Let $\vec{a}$ be represented by $\overrightarrow{OP}$ and $\vec{b}$ be represented by $\overrightarrow{OQ}$ . Let $\angle POQ = \theta$, figure complete the parallelogram OPRQ. Join PQ. Draw QN ⊥ OP.In $\Delta OQN$$\sin \theta = \frac{QN}{OQ} = \frac{QN}{b}$$QN = b \sin \theta$Now, by definition,$|\vec{a} \times \vec{b}| = a b \sin \theta = (OP)(QN)$$= \frac{2(OP)(QN)}{2}=2\times$ area of $\Delta OPQ$∴ area of ΔOPQ $= \frac{1}{2} |\vec{a} \times \vec{b}|$, which was to be proved.