Question

已知A(t-1,1-t,t),B(2,t,t),则AB的绝对值的最小值为多少？

Answer 1

解：|AB| = √[(t – 1 – 2)2 + (1 – t – t)2 + (t – t)2] = √[(t – 3)2 + (1 – 2t)2] = √[t2 – 6t + 9 + 1 – 4t + 4t2] = √(5t2 – 10t + 10) = √[5(t – 1)2 + 5]，所以当且仅当t = 1时，|AB|的最小值为√5 。