A normal vector to a curve at a particular point is a vector perpendicular to the tangent vector of the curve at that point (also called a gradient), for an implicit 2D function in the form $$f(x,y) = 0$$ the 2D gradient is

$\nabla f(x,y) = \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right )$

For an implicit 3D function the normal is the vector perpendicular to the surface, the surface normal at a point $$\mathbf{p}$$ is given by the gradient of the implicit function

$\mathbf{n} = \nabla f (\mathbf{p}) = \left ( \frac{\partial f(\mathbf{p})}{\partial x}, \frac{\partial f(\mathbf{p})}{\partial y}, \frac{\partial f(\mathbf{p})}{\partial z} \right )$

For a plane we know that the dot product of the normal $$\mathbf{n}$$ and any vector that lies in the plane is zero, therefore we can model a plane as the following implicit equation

$(\mathbf{p} - \mathbf{a}) \cdot \mathbf{n} = 0$

Where $$\mathbf{p}$$ and $$\mathbf{a}$$ are any two points lying on the plane, sometimes we want the equation of a plane through points $$\mathbf{a, b, c}$$, the normal can be found by taking the cross product of any two vectors on the plane

$\mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})$

## Transforming normal vectors

Normal vectors do not transform the way we would like when they're multiplied by a transformation matrix, if the points on a surface are transformed by the transformation matrix $$\mathbf{M}$$, a vector $$\mathbf{t}$$ tangent to the surface will still be tangent to the transformed surface, however a surface normal vector $$\mathbf{n}$$ may not be normal to the transformed surface

For example when a transformation matrix $$\mathbf{M} = \mathbf{H_x}(s)$$ that skews points toward the $$x$$ axis multiplies the normal vector $$\mathbf{n}$$, the resulting vector $$\mathbf{Mn}$$ is not normal to the surface, we would like to find a transformation matrix $$\mathbf{N}$$ so that $$\mathbf{Nn}$$ is indeed the surface normal

To find the value of $$\mathbf{N}$$ we start from the fact that the normal $$\mathbf{n}$$ and the tangent $$\mathbf{t}$$ are perpendicular

$\mathbf{ n \cdot t } = 0$

Expressed as a matrix multiplication

$\begin{equation} \label{perpendicular} \mathbf{n}^T \mathbf{t} = 0 \end{equation}$

After the transformation they're still perpendicular so

$(\mathbf{Nn})^T \mathbf{Mt} = 0$

Applying the transpose

$\begin{equation} \label{post-transformation} \mathbf{n}^T \mathbf{N}^T \mathbf{Mt} = 0 \end{equation}$

Relating \eqref{post-transformation} with \eqref{perpendicular} we see that the only way that both equations hold true is that

$\mathbf{N}^T \mathbf{M} = \mathbf{I}$

The value of $$\mathbf{N}$$ is then

\begin{align*} \mathbf{N}^T \mathbf{M} &= \mathbf{I} \\ \mathbf{N}^T \mathbf{MM}^{-1} &= \mathbf{IM}^{-1} \\ \mathbf{N}^T &= \mathbf{M}^{-1} \\ \mathbf{N} &= (\mathbf{M}^{-1})^T \end{align*}

References
• Shirley, P. and Ashikhmin, M. (2005). Fundamentals of computer graphics. Wellesley, Mass.: AK Peters.
• Online Graphics Transforms 2: Normals. (2016). [online] YouTube. Available at: https://www.youtube.com/watch?v=fK45BV7QJe0 [Accessed 8 Mar. 2016].