|ka+b+c|=√[(ka+b+c)·(ka+b+c)]=√(k^2|a|^2+|b|^2+|c|^2+2ka·b+2b·c+2kc·a)=√(k^2+1+1-k-1-k)
=√(k^2-2k+1)=|k-1|>1
解得k>2或k