Vector Reflection (2D Derivation)

Troy Williams

2021-06-26

Vector Reflection (2D Derivation)

This notebook will work through the explanation of determining the 2D vector reflection from a surface. I had a problem where I needed to determine the reflected vector from an incident vector in two dimensions. There are a lot of pages out there with good explanations. But there is a lot of seemingly conflicting information that caused me to ask some questions and spawn this article. This source is quite nice. It walks you through the steps and develops a valid relationship:

\[ R = 2 \left( \vec{N} \cdot \vec{L} \right) \vec{N} − \vec{L} \]

Unfortunately, the formula is derived with both the incident vector and reflected vector pointing away from the surface. The system diagram can be seen in fig. 1:

Figure 1: Basic system from here

I don’t care for this derivation. I understand why this convention was chosen, it makes derivation easy. To use this equation, you have to take the negative of the incident vector. To me it seems wrong, why should I have to negate the input even before it is used in a calculation. This isn’t the end of the world, but it is something to be aware of.

I wanted something that modeled the system directly with the incident vector pointing at the surface it would hit. See fig. 2 for details:

Figure 2: Incident vector, I, reflecting from surface with normal N as vector R.

I wanted something that would not require changing any of the inputs. I wanted something that could use the inputs directly. Here are the components that are illustrated in fig. 2:

This source on Stackexchange, lists the formula that is suited for the system I am interested in. I wanted to understand where this formula comes from and how it is derived. The rest of the article works through that process. If we examine the system above, we can see that it is symmetrical, we know this because the angle of incidence must equal the angle of reflection.

Projection

If we project \(\vec{I}\) to the normal vector, \(\vec{N}\) we will obtain \(\vec{N'}\) (fig. 3).

A key fact here is, by symmetry, we expect that projecting \(\vec{R}\) to \(\vec{N}\) will produce \(-\vec{N'}\). This took me a while to build an intuition for. It took a careful examination of the problem to grasp it. The projection of \(\vec{I}\) or \(\vec{R}\) maintains their direction, pointing towards the reflection plane or away from the plane. This is the key in the difference of the formula derivations from the previously mentioned sources.

Ride the Point

If you imagine you are projecting particles at the surface and they obey Snell’s law and reflect off the surface (ignoring collisions and energy loss, etc.). We want to predict the location of this particle after it is reflected. Boiling it down, we want to predict the reflected direction vector. For this system, if we have a point at the tail of \(\vec{I}\) (fig. 3) and we want to move it to the tip of \(\vec{R}\) how do we do that?

NOTE: See appendix 1 for details about the assumption of projecting particles along with vector addition and subtraction.

From the diagrams below (fig. 3 and fig. 4), if we project \(\vec{I}\) onto \(\vec{N}\) we get \(\vec{N'}\)

\[ \vec{N'} = \left ( \vec{I} \cdot \vec{N} \right) \vec{N} \]

We can solve for \(\vec{U}\):

\[ \vec{U} = \vec{I} - \vec{N'}\]

Figure 3: Vector U defined.

By symmetry, we assume that the vector from the tail of \(\vec{N'}\) to the tip of \(\vec{R}\) is equal to \(\vec{U}\):

\[ \vec{U} + \vec{N'} = \vec{R} \]

\[ \vec{U} = \vec{R} - \vec{N'}\]

Figure 4: The complete system, how to get from the incident vector to the reflected vector.

This means:

\[ \vec{I} + \vec{R} = 2\vec{U} \]

Solving for \(\vec{R}\):

\[ \vec{R} = 2\vec{U} - \vec{I}\]

Expand and simplify:

\[ \vec{R} = 2 \left( \vec{I} - \left ( \vec{I} \cdot \vec{N} \right) \vec{N} \right) - \vec{I}\]

\[ \vec{R} = 2\vec{I} - 2\left ( \vec{I} \cdot \vec{N} \right) \vec{N} - \vec{I}\]

\[ \vec{R} = \vec{I} - 2\left ( \vec{I} \cdot \vec{N} \right) \vec{N} \](1)

Conclusion

We can see the equation we need is eq. 1. It defines how to construct the reflection vector from the incident vector. The derivation was straight forward (once I had a better grasp of some of the basics). I felt like I should have had a better grasp when I learned these years ago. I might have, but in the ensuing years, that knowledge was lost.

Appendix

Vector Addition and Subtraction

As an aside, I took a quick refresher of the basics of vector addition and subtraction. It is a seemingly simple topic that every first year science student is acquainted with. I hadn’t thought much about the process or what it really meant since those days. I took it for granted that I knew the topic. This source contains a nice refresher. I went back to first principles to develop a better intuition for the issue.

In essence you are trying to form a parallelogram between the vectors you are adding, with the diagonal representing the new vector. Something like this:

\[ \vec{W} = \vec{U} + \vec{V} \]

Figure 5: Vector Addition - Parallelogram

\[ \vec{W} = \vec{U} - \vec{V} \]

Figure 6: Vector Subtraction

Visually, I had trouble with this situation:

Figure 7: Basic System with two vectors pointing at each other.

Forming the parallelogram didn’t help with the intuition. Something simple like this can cause a lot of grief when working on advanced topics. How do we reason about \(\vec{W}\) given that \(\vec{U}\) and \(\vec{V}\) point towards the same point.

Given \(\vec{U}\) and \(\vec{V}\) how do we get \(\vec{W}\). It took me a while to really understand what is happening. There are two approaches a visual approach and an algebraic approach.

In the following diagram, if we play with it a little we can see that:

\[\vec{W} = \vec{U} - \vec{V}\]

Figure 8:

What helped me is to visualize a point at the tail of \(\vec{U}\) and to translate that point to the tip of \(\vec{W}\). How do we get there from here? Take the point and slide it along the direction vector it is pointing to the tip of \(\vec{U}\). Now slide it along \(\vec{V}\), but we can’t. It is pointing in the wrong direction. We can however slide it along \(-\vec{V}\). That brings us to \(\vec{W}\).

That intuition helped me reason about the vector addition and subtraction. It seems we are pushing points around. It gave me a deeper understanding of the process in a way that I hadn’t comprehended in the past. I am very much a person that would simply look up the formula and apply it. I suppose that is my engineering training taking hold. In these past years I have forced my self to understand the mathematics I work with on a deeper level, understanding the most basic building blocks. It has led down some deep rabit holes.

Alternatively, there is another, simpler approach. We can take the algebraic approach. Observe that:

\[\vec{U} = \vec{W} + \vec{V}\]

Rearranging and solving for \(\vec{W}\):

\[\vec{W} = \vec{U} - \vec{V}\]

The algebra is easier, but I think it is still a good idea to have an intuition about the nature of the problem and what the mathematics is trying to solve.

NOTE: The algebraic approach works for two dimensional and three dimensional vectors. It should holds for n-dimensional vectors (I haven’t proven it to myself, I would defer to people how are much better educated in this regard). The intuition may not hold true in higher dimensions - I make no claim to that. I am simply presenting what has helped me.

Dot Product

Wikipedia has an excellent definition of the dot product. It is correct and very informative. But I had a gap in a very fundamental aspect of it. The dot product, scalar product or inner product is defined for vectors \(\vec{a}\) and \(\vec{b}\):

\[ \vec{a} \cdot \vec{b} = \sum_{i=1}^{n} a_i b_i = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \]

I was always taught that the dot product is the projection of the \(\vec{a}\) onto the \(\vec{b}\). I always thought there was an implied order, that is, you could project \(\vec{a}\) onto \(\vec{b}\) and \(\vec{b}\) onto \(\vec{a}\). If you look at the Wikipeida page, it seems to imply this. However, that thought isn’t supported by the mathematics.

The dot product is commutative - a very important notion:

\[ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \]

I am not sure why this was stuck in my head for so long. I had to work through it, reconcile and eliminate that faulty thinking. Looking at fig. 9, we can see that the dot product is typically representing one vector projected onto another. I think this is the source, seeing diagrams and definitions that explicitly mention it. If we really think about it, what the dot product represents is the shortest distance we would travel along a vector (among many other notions that may not be related to geometric distance). If b, were shorter than a, we would have the diagram reversed. I think this is similar to the issue in algebra labeling the unknown quantity as x.

Figure 9: vector a projected onto vector b (attribution)