find the point on the plane

Question

find the point on the plane 20x+15y+12z = 120 that is closest to the sphere x^2+y^2+z^2=1

Answer 1

Minimizing a distance is the same as minimizing the squared distance. Also, minimizing a function f, is the same as minimizing f - c where c is any constant c. These observations imply that is equivalent to just minimize the squared distance from the center of the sphere. That is, minimizing x² + y² + z² given that 20x+15y+12z = 120.

 

There are two good ways to do this.

 

Method 1: Lagrange Multipliers.

 

f(x) = x² + y² + z²

g(x) = 20x+15y+12z

 

The system g=120 ; ∇f = λ∇g is as follows.

 

20x+15y+12z = 120

2x = 20λ

2y = 15λ

2z = 12λ

 

If λ=0, then the latter three equations imply x=y=z=0 which doesn't satisfy the first equation. Thus λ≠0.

 

Therefore the latter three equations imply y = 3x/4 and z = 3x/5. Plug this in.

 

20x + 15(3x/4) + 12(3x/5) = 120

 

20x + (45/4)x + (36/5)x = 120

 

400x + 225x + 144x = 2400

 

769x = 2400

 

x = 2400/769

 

y = 1800/769

 

z = 1440/769

 

Method 2: Vectors.

 

The normal vector to the plane is n=⟨20,15,12⟩, so the parametric equation to the orthogonal line through the origin is

 

x = 20t ; y = 15t ; z = 12t

 

These three equations together with 20x+15y+12z = 120 are the same system as with the Lagrange Multipliers if you set t = λ/2, so you will get the same solution.