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已知向量a=(1,1),b=(1,0),c满足a·c等于0,且|a|=|c|,b·c>0.

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已知向量a=(1,1),b=(1,0),c满足a·c等于0,且|a|=|c|,b·c>0.
(1)求向量c
(2)若映射f:(x,y)→(x`,y`)=xa+yb,求映射 f下点(1,2)的原象.
以上 a b c 都为向量!
(1)
|a| = √2
let c=(x,y)
=> x^2 +y^2 =2 (1)
a.c =0
(1,1)(x,y) =0
=> x+y = 0 (2)
b.c >0
(1,0)(x,y)>0
=> x > 0
sub (2) into (1)
2x^2 = 2
x = 1 or -1( rejected , x>0)
y = -1
c ( 1,-1)
(2)
f:(x,y)->(x',y') = xa+yb
xa+yb = (1,2)
x(1,1) + y(1,0) = (1,2)
(x+y, x) =(1,2)
=> x+y =1
and x =2
=> y = -1
1,2)的原象 = ( 2, -1)