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Use Lagrange multipliers to find the shortest distance?

Use Lagrange multipliers to find the shortest distance from the point (2, 0, -3) to the plane x+y+z=1.

(x-2)^2+y^2+(z+3)^2

g=x+y+z=1

<2(x-2), 2y, 2(z+3)>=λ<1, 1, 1>

2(x-2)=1λ

2y=1λ

2(z+3)=1λ

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
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To make it easier, you can minimize the squared distance.

The squared distance is f(x,y,z) = (x - 2)2 + y2 + (z + 3)2

For the constraint, you can take g(x,y,z) = x + y + z = 1, as you already did.

This way ∇f = λ∇g to get the other three equations, which you already computed. Here they are simplified, followed by the constraint.
2x - 4 - λ = 0

2y - λ = 0

2z + 6 - λ = 0

x + y + z = 1

We can eliminate λ using the second equation in which λ = 2y. You get

2x - 2y = 4

-2y + 2z = -6

x + y + z = 1

Of these two equations, add the first two together and divide both sides by two.

x - 2y + z = -1

Subtract this equation from the constraint.

3y = 2

y = 2/3

Now substitute, this in.

2x - 2(2/3) = 4

x = (4 + 4/3) / 2 = 8/3

-2(2/3) + 2z = -6

z = (-6 + 4/3) / 2 = -7/3

You get <8/3, 2/3, -7/3>.

It's distance from <2,0,-3> is

√[(8/3-2)2+(2/3)2+(-7/3+3)2] = (2/3)√3