$3$ points $ABC$ on a plane, $O$ as origins. $OA = \vec a$, $OB= \vec b$, $OC = \vec c$. Point $M$ inside $\triangle ABC$. And $\triangle MAB : \triangle MBC : \triangle MCA = 2 : 3 : 5$.
A straight line $BM$ intersect side $AC$ at $N$. Express $OM$ in terms of vector $a,b,c$.
Can you give me some hint? I have been thinking, what i got is, $BM : MN = 1 : 1.$ $AB : BC = 2 : 3$ $\vec AB = OB - OA = \vec b - \vec a$ $\vec CB = OB - OC = \vec b - \vec c$
$\vec OM = OB + BM$
$\vec BM = \frac 12 BN$ Then, i have difficulity in expressing $BN$ in terms of vector $a,b,c$.
Note that from the given $\triangle MAB : \triangle MBC : \triangle MCA = 2 : 3 : 5$, the following relationship can be derived,
$$\vec{AN} = \frac25 \vec{AC} =\frac25 (\vec{BC} - \vec{BA})$$
Then,
$$\vec{BN} = \vec{AN} -\vec{AB} = \frac25 (\vec{BC} - \vec{BA}) +\vec{BA} = \frac25 \vec{BC} +\frac35 \vec{BA} $$
where $\vec{BC} =\vec{c} - \vec{b} $ and $\vec{BA} =\vec{a} - \vec{b} $. Thus,
$$\vec{OM}=\vec{BM}+\vec{OB}=\frac12\vec{BN}+\vec{b} = \frac3{10}\vec{a} + \frac12\vec{b} + \frac15\vec{c}$$