Determine the point on the plane x 5y z 6 that is closest

Determine the point on the plane x - 5y + z = 6 that is closest to the point (1, 1, 1). And the distance. Redo Problem 1 using a Lagrange multiplier.

We need to minimise

f(x,y,z)=(x-1)^2+(y-1)^2+(z-1)^2

Constraint is g(x,y,z)=x-5y+z-6

Lagrangian is

L=f+tg

L_x=0=2(x-1)+t=0 , x=-t/2+1

L_y=2(y-1)-5t=0, y=5t/2+1

L_z=2(z-1)+t=0, z=-t/2+1

x-5y+z-6=0

-t/2+1-5(5t/2+1)-t/2+1-6=0

-t-4-25t/2-5=0

-27t/2-9=0

-3t/2=1

t=-2/3

So the point is

x=-t/2+1=4/3

y=5t/2+1=-2/3

z=-t/2+1=4/3

Poitn is (4/3,-2/3,4/3)