已知a向量等于（1,1）b向量等于（0,-2）,若ka-b与a+b夹角为120度,则k等于?a,b均为向量

a² = 1 + 1 = 2
b² = 0 + 4 = 4
a · b = 0 - 2 = - 2
|ka-b|=|(k,k+2)|=√[k²+(k+2)²]=√(2k²+4k+4)
|a+b|=|(1,-1)|=√2
θ=120度
(ka - b) · (a + b)
= ka² + ka·b - b·a - b²
= ka² + (k-1) a·b - b²
= 2k - 2(k-1) - 4
= - 2
= |ka-b|·|a+b|·cosθ
整理得 (k+1)²=3
k= -1±√3