Vector Formalism in Introductory Physics IV: Unwrapping Cross Products Geometrically

TL;DR: Vector cross products are not like products of real numbers, for which there is an inverse operation to “undo” multiplication. I don’t think we should introduce cross products as a form of “multiplication” in introductory physics courses because it may reinforce the urge to “divide by a vector.” A better approach may be to formally present ways of unwrapping cross products consistent with Gibbsian vector algebra. But first, a geometric understanding is necessary.

In the last post, I showed how to unwrap dot products. In this post, I will show how to unwrap cross products. As with dot products, we will see that there is no unique solution. There is an additional constraint on cross products that does not exist with dot products, and that is direction.

So what is a cross product? It is an operation that takes two vectors and produces a new object that we sometimes mistakenly call a vector. Actually, it is a pseudovector. It has the obligatory magnitude and direction (which, as you should understand by now, do not define what is and is not a vector) but it has the additional property that it behaves a certain way under a coordinate inversion. I will not address that in this post however. As with dot products, we begin by introducing some geometry which helps with visualization.

Consider two vectors \mathbf{x} and \mathbf{a} with the former being the unknown, their cross product, and the resulting object.

    \[ \mathbf{x}\times\mathbf{a} = \mathbf{c} \]

We want to find all vectors \mathbf{x} that satisfy this equation. The problem is that there are infinitely many of them.

    \[\begin{aligned} \left\lVert\mathbf{x}\right\rVert\widehat{\mathbf{x}}\times\left\lVert\mathbf{a}\right\rVert\widehat{\mathbf{a}} &=\left\lVert{\mathbf{c}\right\rVert\widehat{\mathbf{c}} \\ \left\lVert\mathbf{x}\right\rVert\left\lVert\mathbf{a}\right\rVert\left(\widehat{\mathbf{x}}\times\widehat{\mathbf{a}}\right) &=\left\lVert{\mathbf{c}\right\rVert\widehat{\mathbf{c}} \\ \left\lVert\mathbf{x}\right\rVert\left\lVert\mathbf{a}\right\rVert\sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}}\widehat{\mathbf{c}} &=\left\lVert{\mathbf{c}\right\rVert\widehat{\mathbf{c}} \\ \left\lVert\mathbf{x}\right\rVert\left\lVert\mathbf{a}\right\rVert\sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}}\widehat{\mathbf{c}}\bullet\widehat{\mathbf{c}} &=\left\lVert{\mathbf{c}\right\rVert\widehat{\mathbf{c}}\bullet\widehat{\mathbf{c}} \\ \left\lVert\mathbf{x}\right\rVert\left\lVert\mathbf{a}\right\rVert\sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} &= \left\lVert{\mathbf{c}\right\rVert \end{aligned}\]

While \left\lVert\mathbf{c}\right\rVert and \left\lVert\mathbf{a}\right\rVert are fixed, different combinations of \left\lVert\mathbf{x}\right\rVert and \sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} give values equal to \dfrac{\left\lVert\mathbf{c}\right\rVert}{\left\lVert\mathbf{a}\right\rVert}. Let’s look at some geometric implications.

If we pick a value of \sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} then there are vectors \mathbf{x} with a magnitude, indeed THE magnitude, \dfrac{\left\lVert\mathbf{c}\right\rVert}{\left\lVert\mathbf{a}\right\rVert \sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}}} that satisfies the equation. These vectors lie in the plane perpendicular to \mathbf{c}. Picking a different value of \sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} produces a different vector in that plane.

If we pick a value of \left\lVert\mathbf{x}\right\rVert then there are two vectors making an angle consistent with \sin\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} having a value \dfrac{\left\lVert\mathbf{c}\right\rVert}{\left\lVert\mathbf{a}\right\rVert\left\lVert\mathbf{x}\right\rVert}.

All solutions are constrained to lie in the plane perpendicular to \hat{\mathbf{c}}, and notice that the tip of all the solution vectors lie on a line. This suggests that unwrapping a cross product might be related to the equation of a line.

By inspection, we can see that

    \[\mathbf{x} = \dfrac{\mathbf{a}\times\mathbf{c}}{\left\lVert\mathbf{a}\right\rVert^2}\]

must be a solution.

But this is not the most general solution, because we could also add an arbitrary vector that is parallel to \mathbf{a}. Getting an arbitrary vector parallel to \mathbf{a} is as simple as forming a scalar multiple \lambda\mathbf{a} where \lambda is an arbitrary scalar. This seems directly analogous to adding a constant of integration to an indefinite integral, something which introductory physics students seem content to accept.

Therefore the most general solution is

    \[\mathbf{x} = \dfrac{\mathbf{a}\times\mathbf{c}}{\left\lVert\mathbf{a}\right\rVert^2}+\lambda\mathbf{a}\]

It is probably intuitive that unwrapping a cross product leads to another cross product, but what is the significance of the factor of \left\lVert\mathbf{a}\right\rVert^2 in the denominator? I will explain this in a future post.

By the way, we can check this solution simply by crossing each side with \mathbf{a} and using the BAC-CAB identity.

    \[\begin{aligned} \mathbf{x}\times\mathbf{a} &= \left(\dfrac{\mathbf{a}\times\mathbf{c}}{\left\lVert\mathbf{a}\right\rVert^2}+\lambda\mathbf{a}\right)\times\mathbf{a} \\ &=\dfrac{\left(\mathbf{a}\times\mathbf{c}\right)\times\mathbf{a}}{\left\lVert\mathbf{a}\right\rVert^2} + 0 \\ &=\dfrac{\left(\mathbf{a}\bullet\mathbf{a}\right)\mathbf{c}-\left(\mathbf{c}\bullet\mathbf{a}\right)\mathbf{a}}{\left\lVert\mathbf{a}\right\rVert^2} \\ &= \mathbf{c} \end{aligned}\]

So this somewhat convoluted expression really does satisfy the original equation and constitutes a solution thereof.

Here is a GlowScript program written with Trinket that illustrates the geometry, and visualization thereof, involved in unwrapping a cross product. Embedding with Trinket lets users modify the code, run it to see the result of the modification, and then restore the original code. Experiment with it yourself.

In researching the literature for this post, I came across an interesting notational problem. Specifically, when we write \widehat{\mathbf{x}}\times\widehat{\mathbf{a}}, by definition we must account for the angle between the two vectors if they are not orthogonal (in which case accounting for the angle simply gives a factor of 1). If \widehat{\mathbf{x}} and \widehat{\mathbf{a}} are not orthogonal, and if we are only interested in the direction of the cross product, we need a way to notate the direction. In other words, we need to notate the direction of the cross product without implying the need to account for the angle between the vectors. Does that mean we need to write something like \widehat{\widehat{\mathbf{x}}\times\widehat{\mathbf{a}}}? I don’t know, but it’s something I’m thinking about. Clear and unambiguous notation is important, especially in introductory physics.

In the next post, I will show how to unwrap both dot and cross products algebraically and will explain why both must be taken into consideration. I have never seen this explained before so I hope to make it worth your time to read.

As always, feedback is welcome. I am especially interested in feedback on the clarity of my writing.

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