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

thanks

- Thread starter Icosahedron
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- #1

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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

HallsofIvy

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Decompose [itex]\vec{v}[/itex] into components parallel to and orthogonal to [itex]\vec{a}[/itex].

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

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

Now, reflecting [itex]\vec{v}[/itex] along an axis orthogonal to [itex]\vec{a}[/itex] gives a new vector having the same component

[itex]\vec{v}-(\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a} -\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a}[/itex]

[itex]= \vec{v}-2(\vec{a}\cdot\vec{v}/|\vec{a}|^2)\vec{a}[/itex]

- #3

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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 . (**v****a**) / |**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 (**v****a**) should give a clue.)

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

Edit: Oh, I was pwned again. :)

Let

The drawing should show an isosceles triangle, bisected by the reflection axis, where the two equally-sized sides are |

Now, the expression 2 . |

(This sign is the missing part that I was lazy to complete - I'd guess it has to do with

Doing the multiplication above (and correcting the sign), you should get

Edit: Oh, I was pwned again. :)

Last edited:

- #4

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thanks a lot!!

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