1：已知a.b.c是单位向量,且a*b=0则(a-c)*(b-c)的最小值=?2：已知平面向量a.b.c满足｜a｜=1&

(1)
|a+b|^2=|a|^2+|b|^2 +2a.b = 2
=|a+b|=√2
(a-c).(b-c)
= a.b+|c|^2-(a+b).c
= 1- (a+b).c
min (a-c).(b-c) at c//(a+b)
min(a-c).(b-c)
= 1-|a+b||c|
= 1- √2
(2)
|a|=1,|b|=2,|c|=3
let a,b夹角=b,c夹角=c,a夹角=x
=> x = 2π/3
a.b= |a||b|cos2π/3 = -1
b.c =|b||c|cos2π/3 = -3
c.a =|c||a|cos2π/3 = -3/2
|a+b+c|^2
=|a|^2+|b|^2+|c|^2+2(a.b+b.c+c.a)
=1+4+9 -(2+6+3)
=3
|a+b+c|=√3
(3)
c=xa+yb
(-3,-6) =(x-4y,2x+4y)
=>
x-4y =-3 (1) and
2x+4y=-6 (2)
(1) +(2)
3x=-9
x=-3
=> y=0