Vector Formalism in Introductory Physics III: Unwrapping Dot Products Geometrically

TL;DR: Vector dot 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 dot 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 dot products consistent with Gibbsian vector algebra. But first, a geometric understanding is necessary.

By the time students take introductory physics, they have hopefully mastered the arithmetic of real numbers. They can add, subtract, multiply, and divide them and solve equations in terms of real numbers. Let’s consider multiplication. Students know that when they perform any of these operations with real numbers, two real numbers are needed as operands. Symbolic letters, variables, are frequently used to represent these operands. Solving simple equations with real numbers and variables that represent them involves performing operations that “undo” the operations in the original equation with the intent of isolating the “unknown” variable and equating it to some arithmetic combination of real numbers to solve the equation. Students also ostensibly know that addition and subtraction “undo” each other while multiplication and division also “undo” each other. Lastly, students know that combining any two real numbers with any of these four operations gives a real number in return. That was a mouthful, but students can do it with little difficulty in most cases.

Now, in introductory physics courses we introduce vectors with the sloppy enough definition of “something with magnitude and direction” which of course isn’t quite true (or true enough for my tastes). Then, we blithely state that there are two ways to “multiply vectors.” One way gives a vector, which seems in line with multiplying real numbers because multiplying two real numbers gives a real number. The other way of multiplying vectors gives a real number, which is not at all like multiplying real numbers. I don’t know that students reflect on this deeply, and indeed I don’t know that most instructors do either. I think it’s important to do so, though, because it’s the first sign that something is either amiss, or at least different, about vector arithmetic and vector algebra. We then perhaps, but not always in my experience, state that “vector division isn’t defined” without, again in my experience, giving any explanation for why this must be the case. Note that I said “must be the case” rather than “is the case.” Before moving on, I can think of a third way of multiplying vectors that gives neither a vector nor a real number; it gives a second rank tensor. I will save this for another post in this series.

So what is a dot product? It is an operation that takes two vectors and produces a real number, a scalar. There are many ways this operation can be defined, so we have to get more specific and that involves introducing geometry.

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

    \[ \mathbf{x}\bullet\mathbf{a} = b \]

For most of this post, I will assume b>0. We want to find all vectors \mathbf{x} that satisfy this equation. The problem is that there are infinitely many of them.

    \[\begin{aligned} \left\lVert\mathbf{x}\right\rVert\widehat{\mathbf{x}}\bullet\left\lVert\mathbf{a}\right\rVert\widehat{\mathbf{a}} &= b \\ \left\lVert\mathbf{x}\right\rVert\left\lVert\mathbf{a}\right\rVert\widehat{\mathbf{x}}\bullet\widehat{\mathbf{a}} &= b \\ \left\lVert\mathbf{x}\right\rVert\left\lVert\mathbf{a}\right\rVert\cos\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} &= b \\ \left\lVert\mathbf{x}\right\rVert\cos\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} &= \dfrac{b}{\left\lVert\mathbf{a}\right\rVert} \end{aligned}\]

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

If we pick a value of \cos\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} then there are vectors \mathbf{x} with a magnitude, indeed THE magnitude, \dfrac{b}{\left\lVert\mathbf{a}\right\rVert \cos\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}}} that satisfies the equation. These vectors can be arranged to form a cone. Picking a different value of \cos\theta_{\widehat{\mathbf{x}},\widehat{\mathbf{a}}} produces a different cone.

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

By inspection, we can see that

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

must be a solution. Note that this is merely one particular multiple of \mathbf{a}.

But this is not the most general solution, because we could also add an arbitrary vector that is perpendicular to \mathbf{a}. Getting an arbitrary vector perpendicular to \mathbf{a} is as simple as forming the cross product \mathbf{a}\times\mathbf{f} where \mathbf{f} is an arbitrary vector. 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

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

An obvious issue here is the presence of the cross product. Why is it there, and what is its geometric significance? That will become apparent later, but for now, it is part of what may be considered, in a very broad sense, unwrapping a dot product. In other words, it’s a consequence of “dividing” a dot product by a vector.

By the way, we can check this solution simply by dotting each side with \mathbf{a}.

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

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 dot 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.

Up to now, I have assumed that b>0 but let’s consider the implications of letting b<0. A negative dot product means that the two operands have antiparallel components instead of parallel components. Geometrically, these vectors will also form a cone, sharing its axis and vertex with the cone described above. The cones form a double cone. See if you can modify the GlowScript program above to visualize this double cone.

I will close with a rather provocative thought. How would you feel about defining the division of a dot product by one of the two operands, thereby solving for the other operand, as simply the expression which when dotted with the known operand satisfies the original equation?

    \[\begin{aligned} \mathbf{x}\bullet\mathbf{a} &= b \\ \frac{\mathbf{x}\bullet\mathbf{a}}{\mathbf{a}} &\defines \dfrac{b}{\left\lVert\mathbf{a}\right\rVert^2}\mathbf{a} + \mathbf{a}\times\mathbf{f} \\ &\defines \dfrac{b}{\left\lVert\mathbf{a}\right\rVert^2}\mathbf{a} - \mathbf{f}\times\mathbf{a} \\ &\defines \left(\dfrac{b}{\left(\_\right)\bullet\left(\_\right)}\left(\_\right) - \mathbf{f}\times\left(\_\right)\right) \mathbf{a} \\ \therefore \mathbf{x} &\defines \left(\dfrac{b}{\left(\_\right)\bullet\left(\_\right)}\left(\_\right) - \mathbf{f}\times\left(\_\right)\right) \mathbf{a} \end{aligned}\]

Note that on the last line, I “factored” out the known vector \mathbf{a} (while knowingly committing an egregious error in the denominator of the first term), leaving an expression with placeholders that can be considered an operator that operates on \mathbf{a} and thereby effectively “dividing” the given dot product by \mathbf{a} to give the unknown vector \mathbf{x}. This operator unwraps the dot product. Remember that \mathbf{f} is any arbitrary vector, and I will return to this point in a future post. Also in a forthcoming post, I will revisit unwrapping dot products, but from an algebraic perspective.

As always, feedback is welcome.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.