## Calculus (3rd Edition)

$⟨0,-\frac{9}{2},0⟩$
By using the right-hand rule for the cross product of the vectors, we see that the direction of $\textbf{v}\times\textbf{w}$ is along the negative y-axis as the vectors are in the xz-plane. So, the x-component and z-component are both 0 and the magnitude of $\textbf{v}\times\textbf{w}$ is nothing but the component along the negative y-axis. $||\textbf{v}\times\textbf{w}||=||\textbf{v}||\,||\textbf{w}||\,\sin\theta=3\times3\times\sin\frac{\pi}{6}=\frac{9}{2}$ Component along y-axis =$-\frac{9}{2}$ Therefore, the required vector is $\textbf{v}\times\textbf{w}=⟨0, -\frac{9}{2},0⟩$