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

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- Sep 26th 2008, 07:41 PMArythUnit Normal
Show that a unit normal to the surface $\displaystyle x^3y^3 + y - z + 2 = 0$ is given by $\displaystyle \bold{n} = \frac{1}{\sqrt{2}}(\bold{j} - \bold{k})$

- Oct 18th 2008, 05:08 AMRebesques
Start by finding a normal to the point (that you have not mentioned (Thinking)), and normalize, dividing by its length.

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