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

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

- 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].