题目

已知向量a=(0,1),向量b=(-√3/2,-1/2),向量c=(√3/2,1/2),xa+yb+zc=(1,1),则x^2+y^2+z^2的最小值为

提问时间：2020-07-24

答案

xa+yb+zc=(1,1)
x(0,1)+y(-√3/2,-1/2)+z(√3/2,1/2)=(1,1)
=>
(-√3/2)y +(√3/2)z=1 (1) and
x-(1/2)y+(1/2)z = 1 (2)
from (1)
z-y = 2√3/3
from (2)
z-y = 2(1-x)
2√3/3 = 2(1-x)
x= 1- √3/3
S = x^2+y^2+z^2
= (1- √3/3)^2 + (z- 2√3/3)^2 +z^2
S' = 2(z- 2√3/3)+2z =0
z= √3/3
S'' >0 ( min)
min S = (1- √3/3)^2 +(√3/3-2√3/3)^2+(√3/3)^2
= 1+1/3 - 2√3/3 + 1/3+1/3
= 2-2√3/3