Show that the area of the triangle contained between the vectors $\text{a}$ and $\text{b}$ is one half of the magnitude of $\text{a}\times \text{b}$.

Consider two vectors OK = vector $|a|$and OM = vector$|b|$, inclined at an angle $\theta$ as shown in the following figure.

In $△OMN,$ we can write the relation:

$sin\theta =\frac{MN}{OM}=\frac{MN}{\left|\vec{b}\right|}$

$⟹MN=\left|\vec{b}\right|sin\theta$

$\left|\vec{a}\times \vec{b}\right|=\left|\vec{a}\right|\left|\vec{b}\right|sin\theta$

             $$\begin{gathered} =OK\times MN \\ =2\times \frac{1}{2}\times OK\times MN \end{gathered}$$

             $=2\times$ Area of $△OMK$

$⟹$ Area of $△$ $OMK=\frac{1}{2}\times \left|\vec{a}\times \vec{b}\right|$