If $v \times w = \langle 1,-3,4\rangle$ and the dot product of $v$ and $w$ is $6$, find $\tan\theta$ is the angle between $v$ and $w$.
Not sure what direction is proper to head in. I initially though of using the $\cos(\theta)$ formula of the dot product of $v$ and $w$ over the magnitude of $v$ and $w$ (but am unsure how to directly find the magnitude given just the cross product and dot product). I believe I may just be going in the wrong direction, but could not find any related formulas other than the $\sin/\cos$ ones typically used for finding $\theta$ between vectors.
Best Answer
$|v\times w| = |v||w|\sin \theta = \sqrt{1^2+(-3)^2+4^2} = \sqrt{26}$, and
$v\cdot w = |v||w|\cos \theta = 6 \to \tan \theta = \dfrac{\sqrt{26}}{6} \to \theta = \tan^{-1}\left(\frac{\sqrt{26}}{6}\right)$