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The position vectors of the points $\mathrm{A}, \mathrm{B}$ are $\vec{a}, \vec{b}$ respectively. If the position vector of the point $C$ is $\frac{\vec{a}}{2}+\frac{\vec{b}}{3}$, then
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The correct answer is:
$C$ lies inside $\triangle \mathrm{OAB}$
$\overrightarrow{O C}=\frac{\vec{a}}{2}+\frac{\vec{b}}{3}$
Let $D$ is mid point of $O A$,
$\overrightarrow{O D}=\frac{\vec{a}}{2}$
Let $E$ is a point dividing $O B$ in $1: 2$
$\therefore \quad \overrightarrow{O E}=\frac{\vec{b}}{3}$
Clearly $\overrightarrow{O C}=\overrightarrow{O D}+\overrightarrow{O E}$
Which will be along diagonal $\overrightarrow{O C}$ of parallelogram $O D C E$ which is clearly inside $\triangle O A B$.
$\therefore \quad C$ lies inside $\triangle O A B$.
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