Use Lagrange Multipliers to solve the problem. Find the distance from the point (1,-1,2) to the plane x+y-z=3.

Question

I do not understand the steps for the Lagrange Multipliers

Answer 1

Distance squared to the point (1,-1,2): d²=(x-1)²+(y+1)²+(z-2)²

Let f=(x-1)²+(y+1)²+(z-2)²  (the fcn. we wish to minimize) we only care about points on the plane so let the constraint be x+y-z=3

so that g=x+y-z-3. 

Then: ∇f=λ∇g  --->  <2(x-1), 2(y+1),2(z-2)>=λ<1,1,-1>,

this plus the constraint (x+y-z=3) gives 4 eqs. in 4 unknowns.

 

2(x-1)= λ,   2(y+1)= λ,   2(z-2)=-λ   give (by equating the λs)

2(x-1)=2(y+1) --> y=x-2   and   x-1=-(z-2)  --> z=3-x

Insert these into the constraint and solve for x:

x+(x-2)-(3-x)=3 --> x=8/3  so y=8/3-2=2/3 and z=3-8/3=1/3

so the point (8/3,2/3,1/3) is the point on the plane closest

to the point (1,-1,2).

Insert this back into the distance formula: d²=(x-1)²+(y+1)²+(z-2)²

so d²=3(5/3)²  so d=5√3/3 .

 

You can check (8/3,2/3,1/3) by looking at the straight line

through the point (1,-1,2) and perpendicular to the plane.

Note: x+y-z=3 gives  N=<1,1,-1> for a normal vector, so the line (ie, r=P₀+tN) would be given by the eqs. x=1+t, y=-1+t, z=2-t .

Just substitute these into x+y-z=3 and solve for t  for the point.