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