Normal (geometry)

For a convex polygon (such as a triangle), a surface normal can be calculated as the vector cross product of two (non-parallel) edges of the polygon.

For a plane given by the equation ${\displaystyle ax+by+cz+d=0,}$ the vector ${\displaystyle \mathbf {n} =(a,b,c)}$ is a normal.

For a plane whose equation is given in parametric form

$${\displaystyle \mathbf {r} (s,t)=\mathbf {r} _{0}+s\mathbf {p} +t\mathbf {q} ,}$$

where ${\displaystyle \mathbf {r} _{0}}$ is a point on the plane and ${\displaystyle \mathbf {p} ,\mathbf {q} }$ are non-parallel vectors pointing along the plane, a normal to the plane is a vector normal to both ${\displaystyle \mathbf {p} }$ and ${\displaystyle \mathbf {q} ,}$ which can be found as the cross product ${\displaystyle \mathbf {n} =\mathbf {p} \times \mathbf {q} .}$

If a (possibly non-flat) surface ${\displaystyle S}$ in 3D space ${\displaystyle \mathbb {R} ^{3}}$ is parameterized by a system of curvilinear coordinates ${\displaystyle \mathbf {r} (s,t)=(x(s,t),y(s,t),z(s,t)),}$ with ${\displaystyle s}$ and ${\displaystyle t}$ real variables, then a normal to S is by definition a normal to a tangent plane, given by the cross product of the partial derivatives

$${\displaystyle \mathbf {n} ={\frac {\partial \mathbf {r} }{\partial s}}\times {\frac {\partial \mathbf {r} }{\partial t}}.}$$

If a surface ${\displaystyle S}$ is given implicitly as the set of points ${\displaystyle (x,y,z)}$ satisfying ${\displaystyle F(x,y,z)=0,}$ then a normal at a point ${\displaystyle (x,y,z)}$ on the surface is given by the gradient

$${\displaystyle \mathbf {n} =\nabla F(x,y,z).}$$

since the gradient at any point is perpendicular to the level set ${\displaystyle S.}$

For a surface ${\displaystyle S}$ in ${\displaystyle \mathbb {R} ^{3}}$ given as the graph of a function ${\displaystyle z=f(x,y),}$ an upward-pointing normal can be found either from the parametrization ${\displaystyle \mathbf {r} (x,y)=(x,y,f(x,y)),}$ giving

$${\displaystyle \mathbf {n} ={\frac {\partial \mathbf {r} }{\partial x}}\times {\frac {\partial \mathbf {r} }{\partial y}}=\left(1,0,{\tfrac {\partial f}{\partial x}}\right)\times \left(0,1,{\tfrac {\partial f}{\partial y}}\right)=\left(-{\tfrac {\partial f}{\partial x}},-{\tfrac {\partial f}{\partial y}},1\right);}$$

or more simply from its implicit form ${\displaystyle F(x,y,z)=z-f(x,y)=0,}$ giving ${\displaystyle \mathbf {n} =\nabla F(x,y,z)=\left(-{\tfrac {\partial f}{\partial x}},-{\tfrac {\partial f}{\partial y}},1\right).}$ Since a surface does not have a tangent plane at a singular point, it has no well-defined normal at that point: for example, the vertex of a cone. In general, it is possible to define a normal almost everywhere for a surface that is Lipschitz continuous.

A vector field of normals to a surface

The normal to a (hyper)surface is usually scaled to have unit length, but it does not have a unique direction, since its opposite is also a unit normal. For a surface which is the topological boundary of a set in three dimensions, one can distinguish between the inward-pointing normal and outer-pointing normal. For an oriented surface, the normal is usually determined by the right-hand rule or its analog in higher dimensions.

If the normal is constructed as the cross product of tangent vectors (as described in the text above), it is a pseudovector.

When applying a transform to a surface it is often useful to derive normals for the resulting surface from the original normals.

Specifically, given a 3×3 transformation matrix ${\displaystyle \mathbf {M} ,}$ we can determine the matrix ${\displaystyle \mathbf {W} }$ that transforms a vector ${\displaystyle \mathbf {n} }$ perpendicular to the tangent plane ${\displaystyle \mathbf {t} }$ into a vector ${\displaystyle \mathbf {n} ^{\prime }}$ perpendicular to the transformed tangent plane ${\displaystyle \mathbf {Mt} ,}$ by the following logic:

Write n′ as ${\displaystyle \mathbf {Wn} .}$ We must find ${\displaystyle \mathbf {W} .}$

$${\displaystyle {\begin{alignedat}{5}W\mathbb {n} {\text{ is perpendicular to }}M\mathbb {t} \quad \,&{\text{ if and only if }}\quad 0=(W\mathbb {n} )\cdot (M\mathbb {t} )\\&{\text{ if and only if }}\quad 0=(W\mathbb {n} )^{\mathrm {T} }(M\mathbb {t} )\\&{\text{ if and only if }}\quad 0=\left(\mathbb {n} ^{\mathrm {T} }W^{\mathrm {T} }\right)(M\mathbb {t} )\\&{\text{ if and only if }}\quad 0=\mathbb {n} ^{\mathrm {T} }\left(W^{\mathrm {T} }M\right)\mathbb {t} \\\end{alignedat}}}$$

Choosing ${\displaystyle \mathbf {W} }$ such that ${\displaystyle W^{\mathrm {T} }M=I,}$ or ${\displaystyle W=(M^{-1})^{\mathrm {T} },}$ will satisfy the above equation, giving a ${\displaystyle W\mathbb {n} }$ perpendicular to ${\displaystyle M\mathbb {t} ,}$ or an ${\displaystyle \mathbf {n} ^{\prime }}$ perpendicular to ${\displaystyle \mathbf {t} ^{\prime },}$ as required.

Therefore, one should use the inverse transpose of the linear transformation when transforming surface normals. The inverse transpose is equal to the original matrix if the matrix is orthonormal, that is, purely rotational with no scaling or shearing.

Let V be the variety defined in the 3-dimensional space by the equations

$${\displaystyle x\,y=0,\quad z=0.}$$

This variety is the union of the ${\displaystyle x}$-axis and the ${\displaystyle y}$-axis.

At a point ${\displaystyle (a,0,0),}$ where ${\displaystyle a\neq 0,}$ the rows of the Jacobian matrix are ${\displaystyle (0,0,1)}$ and ${\displaystyle (0,a,0).}$ Thus the normal affine space is the plane of equation ${\displaystyle x=a.}$ Similarly, if ${\displaystyle b\neq 0,}$ the normal plane at ${\displaystyle (0,b,0)}$ is the plane of equation ${\displaystyle y=b.}$

At the point ${\displaystyle (0,0,0)}$ the rows of the Jacobian matrix are ${\displaystyle (0,0,1)}$ and ${\displaystyle (0,0,0).}$ Thus the normal vector space and the normal affine space have dimension 1 and the normal affine space is the ${\displaystyle z}$-axis.