已知|a|=2√2,|b|=2 a*b=2 c满足(a-c)*(b-c)=0 则|c|的最小值为 abc上都带箭头 是向量.

|a|=2√2,|b|=2 a b=2
|a+b|^2
= (a+b).(a+b)
= |a|^2+|b|^2+2a.b
= 4+8+ 4
=16
=> |a+b| =4
设c、(a+b) 夹角=β
(a-c).(b-c) =0
a.b -a.c- b.c +|c|^2 =0
|c|^2 - c.(a+b) + 2 = 0
|c|^2 - |c||a+b|cosβ + 2 =0
|c|^2 - 4|c|cosβ + 2 =0
|c| = 2+√(4(cosβ)^2-2) or 2-√(4(cosβ)^2-2)
= 2+√(2cos2β) or 2- √(2cos2β)
min|c| when cos2β = 1
min|c| = 2 - √2