1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proving A Result About the Cross Product
Loading...