Question

Let $O$ be an interior point of triangle $ABC,$ such that $2\overrightarrow{\mathrm{OA}}+3\overrightarrow{\mathrm{OB}}+4\overrightarrow{\mathrm{OC}}=0,$ then the ratio of the area of $\Delta ABC$ to the area of $\Delta AOC$ is

• A. $3:1$
• B. $3:2$
• C. $2:1$
• D. $4:3$

Answer 1

Answer: (A)

Let position vectors of $O,A,B,C$ are $\overrightarrow{O},\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ respectively.
$2\overrightarrow{\mathrm{OA}}+3\overrightarrow{\mathrm{OB}}+4\overrightarrow{\mathrm{OC}}=0$
$\Rightarrow 2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}=0$
Area of $\Delta AOC=\frac{1}{2}\left|\overrightarrow{\mathrm{OA}}\times \overrightarrow{\mathrm{OC}}\right|=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{c}\right|$
Area of $\Delta ABC=\frac{1}{2}\left|\overrightarrow{\mathrm{BA}}\times \overrightarrow{\mathrm{BC}}\right|=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|$
$=\frac{1}{2}\left|\overrightarrow{a}\times \left(\frac{-4\overrightarrow{c}-2\overrightarrow{a}}{3}\right)+\left(\frac{-4\overrightarrow{c}-2\overrightarrow{a}}{3}\right)\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|$
$=\frac{1}{2}\left|\frac{4}{3}\overrightarrow{c}\times \overrightarrow{a}+\frac{2}{3}\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{c}\times \overrightarrow{a}\right|$
$=\frac{3}{2}|\overrightarrow{c}\times \overrightarrow{a}|$
$\frac{Areaof\Delta ABC}{Areaof\Delta AOC}=\frac{3}{1}$