如图,在△ABO中,向量OC=1/4向量OA,向量OD=1/2向量OB,AD与BC相交于点M,设向量OA=向量a,向量O

设DM=λDA,CM=υCB
OM=OD+DM=1/2OB+λDA=1/2OB+λ(OA-OD)=1/2OB+λ(OA-1/2OB)=λOA+(1-λ)/2OB
OM=OC+CM=1/4OA+υCB=1/4OA+υ(OB-OC)=1/4OA+υ(OB-1/4OA)=(1-υ)/4OA+υOB
∴λOA+(1-λ)/2OB=(1-υ)/4OA+υOB
λ=(1-υ)/4
υ=(1-λ)/2
λ=1/7,υ=3/7
OM=1/7OA+6/7OB
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