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已知a向量=(41,-1)b向量=(-1,1)若向量c满足(a-c)(b-c)=0求c的模和a-c的模的最小值
人气:467 ℃ 时间:2020-06-03 07:17:01
解答
设c的坐标为 (x,y)
则 (a-c)(b-c)=a*b-(a+b)*c+c*c=0
=>-41-1-(40,0)*(x,y)+(x^2+y^2)=0 =>
x^2+y^2-41x=42
=>(x-20)^2+y^2=442
设 x=r*cost+20.y=r*sint r=442^0..5
则 c*c=x^2+y^2=r^2*7(sint)^2+(rcost+20)^2=r^2+400+40*r*cost>=r^2+400-40*r=(r-20)^2
=> cost=-1 时 min{|c|}=r-20=442^2-20
|c-a|^2=(rcost+20-41)^2+(rsint+1)^2=r^2+21^2-42*r*cost+1+2*r*sint
=r^2+21^2+1+2r*(sint-21cost)>=r^2+21^2+1-2r*(1+21^2)^0.5
=442+441+1-2*442^0.5*(442)^0.5=0
min{|a-c|}=0
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