∵(b-c)*a=ab-ac= |a|.|b|.cos120°- |a|.|c|.cos120°=
1×1×cos120°- 1×1×cos120° = 0
∴ 向量a垂直(b-c)
∵作图分析可知：b+c=-a
∴|ka+b+c|=|(k-1)a|
∵a的模为1
∴|k-1|＞1,
∴k2