Respuesta :

LammettHash
Lagrangian:

[tex]L(x,y,z,\lambda)=x^2+(y-2)^2+(z-5)^2+\lambda(x-2y+3z-6)[/tex]

where the function we want to minimize is actually [tex]\sqrt{x^2+(y-2)^2+(z-5)^2}[/tex], but it's easy to see that [tex]\sqrt{f(\mathbf x)}[/tex] and [tex]f(\mathbf x)[/tex] have critical points at the same vector [tex]\mathbf x[/tex].

Derivatives of the Lagrangian set equal to zero:

[tex]L_x=2x+\lambda=0\implies x=-\dfrac\lambda2[/tex]
[tex]L_y=2(y-2)-2\lambda=0\implies y=2+\lambda[/tex]
[tex]L_z=2(z-5)+3\lambda=0\implies z=5-\dfrac{3\lambda}2[/tex]
[tex]L_\lambda=x-2y+3z-6=0[/tex]

Substituting the first three equations into the fourth gives

[tex]-\dfrac\lambda2-2(2+\lambda)+3\left(5-\dfrac{3\lambda}2\right)=6[/tex]
[tex]11-7\lambda=6\implies \lambda=\dfrac57[/tex]

Solving for [tex]x,y,z[/tex], we get a single critical point at [tex]\left(-\dfrac5{14},\dfrac{19}7,\dfrac{55}{14}\right)[/tex], which in turn gives the least distance between the plane and (0, 2, 5) of [tex]\dfrac5{\sqrt{14}}[/tex].
yoodyannapolis

The point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5) is[tex](\frac{-3}{14} ,\frac{24}{7},\frac{61}{14})[/tex].

Let (x,y,z) be the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).Then

[tex]f=x^{2}+(y -2)^{2} +(z-5)^{2} \\\\g=x -2y + 3z - 6[/tex]

The gradient of f and g are

Δ[tex]f=(2x,2(y-3),2(z-5))\\[/tex]=

Δ[tex]g=(1,-2,3)[/tex]

Setting Δf=kΔg

[tex](2x,2(y-3),2(z-5))=[/tex]k(1,-2,3)

2x=k,2(y-3)=-2k,2(z-5)=3k

Solving we get the point [tex](\frac{-3}{14} ,\frac{24}{7},\frac{61}{14})[/tex]

Therefore the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5) is[tex](\frac{-3}{14} ,\frac{24}{7},\frac{61}{14})[/tex].

Learn more:https://brainly.in/question/25284663

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