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Homework Help: Proving A Result About the Cross Product

  1. Jun 16, 2014 #1
    Here is the claim I am trying to prove: Suppose we have two vectors [itex]\mathbf{r}[/itex] and [itex]\mathbf{s}[/itex]. I would like to show that there are only two directions in which the resultant vector of the cross product [itex]\mathbf{r} \times \mathbf{s}[/itex] can point, parallel and antiparallel.

    How might one prove this? Could someone proffer a few hints?
     
  2. jcsd
  3. Jun 16, 2014 #2
    How did you define the cross product ##\mathbf{r}\times \mathbf{s}##?

    And isn't the cross product ##\mathbf{r}\times \mathbf{s}## a well-defined vector? As such, it only points in one direction? Why would it point in two direction? And what is parallel and anti-parallel? Parallel to what?
     
  4. Jun 16, 2014 #3
    I'm not sure what you mean. There's only one direction that [itex]\mathbf{r} \times \mathbf{s}[/itex] points; it's in the direction of [itex]\mathbf{r} \times \mathbf{s}[/itex]! :biggrin:

    Seriously though, perhaps you want to prove that [itex]\mathbf{r} \times \mathbf{s}[/itex] lies in the 1 dimensional space perpendicular to the plane containing [itex]\mathbf{r}[/itex] and [itex]\mathbf{s}[/itex]?
     
  5. Jun 16, 2014 #4
    This, I suppose, is closer to what I want to demonstrate.
     
  6. Jun 16, 2014 #5
    But then we need to know how you defined ##\mathbf{r}\times\mathbf{s}##.
     
  7. Jun 16, 2014 #6
    Well, the definition Taylor provides in his Classical Mechanics text is

    [itex]\mathbf{r} \times \mathbf{s} = \mathbf{p}[/itex]

    where the three orthogonal components of [itex]\mathbf{p}[/itex] are found by computing

    [itex]p_x = r_ys_z - r_zs_y[/itex]

    [itex]p_y = r_zs_x - r_xs_z[/itex]

    [itex]p_z = r_xs_y - r_ys_x[/itex]
     
  8. Jun 16, 2014 #7
    If your definition of ##\mathbf{r} \times \mathbf{s}## is by a formula, it would be a good idea to check that ##\mathbf{r} \times \mathbf{s}## is perpendicular to both ##\mathbf{r}## and ##\mathbf{s}##.
     
  9. Jun 16, 2014 #8
    So you need to show that ##\mathbf{r}## is orthogonal to ##\mathbf{r}\times \mathbf{s}## (and the same for ##\mathbf{s}##). You can do this with dot products. Indeed, it suffices to show

    [tex]\mathbf{r}\cdot (\mathbf{r}\times \mathbf{s}) = 0[/tex]

    Can you do this?
     
  10. Jun 16, 2014 #9
    Yes, I can do this, and actually did this earlier today. I simply defined a coordinate system such that the x-axis of this coordinate system coincided with [itex]\mathbf{r}[/itex] ([itex]\mathbf{r}[/itex] acts as a basis vector) Doing this, easily computed the components of [itex]\mathbf{p}[/itex] and took the inner product of this with [itex]\mathbf{r}[/itex], which lead to zero.
     
  11. Jun 16, 2014 #10
    Why not just do the algebra?

    ##\mathbf{r}\cdot (\mathbf{r}\times \mathbf{s}) =r_x(r_ys_z-r_zs_y) + \ldots##
     
  12. Jun 16, 2014 #11
    Joffan, I did do the algebra earlier today. In this post, I just summarized what I did in English.
     
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