Proving A Result About the Cross Product

[Bashyboy]

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?

 

[micromass]

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?

 

[MisterX]

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]!

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]?

 

[Bashyboy]

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]?

This, I suppose, is closer to what I want to demonstrate.

 

[micromass]

This, I suppose, is closer to what I want to demonstrate.

But then we need to know how you defined ##\mathbf{r}\times\mathbf{s}##.

 

[Bashyboy]

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]

 

[Quesadilla]

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}##.

 

[micromass]

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?

 

[Bashyboy]

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.

 

[Joffan]

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.

Why not just do the algebra?

##\mathbf{r}\cdot (\mathbf{r}\times \mathbf{s}) =r_x(r_ys_z-r_zs_y) + \ldots##

 

[Bashyboy]

Joffan, I did do the algebra earlier today. In this post, I just summarized what I did in English.