Unit Normal

**Aryth** · September 26th 2008, 07:41 PM

Show that a unit normal to the surface $x^3y^3 + y - z + 2 = 0$ is given by $\bold{n} = \frac{1}{\sqrt{2}}(\bold{j} - \bold{k})$

**Rebesques** · October 18th 2008, 05:08 AM

Start by finding a normal to the point (that you have not mentioned

), and normalize, dividing by its length.

In general, the surface given by $f(x,y,z)=0$ will have the vector $\nabla f(x_0,y_0,z_0)$ as normal at $f(x_0,y_0,z_0)$ .

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