Example Find a unit normal vector ${\bf n}$ of the surface $x^2y + 2xz = 4$ at the point $(2,-2,3)$

Answer Note that $\nabla \phi$ at the point $P$ is diagonal to the surface $\phi(x,y,z) = x^2 y + 2xz = 4$. Then $\boldsymbol{n}$ is

$\displaystyle \boldsymbol{n} = \frac{(\nabla \phi)_{P}}{\vert\nabla \phi\vert _{P}}$

ここで, $(\nabla \phi)_{P} = (2xy +2z)\boldsymbol{i} + x^2 \boldsymbol{j} + 2x\boldsymbol{k})\mid_{(2,-2,3)} = -2\boldsymbol{i} + 4\boldsymbol{j} + 4\boldsymbol{k}$. Thus,

$\displaystyle \boldsymbol{n} = \frac{-2\boldsymbol{i} + 4\boldsymbol{j} + 4\bol...
... + 16 + 16}} = \frac{1}{3}(-\boldsymbol{i} + 2\boldsymbol{j} + 2\boldsymbol{k})$