Question

已知P是△ABC内一点,且满足向量PA+2PB+3PC=0,记△ABP，△

BCP，△ACP的面积依次为S1.S2.S3，则它们之比为？

Answer 1

PA+2PB+3PC=0
(PA+2PB+3PC)xPA=0xPA (x : cross product )
2PBxPA+3PCxPA=0
2|PBxPA|=3|PCxPA|
S1=(3/2)S3 (1)
also
PA+2PB+3PC=0
(PA+2PB+3PC)xPB=0
|PAxPB|=3|PCxPB|
S1=3S2 (2)
from (1) ,(2)
S1:S2:S3= 1:3: 3/2=2:6:3

追问

出错了，检查，没有这个选项啊...

追答

PA+2PB+3PC=0
(PA+2PB+3PC)xPA=0xPA (x : cross product )
2PBxPA+3PCxPA=0
2|PBxPA|=3|PCxPA|
S3=(2/3)S1 (1)
also
PA+2PB+3PC=0
(PA+2PB+3PC)xPB=0
|PAxPB|=3|PCxPB|
S2=(1/3)S1 (2)
from (1) ,(2)
S1:S2:S3= 1:1/3: 2/3=3:1:2

Answer 2

建系取特殊点，P(0，0),B(1，0),C(0，1),通过条件计算出A的坐标，计算面积之比为3:1:2