# 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 $\large\frac{\pi}{4}$