Question

已知非零向量a，b的夹角为60°，且|a|=|b|=2，若向量c满足（a-c）*（b-c）=0，求|c|的最大值

Answer 1

let (a+b),c 夹角 =x
|a+b|^2 =(a+b).(a+b)
= |a|^2 +|b|^2 +2|a||b|cos60° = 12
|a+b| = 2√3
(a-c).(b-c)=0
a.b -a.c-b.c + |c|^2 =0
2 - (a+b).c + |c|^2 =0
2 - |a+b||c|cosx + |c|^2 =0
2- 2√3|c|cosx +|c|^2 =0
|c| = [2√3cosx +√(12(cosx)^2- 8)] / 2 or [2√3cosx -√(12(cosx)^2- 8)] / 2
max|c| at cosx = 1
max |c| = √3+ 1