问题
填空题
已知
|
答案
∵
=(1-t,1-t,t),a
=(2,t,t),b
∴向量
-b
=(1+t,2t-1,0)a
可得向量
-b
的模|a
-b
|=a
=(1+t)2+ (2t-1)2+02 5t2-2t+2
∵5t2-2t+2=5(t-
)2+1 5 9 5
∴当且仅当t=
时,5t2-2t+2的最小值为1 5 9 5
所以当t=
时,|1 5
-b
|的最小值是a
=9 5 3 5 5
故答案为:3 5 5