Question

已知a向量=(1,0),b=(0,1),若向量c=(m,n)满足(a-c)*(b-c)=0,则点(m,n)到直线x+y+

Answer 1

a-c=(1-m,-n) ,b-c=(-m,1-n)

(a-c)*(b-c)=0===>-m(1-m)-n(1-n)=0===>m^2+n^2=m+n

点(m,n)到直线x+y+1=0的距离d=|m+n+1|/√2=|m^2+n^2+1|/√2>=|2mn+1|/√2

(当且仅当m=n时取到等号）此时m=n=0或m=n=1

m=n=0时dmin=√2/2

m=n=1时dmin=3√2/2