## Precalculus (10th Edition)

The magnitude of the cross product: $||u\times v||=||u||\cdot||v||\sin{C}.$ Hence the statement is true. By definition the area of a triangle: $K=\frac{uv\cdot \sin{(C)}}{2}$, where $C$ is the included angle of $u$ and $v$. We can compute the area of the parallelogram by adding the areas of the two triangles that make the parallelogram. Which here (according to the picture) is: $\frac{uv\cdot \sin{(C)}}{2}+\frac{uv\cdot \sin{(C)}}{2}=uv\cdot \sin{(C)}=||u||\cdot||v||\cdot \sin{(C)}.$ Therefore the statement is true.