# Reflection of vectors

1. Apr 27, 2008

### Icosahedron

Could someone explain how the reflection formula comes about?

R_a v = v - 2a(va)/ (aa) , where v and a are vectors and v is relected along the axis orthogonal to a

thanks

2. Apr 27, 2008

### HallsofIvy

Staff Emeritus
I presume that by "va" and "aa" you mean the dot product.

Decompose $\vec{v}$ into components parallel to and orthogonal to $\vec{a}$.

Recall that $\vec{a}\cdot\vec{v}$ can be defined as "$|\vec{a}||\vec{v}| cos(\theta)$ where $\theta$ is the angle between $\vec{a}$ and $\vec{v}$.

From simple trigonometry, the component of $\vec{v}$ parallel to $\vec{a}$ is $|\vec{v}|cos(\theta)= \vec{a}\cdot\vec{v}/|\vec{a}|$. To get a "vector projection", a vector in the direction of $\vec{a}$ with that length, multiply by the unit vector in the direction of $\vec{a}$, $\vec{a}/|\vec{a}|$. That gives $(\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a}$ as the "vector component of $\vec{v}$ parallel to $\vec{a}$". Since the components parallel and perpendicular to $\vec{a}$ must add to $\vec{v}$, the component perpendicular to $\vec{a}$ is $\vec{v}-(\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a}$.

Now, reflecting $\vec{v}$ along an axis orthogonal to $\vec{a}$ gives a new vector having the same component perpendicular to $\vec{a}$ but with the component parallel reversed:
$\vec{v}-(\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a} -\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a}$
$= \vec{v}-2(\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a}$

3. Apr 27, 2008

### dodo

I have an incomplete answer, but maybe you can finish it.

Let v' be the reflection of v over some axis. If you try to draw a vector s such that v = v' + s, using the triangle method to add the vectors v and s, you'll see that s is orthogonal to the reflection axis, and thus colinear to a.

The drawing should show an isosceles triangle, bisected by the reflection axis, where the two equally-sized sides are |v| and |v'|, the third side being of size |s|. From this triangle you should be able to calculate the magnitude of s as equal to 2 . |v| . cos theta, where "theta" is the angle between v (or v') and the normal to the reflection axis. (Try it on the drawing!)

Now, the expression 2 . |v| . cos theta is equal in magnitude to 2 . (va) / |a|. Multiplying this quantity by an unitary vector in the direction of a, namely a / |a|, you should get your vector s... or almost. There is a +/-1 factor not accounted yet for: has s the same direction of a or the opposite?

(This sign is the missing part that I was lazy to complete - I'd guess it has to do with v being or not on the same side of the axis than a; examining the sign of (va) should give a clue.)

Doing the multiplication above (and correcting the sign), you should get s = 2 (va) . a / |a|^2 = 2 (va) . a / (aa), and when replacing this expression for s into v' = v - s, you get your reflection formula.

Edit: Oh, I was pwned again. :)

Last edited: Apr 27, 2008
4. Apr 27, 2008

### Icosahedron

thanks a lot!!