5, 2013 by thanvigandhi_1

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