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Vector multiplication question

  1. Jan 17, 2009 #1

    Nabeshin

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    I'm not sure where to put this question, since it deals with both physics and math, so I figured here would be a good starting point.

    In the book of astrodynamics I'm currently reading, I came across this expansion:

    [tex](\vec{r}\times\vec{v})\times\vec{r}=[\vec{v}(\vec{r}\cdot\vec{r})-\vec{r}(\vec{r}\cdot\vec{v})][/tex]

    Can anyone explain how this result is arrived at? If any physical significance is needed, r is a position vector and v its derivative with respect to time, the velocity vector.
     
  2. jcsd
  3. Jan 17, 2009 #2
    Well, the standard way is just to use the vector triple product (Lagranges formula, see wikipedia):

    [tex]\vec{a}\times (\vec{b}\times \vec{c})= \vec{b}(\vec{a}\cdot\vec{c}) - \vec c (\vec a\cdot \vec b)[/tex]
    (also know as the BAC CAB rule)

    This relation can easily be proved, by calculating both sides (just insert [tex]\vec {a} = (a_1,a_2,a_3)[/tex] and so on).

    In your case we have, [tex]\vec a = \vec c \equiv \vec r[/tex] and [tex]\vec b \equiv \dot{\vec r} = \vec v[/tex].
     
  4. Jan 17, 2009 #3

    Nabeshin

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    Ah, thank you very much :) Simply a formula I was not familiar with.
     
  5. Jan 18, 2009 #4

    HallsofIvy

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    ??? The "triple product" is just the same as the product asked about so I don't see how that answers your question, except to say "yes, it is a well known calculation". Of course, the way to see that it is true is to use some general <a, b, c>, <x, y, z> for [itex]\vec{r}[/itex] and [itex]\vec{v}[/itex].

    For a little more "intuitive" insight, we know that [itex]\vec{r}\times\vec{v}[/itex] is perpendicular to both [itex]\vec{r}[/itex] and [itex]\vec{v}[/itex] so the result of any vector cross [itex]\vec{r}\times\vec{v}[/itex] must be perpendicular to that vector and so in the plane spanned by [itex]\vec{r}[/itex] and [itex]\vec{v}[/itex] and can be written as a linear combination of them: [itex]a\vec{r}+ b\vec{v}[/itex]. But [itex](\vec{r}\times\vec{v})\times \vec{r}[/itex] must also be perpendicular to [itex]\vec{r}[/itex] so we must subtract off any projection onto [itex]\vec{r}[/itex]. That is the "[itex]\vec{r}(\vec{r}\cdot\vec{v}[/itex]" part.
     
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