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Find an equation of the plane tangent to the surface
x^3 + 2 y^3 + z^3 = 4 x y z
at the point (1,1,1)
An archived instance of a Calc III forum
A tangent plane
skhadka
Find an equation of the plane tangent to the surface
x^3 + 2 y^3 + z^3 = 4 x y z
at the point (1,1,1) = (x_0,y_0,z_0)
x^3 + 2 y^3 + z^3 - 4 x y z = 0
∂F/∂x = 3x^2-4yz
∂F/∂y = 6y^2-4xz
∂F/∂z = 3z^2-4xy
∇ = (3x^2-4yz,6y^2-4xz, 3z^2-4xy)
∇(1,1,1) = (-1,2,-1) = (a,b,c)
Equation of the plane is
a(x-x_0)+b(y-y_0)+c(z-z_0) = 0
-1(x-1)+2(y-1)-1(z-1)=0