Answer
The magnitude of the vector product is $156~m^2$.
Work Step by Step
We know that the scalar product $\vec{A} \cdot \vec{B} = AB~cos(\phi)$, where $\phi$ is the angle between the two vectors.
$AB~cos(\phi) = 112.0~m^2$
$cos(\phi) = \frac{112.0~m^2}{(12.0~m)(16.0~m)}$
$\phi = cos^{-1}(\frac{112.0~m^2}{(12.0~m)(16.0~m)})$
$\phi = 54.31^{\circ}$
We know that the vector product $\vec{A} \times \vec{B} = AB~sin(\phi)$, where $\phi$ is the angle between the two vectors.
We can find the magnitude of the vector product.
$AB~sin(\phi) = (12.0~m)(16.0~m)~sin(54.31^{\circ})$
$AB~sin(\phi) = 156~m^2$
The magnitude of the vector product is $156~m^2$
