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
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|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

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