Vector Algebra Question 7
Question 7 - 2024 (29 Jan Shift 2)
Let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{b}}$, where $\mathrm{O}$ is the origin. If $\mathrm{S}$ is the parallelogram with adjacent sides $\mathrm{OA}$ and $\mathrm{OC}$, then area of the quadrilateral $\mathrm{OABC}$ is equal to
(1) 6
(2) 10
(3) 7
(4) 8
Show Answer
Answer (4)
Solution
Area of parallelogram, $\quad S=|\vec{a} \times \vec{b}|$
Area of quadrilateral $=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC})$
$=\frac{1}{2}{|\overrightarrow{\mathrm{a}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{b}})|+|\overrightarrow{\mathrm{b}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{b}})|}$
$=8|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|$
Ratio $=\frac{8|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|}{|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})| \text { math }}=8$
