Vector Formalism in Introductory Physics V: Two Equations, One Solution

TL;DR: Solving seemingly trivial dot product and cross product equations leads to an astonishing result, namely that they have the same solution, which can be derived both geometrically and algebraically. Establishing this common solution is an important step in motivating formal Gibbsian vector algebra.

In the previous two posts, I demonstrated that the simple dot product equation \mathbf{x}\bullet\mathbf{b}=\lambda and the simple cross product equation \mathbf{x}\times\mathbf{b}=\mathbf{c} have a common solution. Remembering that \mathbf{x} is our unknown vector, and that \lambda, \mathbf{b}, and \mathbf{c} are known quantities the common solution turns out to be

    \[ \mathbf{x} = k \mathbf{b} + \alpha \mathbf{b} \times \mathbf{c} \]

where k and \alpha are proportionality constants. You may want to go back over those posts to refresh your memory (which probably wouldn’t be necessary if I were more diligent in writing these posts). The goal of this post is to solve for the constants k and \alpha.

To solve for k let’s assume \mathbf{x} is parallel to \mathbf{b}. That means \mathbf{x} must be a multiple of \mathbf{b}, namely k\mathbf{b}. Substituting this into the equation \mathbf{x}\bullet\mathbf{b}=\lambda and solving for k gives k=\dfrac{\lambda}{\mathbf{b}\bullet\mathbf{b}}. Remember that \lambda is known and is equal to \mathbf{x}\bullet\mathbf{b}. So the expression for k becomes k = \dfrac{\mathbf{x}\bullet\mathbf{b}}{\mathbf{b}\bullet\mathbf{b}}. With this value of k, we can now write the solution for \mathbf{x} as

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

and we’re done.

Algebraically, we can arrive at the same expression as follows:

    \[\begin{aligned} \mathbf{x} \bullet \mathbf{b} &= \lambda && (1) \\ \left( k \mathbf{b}+\alpha\mathbf{b}\times\mathbf{c} \right) \bullet\mathbf{b} &= \lambda && (2) \\ k\mathbf{b}\bullet\mathbf{b} + \alpha\mathbf{b}\times\mathbf{c}\bullet\mathbf{b} &= \lambda && (3) \\ k\mathbf{b}\bullet\mathbf{b} + 0 &= \lambda && (4) \\ k\mathbf{b}\bullet\mathbf{b} &= \lambda && (5) \\ k\mathbf{b}\bullet\mathbf{b} &= \mathbf{x}\bullet\mathbf{b} && (6) \\ \therefore k &= \frac{\mathbf{x}\bullet\mathbf{b}}{\mathbf{b}\bullet\mathbf{b}} && (7) \end{aligned}\]

To solve for \alpha, let’s assume \mathbf{x}, \mathbf{b} and \mathbf{c} are mutually perpendicular. That means \mathbf{x} must be a multiple of \mathbf{b}\times\mathbf{c}, namely \alpha\mathbf{b}\times\mathbf{c}. Substituting this into the equation
\mathbf{x}\bullet\mathbf{b}=\lambda, expanding the resulting triple cross product, and remembering that the three vectors are mutually perpendicular, the expression for \alpha becomes \alpha = \dfrac{1}{\mathbf{b}\bullet\mathbf{b}}. With this value of \alpha, and remembering that \mathbf{c} = \mathbf{x}\times\mathbf{b}, we can now write the solution for \mathbf{x} as

    \[ \mathbf{x} = \frac{\mathbf{b}\times\left(\mathbf{x}\times\mathbf{b}\right)}{\mathbf{b}\bullet\mathbf{b}} \]

and again we’re done.

Algebraically, we can arrive at the same expression as follows:

    \[\begin{aligned} \mathbf{x} \times \mathbf{b} &= \mathbf{c} && (1) \\ \left( k \mathbf{b}+\alpha\mathbf{b}\times\mathbf{c} \right) \times\mathbf{b} &= \mathbf{c} && (2) \\ k\mathbf{b}\times\mathbf{b} + \alpha\left(\mathbf{b}\times\mathbf{c}\right)\times\mathbf{b} &= \mathbf{c} && (3) \\ 0 + \alpha\left( \left(\mathbf{b}\bullet\mathbf{b}\right)\mathbf{c}-\left(\mathbf{b}\bullet\mathbf{c}\right)\mathbf{b}\right) &= \mathbf{c} && (4) \\ \alpha\left(\left(\mathbf{b}\bullet\mathbf{b}\right)\mathbf{c}-0\right) &= \mathbf{c} && (5) \\ \alpha\left(\mathbf{b}\bullet\mathbf{b} \right)\mathbf{c} &= \mathbf{c} && (6) \\ \alpha\left(\mathbf{b}\bullet\mathbf{b} \right)\mathbf{c}\bullet\mathbf{c} &= \mathbf{c}\bullet\mathbf{c} && (7) \\ \alpha\mathbf{b}\bullet\mathbf{b} &= 1 && (7) \\ \therefore \alpha &= \frac{1}{\mathbf{b}\bullet\mathbf{b}} && (8) \end{aligned}\]

Now we can write the most general solution, accounting for solutions parallel to, and perpendicular to, \mathbf{b} as

    \[ \mathbf{x} = \frac{\mathbf{x}\bullet\mathbf{b}}{\mathbf{b}\bullet\mathbf{b}}\mathbf{b} + \frac{\mathbf{b}\times\left(\mathbf{x}\times\mathbf{b}\right)}{\mathbf{b}\bullet\mathbf{b}} \]

Now, there is a potential objection to the way this solution is written. The whole point is to solve for \mathbf{x}, and students have been taught to “isolate the unknown quantity on one side of the equation” but here we see it on both sides. Is this an error? No! We are dealing with vector quantities, and while \mathbf{x} is indeed an unknown quantity, \mathbf{x}\bullet\mathbf{b} is most certainly a known quantity and so is \mathbf{x}\times\mathbf{b}. In fact, each of these two quantities tells us something specific about how \mathbf{x} and \mathbf{b} are related and serve as geometric constraints. I will return to that point in the next post and show a new way to algebraically obtain this same general solution. I will also address the presence of \mathbf{b}\bullet\mathbf{b} in the general solution as this puzzled me for a long time but it turns out to have a very logical reason for being there. Plus, if you look carefully at the triple cross product you can see I wrote it two ways and I will address that too. Needless to say, a lot of things are going to come together in the next post.

As always, your 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.