若a、b、c均为单位向量,且a·b=0,(a-c)`(b-c)

|a|=|b|=|c|=1
a.b =0
(a-c).(b-c) ≤0
a.b -c.a-b.c +|c|^2 ≤0
a.b<= -1+b.c+c.a
|a+b-c|^2
= (a+b-c).(a+b-c)
=|a|^2+|b|^2 +2(a.b-b.c-c.a) +|c|^2
= 3 +2(a.b-b.c-c.a)
≤ 3 - 2 = 1
max |a+b-c| = 1