Question

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

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|的取值范围是
拜托了，谢谢

Answer 1

|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|<= (√3+1)/2

Answer 2

1