若向量MA,MB,MC的起点与终点M,A,B,C互不重合且无三点共线,O是空间任意一点,

MA,MB,MC成为空间一组基底的关系是
=>MA,MB,MC is linearly independent
let
OM = k1MA+k2MB+k3MC
OA = k1'MA+k2'MB+k3'MC
OB= k1''MA+k2''MB+k3''MC
OC =k1'''MA+k2'''MB+k3'''MC
k1MA+k2MB+k3MC -OM=0
k1MA+k2MB+k3MC -(OA+AM)=0
(k1-1)MA+k2MB+k3MC -OA=0
(k1-1)MA+k2MB+k3MC -(k1'MA+k2'MB+k3'MC)=0
(k1-1-k1')MA+(k2-k2')MB+(k3-k3')MC =0
=> k1= 1+k1' and k2=k2' and k3=k3'
OA = (k1-1)MA+k2MB+k3MC (1)
Similarly
OB= k1''MA + k2''MB+k3''OC
k2=k2''+1 and k1=k1'' and k3=k3''
OB = k1MA+(k2-1)MB+k3MC (2)
OC= k1'''MA + k2'''MB+k3'''OC
k3=k3'''+1 and k1=k1'' and k2=k2''
OC = k1MA+k2MB+(k3-1)MC (3)
(1)+(1/3)*(2) +(2/3)*(3)
OA+1/3OB+2/3OC
= k1MA+k2MB+k3MC = OM
Ans B

已知M、A、N三点不共线，且存在实数x1，x2，使得向量MC=x1向量MA+x2向量MB，若A、B、C三点共线，则x1+x2=？ x1+x2=1