title: i，j是在xOy直角坐标系中的单位向量，已知A、B、C共线，如果? 
desc: i，j是在xOy直角坐标系中的单位向量，已知A、B、共线，如果向量OA= -2i+mj , 向量OB=ni+j , 向量OC=5i–j , 且向量OA垂直于向量OB，求（m，n）是多少?
answer: BA = OA - OB = (-2i+mj)-(ni+j) = -(2+n)i+(m-1)j;
  
  CB = OB-OC = (ni+j) -(5i-j) = (n-5)i + 2j;
  
  A、B、C共线;
  -(n+2):(n-5) = (m-1):2;
  
  向量OA垂直于向量OB
  OA * OB = (-2i+mj)*(ni+j)
          = (-2)n+m = 0
  (n-5)*(m-1)+2(n+2) = 0;
  m = 2n;
  
  2n^2 -9n + 9 = 0;
  n = 3/2, n = 3;
  m = 3,  m = 6;
  
  
  (m，n）是 (3, 3/2) 或 (6, 3).
  
  