因为 b1//b,c1//c,所以 可设
b1 = mb = m*(2,3) = (2m,3m)
c1 = nc = n*(1,1) = (n,n)
a = b1 + c1
(1,-2) = (2m,3m) + (n,n)
(1,-2) = (2m+n,3m+n)
1 = 2m + n
-2 = 3m + n
m = -3
n = 7
b1 = (-6,-9)
c1 = (7,7)
a = (1,-2) = (-6,-9) + (7,7)
因为 b1//b,c1//c,所以 可设
b1 = mb = m*(2,3) = (2m,3m)
c1 = nc = n*(1,1) = (n,n)
a = b1 + c1
(1,-2) = (2m,3m) + (n,n)
(1,-2) = (2m+n,3m+n)
1 = 2m + n
-2 = 3m + n
m = -3
n = 7
b1 = (-6,-9)
c1 = (7,7)
a = (1,-2) = (-6,-9) + (7,7)