Q. Let $O$ be an interior point of $\triangle ABC$ such that $\overrightarrow{ OA }+2 \overrightarrow{ OB }+3 \overrightarrow{ OC }=0$. Then the ratio of the area of $\triangle ABC$ to the area of $\triangle AOC$ is
Vector Algebra
Solution:
In the diagram, let D and E be the midpoints of the sides AC and BC, respectively. Then we have
$\overrightarrow{ OA }+\overrightarrow{ OC }=2 \overrightarrow{ OD }$ .....(1)
and $2(\overrightarrow{ OB }+\overrightarrow{ OC })=4 \overrightarrow{ OE }$ ....(2)
From equations (1) and (2), we get
$\overrightarrow{ OA }+2 \overrightarrow{ OB }+3 \overrightarrow{ OC }=2(\overrightarrow{ OD }+2 \overrightarrow{ OE })=0$
It follows that $\overrightarrow{ OD }$ and $\overrightarrow{ OE }$ are collinear, and $|\overrightarrow{ OD }|=2|\overrightarrow{ OE }|$.
Consequently, $\frac{ S _{\triangle A E C}}{ S _{\triangle A O C}}=\frac{3}{2}$ and $\frac{ S _{\triangle A E C}}{ S _{\triangle A O C}}=\frac{3 \times 2}{2}=3$
