# 3 points ABC on a plane, O as origins. OA=vec(a) , OB=vec(b) , OC=vec(c) . Point M inside Delta ABC. And Delta MAB: Delta MBC: Delta MCA=2:3:5. A straight line BM intersect side AC at N. Express OM in terms of vector a,b,c.

3 points ABC on a plane, O as origins. $OA=\stackrel{\to }{a}$ , $OB=\stackrel{\to }{b}$, $OC=\stackrel{\to }{c}$ . Point M inside $\mathrm{△}ABC$. And $\mathrm{△}MAB:\mathrm{△}MBC:\mathrm{△}MCA=2:3:5$
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, $BM:MN=1:1.$ $AB:BC=2:3$ $\stackrel{\to }{A}B=OB-OA=\stackrel{\to }{b}-\stackrel{\to }{a}$ $\stackrel{\to }{C}B=OB-OC=\stackrel{\to }{b}-\stackrel{\to }{c}$
$\stackrel{\to }{O}M=OB+BM$
$\stackrel{\to }{B}M=\frac{1}{2}BN$ Then, i have difficulity in expressing $BN$ in terms of vector a,b,c.
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Note that from the given $\mathrm{△}MAB:\mathrm{△}MBC:\mathrm{△}MCA=2:3:5$, the following relationship can be derived,
$\stackrel{\to }{AN}=\frac{2}{5}\stackrel{\to }{AC}=\frac{2}{5}\left(\stackrel{\to }{BC}-\stackrel{\to }{BA}\right)$
Then,
$\stackrel{\to }{BN}=\stackrel{\to }{AN}-\stackrel{\to }{AB}=\frac{2}{5}\left(\stackrel{\to }{BC}-\stackrel{\to }{BA}\right)+\stackrel{\to }{BA}=\frac{2}{5}\stackrel{\to }{BC}+\frac{3}{5}\stackrel{\to }{BA}$
where $\stackrel{\to }{BC}=\stackrel{\to }{c}-\stackrel{\to }{b}$ and $\stackrel{\to }{BA}=\stackrel{\to }{a}-\stackrel{\to }{b}$ . Thus,
$\stackrel{\to }{OM}=\stackrel{\to }{BM}+\stackrel{\to }{OB}=\frac{1}{2}\stackrel{\to }{BN}+\stackrel{\to }{b}=\frac{3}{10}\stackrel{\to }{a}+\frac{1}{2}\stackrel{\to }{b}+\frac{1}{5}\stackrel{\to }{c}$