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Normal Vector

The normal to a Plane specified by

f(x,y,z)=ax+by+cz+d = 0
\end{displaymath} (1)

is given by
{\bf N} = \nabla f = \left[{\matrix{a\cr b\cr c\cr}}\right].
\end{displaymath} (2)

The normal vector at a point $(x_0,y_0)$ on a surface $z = f(x,y)$ is
{\bf N} = \left[{\matrix{f_x(x_0,y_0)\cr f_y(x_0,y_0)\cr -1\cr}}\right].
\end{displaymath} (3)

In the Plane, the unit normal vector is defined by

\hat {\bf N} \equiv {d\hat {\bf T}\over d\phi},
\end{displaymath} (4)

where $\hat {\bf T}$ is the unit Tangent Vector and $\phi$ is the polar angle. Given a unit Tangent Vector
\hat {\bf T} \equiv u_1\hat {\bf x}+u_2\hat {\bf y}
\end{displaymath} (5)

with $\sqrt{{u_1}^2+{u_2}^2}=1$, the normal is
\hat {\bf N} = -u_2\hat {\bf x}+u_1\hat {\bf y}.
\end{displaymath} (6)

For a function given parametrically by $(f(t), g(t))$, the normal vector relative to the point $(f(t), g(t))$ is therefore given by
$\displaystyle x(t)$ $\textstyle =$ $\displaystyle -{g'\over\sqrt{f'^2+g'^2}}$ (7)
$\displaystyle y(t)$ $\textstyle =$ $\displaystyle {f'\over\sqrt{f'^2+g'^2}}.$ (8)

To actually place the vector normal to the curve, it must be displaced by $(f(t), g(t))$.

In 3-D Space, the unit normal is

\hat {\bf N} \equiv {{d\hat {\bf T}\over ds}\over \left\vert...
...\kappa} {d\hat {\bf T}\over ds},
\hrule width 0pt height 5.2pt
\end{displaymath} (9)

where $\kappa$ is the Curvature. Given a 3-D surface $F(x,y,z)=0$,
\hat{\bf n} = {F_x+F_y+F_z\over\sqrt{{F_x}^2+{F_y}^2+{F_z}^2}}.
\end{displaymath} (10)

If the surface is defined parametrically in the form
$\displaystyle x$ $\textstyle =$ $\displaystyle x(\phi,\psi)$ (11)
$\displaystyle y$ $\textstyle =$ $\displaystyle y(\phi,\psi)$ (12)
$\displaystyle z$ $\textstyle =$ $\displaystyle z(\phi,\psi),$ (13)

define the Vectors
{\bf a}\equiv\left[{\matrix{x_\phi\cr y_\phi\cr z_\phi\cr}}\right]
\end{displaymath} (14)

{\bf b}\equiv\left[{\matrix{x_\psi\cr y_\psi\cr z_\psi\cr}}\right].
\end{displaymath} (15)

Then the unit normal vector is
\hat {\bf N} = {{\bf a}\times{\bf b}\over \sqrt{\vert{\bf a}\vert^2\vert{\bf b}\vert^2 -\vert{\bf a}\cdot{\bf b}\vert^2}}.
\end{displaymath} (16)

Let $g$ be the discriminant of the Metric Tensor. Then
{\bf N} = {{\bf r}_1\times{\bf r}_2\over\sqrt{g}} = \epsilon_{ij} {\bf r}^j.
\end{displaymath} (17)

See also Binormal Vector, Curvature, Frenet Formulas, Tangent Vector


Gray, A. ``Tangent and Normal Lines to Plane Curves.'' §5.5 in Modern Differential Geometry of Curves and Surfaces. Boca Raton, FL: CRC Press, pp. 85-90, 1993.

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© 1996-9 Eric W. Weisstein