# How to get a reflection vector?

I'm doing a raytracing exercise. I have a vector representing the normal of a surface at an intersection point, and a vector of the ray to the surface. How can I determine what the reflection will be?

In the below image, I have d and n. How can I get r? Thanks.

• This is question number 5.13 from I.e irodov :p Apr 20, 2021 at 7:23

$$r = d - 2 (d \cdot n) n$$

where $d \cdot n$ is the dot product, and $n$ must be normalized.

• And if my d happens to be pointing the other direction, I need to negate it first, right? Dec 6, 2010 at 18:00
• @user2755 Yes, but you can test this yourself with pencil and paper using simple cases, e.g. $d = [1,-1]; n=[0,1]$ (incoming down and to the right onto a ground plane facing upwards). With this, $r = [1,-1] - 2 \times (-1) \times [0,1] = [1,-1] + 2 \times [0,1] = [1,-1] + [0,2] = [1,1]$. Dec 6, 2010 at 18:06
• I have no idea how this works, but it works, and that's all that matters. Jan 9, 2022 at 18:53
• @rb3652 Here is my understanding. $d\cdot n$ is the length of $d$ along the direction of $n$ (i.e., the "projection" of $d$ onto $n$). Multiplying by $n$ turns this scalar $d\cdot n$ into a vector of length $d$ along the direction $n$. If you substracted this from $d$, you would remove all of the $n$-direction character from $d$ (i.e., it would be perpendicular to $n$ and would be parallel to b in the diagram). Subtracting it again yields a vector that is reflected back since you're effectively adding the $n$-direction character in the opposite direction of $d$ again. Jul 18 at 13:20

Let $\hat{n} = {n \over \|n\|}$. Then $\hat{n}$ is the vector of magnitude one in the same direction as $n$. The projection of $d$ in the $n$ direction is given by $\mathrm{proj}_{n}d = (d \cdot \hat{n})\hat{n}$, and the projection of $d$ in the orthogonal direction is therefore given by $d - (d \cdot \hat{n})\hat{n}$. Thus we have $$d = (d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$ Note that $r$ has $-1$ times the projection onto $n$ that $d$ has onto $n$, while the orthogonal projection of $r$ onto $n$ is equal to the orthogonal projection of $d$ onto $n$, therefore $$r = -(d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$ Alternatively you may look at it as that $-r$ has the same projection onto $n$ that $d$ has onto $n$, with its orthogonal projection given by $-1$ times that of $d$. $$-r = (d \cdot \hat{n})\hat{n} - [d - (d \cdot \hat{n})\hat{n}]$$ The later equation is exactly $$r = -(d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$

Hence one can get $r$ from $d$ via $$r = d - 2(d \cdot \hat{n})\hat{n}$$ Stated in terms of $n$ itself, this becomes $$r = d - {2 d \cdot n\over \|n\|^2}n$$

• Does this hold for vectors of any dimension? 2D, 3D, 4D, etc? Sep 17, 2016 at 19:36
• @NightElfik Abosolutely Feb 26, 2017 at 9:10

I was trying to understand how to calculate the reflection vector and found these answers. I couldn't understand them easily, so I took my time to do it myself, the good thing is that I can now detail it in an ELI5 fashion!

I did develop the formula using the 3 steps shown in the graphic. I describe them bellow. So, the initial situation is $$\vec{a}$$ pointing toward a plane. Then we have the normal $$\vec{n}$$ of unit lenght and we would like to find $$\vec{b}$$

So, the first step is using the dot product to get a vertical vector that will be used in step 2.

With step 1 my partial formula is: $$2\times\left(a+(-\vec{a})\cdot\vec{n}\times{}n\right)$$

mind the change of sign of $$\vec{a}$$ above, we "flipped" it

Then in step 2, I can write: $$-\vec{a}+2\times\left(a+(-\vec{a})\cdot\vec{n}\times{}n\right)$$

Now, I can distribute: $$-\vec{a}+2\times{}\vec{a}+2\times(-\vec{a})\cdot\vec{n}\times{}n$$

Then simplify, and I end up with: $$\vec{a}+2\times(-\vec{a})\cdot\vec{n}\times{}n$$

If you negate a vector in the dot product, you negate the result of the dot product.

$$\vec{a}\cdot\vec{b}=-(-\vec{a})\cdot\vec{b}$$

That means that I can rewrite the formula like this:

$$\vec{a}-2\times(\vec{a})\cdot\vec{n}\times{}n$$

• Thank you. Though the way you used Cross Product's notation as a multiplication notation confused me big time. Sep 13, 2022 at 5:30

Suppose that $$d$$ and $$r$$ have the same magnitude. $$\lVert r \rVert = \lVert d \rVert$$ From the reflection relationship, we have this equality about cross products. $$r \times n \ = \ d \times n \\ \therefore \ \left( r \ - d \right) \times n \ = \ \vec{0}$$ which means $$r \ - d \ = s \ n \\ \therefore \ r \ = \ d \ + s \ n$$ where $$s$$ is a real number.
Taking their squares, we have $$\lVert r \rVert ^2 \ = \ \lVert d \rVert ^2 + \ 2\ s \left( d \cdot n \right) \ + s^2 \ \lVert n \rVert ^2 \\ \therefore \ s \left( s \ \lVert n \rVert ^2 + \ 2 \ (d \cdot n) \right) = 0 \\$$ So $$s \ = 0 \ , - \frac{2 \ (d \cdot n)}{\lVert n \rVert ^2}$$ Since $$s = 0 \$$ means $$\ d \$$ itself, we take the other value and get $$r \ = \ d - \frac{2 \ (d \cdot n)}{\lVert n \rVert ^2} \ n$$

In case you want to rotate about Y axis you can use the following instead. This is mostly useful for computer graphics applications. Note that $$d$$ is assumed to be pointing outward in the equation below (i.e. ignore the direction of $$d$$ in the picture below) and $$n$$ needs to be normalized: $$r = 2 (d \cdot n) n - d$$