3 points ABC on a plane, O as origins. $OA=\overrightarrow{a}$ , $OB=\overrightarrow{b}$, $OC=\overrightarrow{c}$ . Point M inside $\mathrm{\u25b3}ABC$. And $\mathrm{\u25b3}MAB:\mathrm{\u25b3}MBC:\mathrm{\u25b3}MCA=2:3:5$

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, $BM:MN=1:1.$ $AB:BC=2:3$ $\overrightarrow{A}B=OB-OA=\overrightarrow{b}-\overrightarrow{a}$ $\overrightarrow{C}B=OB-OC=\overrightarrow{b}-\overrightarrow{c}$

$\overrightarrow{O}M=OB+BM$

$\overrightarrow{B}M=\frac{1}{2}BN$ Then, i have difficulity in expressing $BN$ in terms of vector a,b,c.

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, $BM:MN=1:1.$ $AB:BC=2:3$ $\overrightarrow{A}B=OB-OA=\overrightarrow{b}-\overrightarrow{a}$ $\overrightarrow{C}B=OB-OC=\overrightarrow{b}-\overrightarrow{c}$

$\overrightarrow{O}M=OB+BM$

$\overrightarrow{B}M=\frac{1}{2}BN$ Then, i have difficulity in expressing $BN$ in terms of vector a,b,c.