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?

Vector d is the ray; n is the normal; t is the refraction; r is the reflection



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

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

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    $\begingroup$ And if my d happens to be pointing the other direction, I need to negate it first, right? $\endgroup$ – Nick Heiner Dec 6 '10 at 18:00
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    $\begingroup$ @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, $r = [1,-1] - 2 \times (-1) \times [0,1] = [1,-1] + 2 \times [0,1] = [1,-1] + [0,2] = [1,1]$. $\endgroup$ – Phrogz Dec 6 '10 at 18:06

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 $$r = d - {2 d \cdot n\over \|n\|^2}n$$

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    $\begingroup$ Does this hold for vectors of any dimension? 2D, 3D, 4D, etc? $\endgroup$ – NightElfik Sep 17 '16 at 19:36
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    $\begingroup$ @NightElfik Abosolutely $\endgroup$ – felknight Feb 26 '17 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.

Steps to compute the rfection vector

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.


That means that I can rewrite the formula like this:



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