Let $$\overrightarrow {OA} = \overrightarrow a ,\,\overrightarrow {OB} = 10\overrightarrow a + 2\overrightarrow b $$       and $$\overrightarrow {OC} = \overrightarrow b ,$$   where $$O,\,A$$  and $$C$$ are noncollinear points. Let $$p$$ denote the area of the quadrilateral $$OABC,$$   and $$q$$ denote the area of the parallelogram with $$OA$$  and $$OC$$  as adjacent sides. Then $$\frac{p}{q}$$ is equal to :

Solution :

$$\eqalign{ & {\text{Here }}\left| {\overrightarrow a \times \overrightarrow b } \right| = q{\text{ and}} \cr & \frac{1}{2}\left| {\overrightarrow b \times \left( {10\overrightarrow a + 2\overrightarrow b } \right)} \right| + {\text{ }}\frac{1}{2}\left| {\overrightarrow a \times \left( {10\overrightarrow a + 2\overrightarrow b } \right)} \right| = p \cr & \therefore \,5\left| {\overrightarrow b \times \overrightarrow a } \right| + \left| {\overrightarrow a \times \overrightarrow b } \right| = p \cr & {\text{or }}6q = p\,\,\,\,\,\,\,\,\,\therefore \,\,\frac{p}{q} = 6. \cr} $$