Find the point on the plane x+2y+3z=13 that is closest to the point (1,1,1). How would you minimize the function?

Isa Trevino 2021-04-06 Answered
Find the point on the plane x+2y+3z=13 that is closest to the point (1,1,1). How would you minimize the function?
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rogreenhoxa8
Answered 2021-04-08 Author has 109 answers

Apply Lagrange Multipliers:
1) Objective: Minimize Function {Distance From Point (1, 1,1)}:
D=((x1)2+(y1)2+(z1)2)12
2) Constraint Surface:
x+2y+3z=13to g(x,y,z)=x+2y+3z13
3) Formulate Lagrange Multiplier Condition (for constant scalarmultiplier "λ"):
downD=λdowng
(12D12[2(x1)i+2(y1)j+2(z1)k] 
=λ[(1)i+(2)j+(3)k]
4) Determine & Solve Simultaneous Equations:
(x1)=(1)λD12
(y1)=(2)λD12
(z1)=(3)λD12
(x1)=y12=z13
[y=2x1] and [z=3x2]
5) Place Solutions (Eq Set #1) Into Constraint Equation &Derive "x":
x+2y+3z=13
x+2[2x1]+3[3x2]=13
x+4x2+9x6=13
14x=21
x=(32)
6) Place "x" Value Into Eq Set #1 To Complete (x, y, z)Solution:
(x=32)
y=2(32)1=(2)
z=3(32)2=52

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Jeffrey Jordon
Answered 2021-10-14 Author has 2064 answers

Given x+2y+3z=13 Point (1,1,1)

The normal vector to the plane is <1,2,3>

The point you seek would have to be multiple of this vector added to (1,1,1)

P=(1,1,1)+c<1,2,3>

=(1+c,1+2c,1+3c)

But this point has to satisfy the plane`s equation.

(1+c)+2(1+2c)+3(1+3c)=13

1+c+2+4c+3+9c=13

14c=13-6

c=714=12

 the point P is (1+12,1+2×12,1+3×12)

=(32,2,52)

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