Proving A Result About the Cross Product

[Bashyboy]

Here is the claim I am trying to prove: Suppose we have two vectors $\mathbf{r}$ and $\mathbf{s}$. I would like to show that there are only two directions in which the resultant vector of the cross product $\mathbf{r} \times \mathbf{s}$ 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 $\mathbf{r} \times \mathbf{s}$ points; it's in the direction of $\mathbf{r} \times \mathbf{s}$!

Seriously though, perhaps you want to prove that $\mathbf{r} \times \mathbf{s}$ lies in the 1 dimensional space perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{s}$?

 

[Bashyboy]

Seriously though, perhaps you want to prove that $\mathbf{r} \times \mathbf{s}$ lies in the 1 dimensional space perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{s}$?

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

$\mathbf{r} \times \mathbf{s} = \mathbf{p}$

where the three orthogonal components of $\mathbf{p}$ are found by computing

$p_x = r_ys_z - r_zs_y$

$p_y = r_zs_x - r_xs_z$

$p_z = r_xs_y - r_ys_x$

 

[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

$$\mathbf{r}\cdot (\mathbf{r}\times \mathbf{s}) = 0$$

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 $\mathbf{r}$ ($\mathbf{r}$ acts as a basis vector) Doing this, easily computed the components of $\mathbf{p}$ and took the inner product of this with $\mathbf{r}$, 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 $\mathbf{r}$ ($\mathbf{r}$ acts as a basis vector) Doing this, easily computed the components of $\mathbf{p}$ and took the inner product of this with $\mathbf{r}$, 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.