## Calculus (3rd Edition)

$\sqrt {3}$
We calculate the magnitude of the cross product as follows: $||\textbf{v}\times\textbf{w}||=||\textbf{v}||\,||\textbf{w}||\sin \theta$ $||\textbf{v}||=2$ and $\theta=\frac{\pi}{6}$ $||\textbf{w}||=\frac{\textbf{v}\cdot\textbf{w}}{||\textbf{v}||\cos\theta}=\frac{3}{2\times\frac{\sqrt 3}{2}}=\sqrt 3$ Then, we have $||\textbf{v}\times\textbf{w}||=2\times\sqrt {3}\times\sin\frac{\pi}{6}=2\times\sqrt {3}\times\frac{1}{2}= \sqrt {3}$