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Reflection of vectors

  1. Apr 27, 2008 #1
    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

  2. jcsd
  3. Apr 27, 2008 #2


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

    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 perpendicular to [itex]\vec{a}[/itex] but with the component parallel reversed:
    [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]
  4. Apr 27, 2008 #3
    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
  5. Apr 27, 2008 #4
    thanks a lot!!
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