|a+b|²=|a|²+2ab+|b|²=1+2×1×1×cos(a,b)+1=1
∴cos(a,b)=-1/2
∴|a-tb|²=|a|²-2tab+|tb|²=1-2t×1×1×(-1/2)+t²=t²+t+1=(t+1/2)²+3/4≥3/4
∴|a-tb|(t∈R)最小值是√3/2