Question

Find unit normal vector to the surface xy^{2}+2yz=8 at the point (3,-2,1)

Answer 1

Given:

$F(x,y,z)=xy^2+2yz-8,\quad M(3,-2,1)$

The unit normal vector to the surface $F(x,y,z)=0$ is given by

$${\bf N}=\frac{\nabla F}{|\nabla F|}$$

$$\nabla F=\left({\bf i}\frac{\partial}{\partial x}+{\bf j}\frac{\partial}{\partial y}+{\bf k}\frac{\partial}{\partial z}\right)(xy^2+2yz-8)$$

$$=\left({\bf i}y^2+{\bf j}(2xy+2z)+{\bf k}2y\right)=4{\bf i}-10{\bf j}-4{\bf k}$$

$$|\nabla F|=\sqrt{4^2+(-10)^2+(-4)^2}=2\sqrt{33}$$

Finally

$${\bf N}=\frac{2{\bf i}-5{\bf j}-2{\bf k}}{\sqrt{33}}$$