Question

在△ABC中,AM=MB,AN=1/3AC,已知BN与CM交于点P,设AB=a,AC=b,用向量a,b表示向量AP.

Answer 1

设BP=xBN,CP=yCM

AC+CP=AB+BP

AC+y(CA+AM)=AB+x（BA+AN)

b+y(-b+1/2a)=a+x(-a+1/3b)

1-y=x/3

1-x=y/2

x=3/5

y=4/5

AP=AB+BP=a+3/5(-a+1/3b)=2/5a+1/5