Check that the point (1,−1,1) lies on the given surface. Then, viewing the surface as a level surface for a function f(x,y,z), find a vector normal to the surface

Question

Check that the point  lies on the given surface. Then, viewing the surface as a level surface for a function , find a vector normal to the surface 

2x^2−2y^2+2z^2=2

vector normal = ?

Answer 1

The equation 2x^2−2y^2+2z^2=2 ( a hyperboloid of one sheet)

can be seen as level surface of the function

F( x,y,z) = 2x^2−2y^2+2z^2

Then ∇ F = [∂F / ∂x ] i + [∂F / ∂y ] j + [∂F / ∂z ] k = 4x i + 4 y j + 4 z k

and evaluated at ( 1, -1, 1) gives us the normal vector of the tangent plane

n = <4, -4, 4> or its scalar multiple n₀ =< (1, -1, 1 >

Then the equation of the tangent plane

is 1 (x-1) -1(y+1) +1(z-1) = 0

x - y +z = 3