Question

已知P为△ABC内一点,向量PA+2向量PB+3向量PC=向量0,则S△PAB：S△PBC：S△PAC=（ ）

Answer 1

△PAB、△PBC、△PAC的面积之比S1:S2:S3

如图：延长PB到B',使PB'＝2PB, 延长PC到C',使PC＝3PC'

则 PA+PB'+PC'＝0, P是ΔAB'C'的重心,则SΔPAB'＝SΔPAC'＝SΔPB'C'＝k

S1=1/2*SΔPAB'=1/2*k,S3=1/3*SΔPAC'=1/3*k

S2=1/2*PB*PC*sin∠BPC

=1/2*1/2PB'*1/3PC'sin∠BPC

=1/6**1/2PB*'PC'sin∠BPC

=1/6*SΔPB'C'

=1/6*k

故 S1:S2:S3=(1/2):(1/6):(1/3)=3:1:2