已知平面向量A,B,C,满足|A|=|B|=1,向量A与B-A的夹角为120度,且(A-C)*(B-C)=0,则|C|的

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  • |a|=|b|=1,a,(b-a)的夹角为120度

    (a-c).(b-c)=0

    To find :|c|

    a.(b-a) = a.b-|a|^2=|a||b-a|cos120度

    =>a.b-1=(-1/2)|b-a|

    (a.b)^2-2a.b+1 = (1/4)(|b|^2+|a|^2-2a.b)

    2(a.b)^2-3a.b+1=0

    (2(a.b)-1)(a.b-1)=0

    a.b =1/2 or 1 (rejected)

    |a+b|^2=|a|^2+|b|^2+2a.b=1+1+1 = 3

    |a+b| =√3

    (a-c).(b-c)=0

    a.b -(a+b).c+|c|^2 =0

    |c|^2+|a+b||c|cosx +a.b =0 ( x =(a+b),c 的夹角)

    2|c|^2+2√3|c|cosx +1 =0

    |c| =[- 2√3cosx +√(12(cosx)^2-8)] /4

    max |c| at cosx = -1

    max |c| = (√3+1)/2

    |c|