Distance from a point to tangent plane

[rac_se]

f(x, y, z) = x² + sin(y) - 2z² = 0 defines a surface in 3 dimensions. First I need to find the equation of the normal line to the surface at point P₀(2pi, 0, 3/2). Then, I need to find the point which is at a distance of 4 from the tangent plane at the point P₀

Equation for the normal line at a point P₀ is given by:
x = x₀ + f_x(P_o)t
y = y₀ + f_y(P_o)t
z = z₀ + f_z(P_o)t

Equation for the tangent plane is given by:
f_x(P₀)(x-x_o) + f_y(P₀)(y-y_o) + f_z(P₀)(z-z_o)

I used the first equation to find the normal line at P₀ and the answer I got is:
x = 2pi-4pi*t
y= t
z = 3/2 - 6t
I know that the distance between a point P₀ on a plane with normal n and a point P is given by the dot product of the vector projection of P₀P and n/|n|. I get the equation for normal n from first part. But, |n| =sqrt(x² + y ² + z²) gives very complicated result. Then, I am not sure how to use the distance formula to find the point. Or, there is some other way to solve this problem.

I hope someone could guide me on how to approach this problem. This is the first time I am posting to this forum. So, please correct me if I made any mistake

Thanks!

 

[vela]

The point P₀ isn't on the surface. Are you sure you copied the problem down correctly?

 

[rac_se]

I am sorry, but yes there is a little mistake, the function that defines the surface is:

f(x, y, z) = -x² + sin(y) - 2z² = 0

I missed the minus sign before x²

 

[vela]

f(P₀) still won't equal 0, so P₀ doesn't lie on the surface.