Show that the area of the triangle formed by the vectors $\vec{a}$ and $\vec{b}$ is one-half of the magnitude of $\vec{a} \times \vec{b}$.

(N/A) Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the sides $\vec{OK}$ and $\vec{OM}$ of a triangle,inclined at an angle $\theta$.
In $\Delta OMN$,where $MN$ is the perpendicular from $M$ to $OK$,we have:
$\sin \theta = \frac{MN}{OM} = \frac{MN}{|\vec{b}|}$
$MN = |\vec{b}| \sin \theta$
The area of $\Delta OMK$ is given by:
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OK \times MN$
Area $= \frac{1}{2} |\vec{a}| |\vec{b}| \sin \theta$
Since the magnitude of the cross product is defined as $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,we substitute this into the area formula:
Area $= \frac{1}{2} |\vec{a} \times \vec{b}|$
Thus,the area of the triangle is one-half the magnitude of the cross product of the two vectors.