Lagrange multiplier

Question

Find the point on the plane 3x+2y+z=13 closest to the point (1,1,1). (Use Lagrange Multiplier)

Answer 1

Let f(x,y,z) = dis from plane to the point(1,1,1) so f= (x-1)² +(y-1)² +(z-1)² subject to the constraint g(x,y,z) where g= 3x +2y + z -13 so f_x = 2(x-1) f_y =2(y-1) f_z =2(z-1) g_x =3 g_y =2 g_z =1. So from the expression grad f =λ grad g this gives 2(x-1)= 3λ 2(y-1)=2λ and 2(z-1)= λ so 2(y-1) =4(z-1) y-1 then equals 2z-2 and z= (y+1)/2 likewise 2(x-1)=6(z-1) so 2x-2 =6z -6 or 2x-2 = 6((y+1)/2) -6 = 3y-3 which gives x=(3y-1)/2 Substituting back into the constraint equation for x and z in terms of y gives y= 2 and x=5/2 and z=3/2. So the point on the plane closet to the point (1,1,1) is ( 5/2,2,3/2).