可以轉(zhuǎn)化為最優(yōu)化問題(在曲面上任取一點,求點到平面距離最小),用拉格朗日乘數(shù)法
d=|x0+2y0+3z0|/√(1+2²+3²)=|x0+2y0+3z0|/√14
目標(biāo)函數(shù)：min f(x0,y0,z0)=14d²=(x0+2y0+3z0)²
約束條件：g(x0,y0,z0)=x0²+2y0-z0=0
構(gòu)造拉格朗日函數(shù)L(x0,y0,z0,λ)=f(x0,y0,z0)+λg(x0,y0,z0)=(x0+2y0+3z0)²+λ(x0²+2y0-z0)
則
∂L/∂x0=2(x0+2y0+3z0)+2λx0=0
∂L/∂y0=4(x0+2y0+3z0)+2λ=0
∂L/∂z0=6(x0+2y0+3z0)-λ=0
∂L/∂λ=x0²+2y0-z0=0
解得
λ=-2(x0+2y0+3z0)=6(x0+2y0+3z0)=0
x0+2y0+3z0=0
x0=y0=z0=0
min f(x0,y0,z0)=14d²=(x0+2y0+3z0)²=0
d=0
即平面和曲面存在交點(0,0,0),最短距離為0萬分感謝................