# If $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $\overrightarrow{a}.\overrightarrow{b}=9$ than find $|\overrightarrow{a} \times \overrightarrow{b}|$

Toolbox:
• $\overrightarrow a.\overrightarrow b=| \overrightarrow a|| \overrightarrow b| \cos \theta$ $\therefore \cos \theta = \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|} \Rightarrow \theta = \cos^{-1} \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|}$
• For two vectors $\overrightarrow a \: and \: \overrightarrow b$, the vector product $\overrightarrow a$ x $\overrightarrow b=|\overrightarrow a||\overrightarrow b| \sin \theta \overrightarrow n$ with $\overrightarrow n \perp$ to $\overrightarrow a \: and \: \overrightarrow b\: and \: \overrightarrow a, \overrightarrow b, \overrightarrow n$ forming a right handed system.
Step 1
$\overrightarrow a.\overrightarrow b = |\overrightarrow a||\overrightarrow b| \cos \theta = (3)(4) \cos \theta$
$\therefore 9 = 12 \cos \theta \: or \: \cos \theta = \large\frac{9}{12} = \large\frac{3}{4}$
Since $\cos \theta > 0, \: \theta$ is acute. $\therefore \sin \theta > 0$
Step 2
$\sin \theta = \sqrt{1-cos^2 \theta} = \sqrt{1-\large\frac{9}{16}} = \large\frac{\sqrt{7}}{4}$
Step 3
$| \overrightarrow a$ x $\overrightarrow b | = | \overrightarrow a | | \overrightarrow b | \sin \theta = (3)(4) \large\frac{\sqrt{7}}{4} = 3\sqrt 7$

edited Jun 21, 2013