(1)求证:(a-b)⊥c;
(2)若|ka+b+c|＞1(k∈R),求k的取值范围.
(1)求证:(a-b)⊥c;
(2)若|ka+b+c|＞1(k∈R),求k的取值范围.
(1)证明：∵|a|=|b|=|c|=1,〈a，b〉=〈b,c〉=〈c,a〉=120°,
∴a·c=|a||c|cos120°=-
,
b·c=|b||c|cos120°=-
.
∴(a-b)·c=a·c-b·c=(-
)-(-
)=0.
∴(a-b)⊥c.
(2)解：由|ka+b+c|＞1，得|ka+b+c|2＞1,即(ka+b+c)2＞1，
∴k2a2+b2+c2+2ka·b+2b·c+2kc·a＞1,
即k2-2k＞0.∴k＜0或k＞2.