当m,n满足什么时,能使a,b,c终点在同一直线上?(O为公共始点,其中c=ma+nb,字母均为向量)

(字母均为向量)
设a,b,c终点分别为A,B,C,则a=OA,b=OB,c=OC
OC=OA+AC=OB+BC (*)
AC应与BC共线,设AC=kBC
由(*) (k-1)BC=OB-OA BC=(OB-OA)/(k-1)
而2OC=OA+AC+OB+BC=OA+OB+(OB-OA)/(k-1)+k(OB-OA)/(k-1)
=(1-(1+k)/(k-1))OA+(1+(1+k)/(k-1))OB
OC=(1-(1+k)/(k-1))OA/2+(1+(1+k)/(k-1))OB/2
则m=(1-(1+k)/(k-1))/2 n=(1+(1+k)/(k-1))/2
所以m+n=1

m+n=1