A Linear Independence Problem from Mrs. Perrett
Back in high school, I had a math teacher called Shirley J. Perrett. (She would introduce herself as “Mrs. \(Pe^2r^2t^2\), but not in that order.”) She had a textbook called A2 Core Maths for Edexcel, by Emanuel, Wood, and Crawshaw. And that textbook had a problem in chapter 15 that none of the students Mrs. Perrett had ever taught had solved without hints.
Well, it turns out that the textbook is available on the Internet Archive, so I thought it would be fun to go back to that problem with a math degree under my belt and see what made it so difficult. I’m pretty sure that it was problem 15E.4, but I’ll present a slightly different version here. The gist of it is that you’re given an arrangement of vectors in the plane that looks like this:
The task is to find the values of the scalars \(m\) and \(n\). Before you go any further, have a play around. The title introduces this as a linear independence problem, but that only comes up near the end.
There are a bunch of different tricks here. The first one is to notice that we can write the position of the point \(Q\) in two different ways. Assuming that \(O\) is the zero point, obviously \(Q\) is pointed to by the vector \((1+n)\mathbf{t}\). But it’s also \(\mathbf{a}+\mathbf{s}\), if we approach it via the point \(A\) instead of directly. So we get our first useful equality: \((1+n)\mathbf{t}=\mathbf{a}+\mathbf{s}\).
There are too many vectors here for us to say anything with any certainty, so let’s try to get rid of some. The second trick is to write everything in terms of just the vectors \(\mathbf{a}\) and \(\mathbf{b}\). In particular, by looking at the outer triangle, we see that \(\mathbf{a}+(1+m)\mathbf{s}=\frac{4}{3}\mathbf{b}\). Similarly, the lower-left triangle gives us \(\mathbf{b}=\mathbf{t}+\frac{1}{2}\mathbf{r}\), and the larger triangle \(OAB\) gives us \(\mathbf{b}=\mathbf{a}+\frac{3}{2}\mathbf{r}\).
If we use these to solve for \(\mathbf{s}\) and \(\mathbf{t}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), and plug back into the first equation, we get \((1+n)\left(\mathbf{b}-\frac{1}{3}\left[\mathbf{b}-\mathbf{a}\right]\right)=\mathbf{a}+\frac{1}{1+m}\left[\frac{4}{3}\mathbf{b}-\mathbf{a}\right]\). Simplifying gives us \(\frac{2}{3}(1+n)\mathbf{b}+\frac{1}{3}(1+n)\mathbf{a}=\frac{4}{3}\frac{1}{1+m}\mathbf{b}+\left[1-\frac{1}{1+m}\right]\mathbf{a}\). We’re starting to get somewhere.
The third and final trick is to use linear independence. In short, the vectors \(\mathbf{a}\) and \(\mathbf{b}\) point in different directions. (If they were parallel to each other, the point \(P\) would not be well-defined. Can you see why?) When two vectors point in different directions, they form a coordinate system for the plane they share. I’m simplifying a complicated concept here, and this way of thinking about it only works for two vectors at a time, but it’s all we need here.
The conclusion is that our equality is actually two equalities: one in the “\(\mathbf{a}\) coordinate”, and one in the “\(\mathbf{b}\) coordinate”. This is because there’s only one way to write any vector in that plane as a combination of \(\mathbf{a}\) and \(\mathbf{b}\). You can’t cancel out a change in the coefficient of \(\mathbf{a}\) just by adding more or less of \(\mathbf{b}\), because they aren’t parallel vectors. This is worth convincing yourself of.
So, if we equate the \(\mathbf{b}\) coefficients, we get \(\frac{2}{3}(1+n)=\frac{4}{3}\frac{1}{1+m}\). If we equate the \(\mathbf{a}\) coefficients, we get \(\frac{1}{3}(1+n)=1-\frac{1}{1+m}\). We’ve got two simultaneous equations in \(m\) and \(n\) now, so we solve them to find the solution: \(m=\frac{2}{3}\) and \(n=\frac{1}{5}\).
Looking at it now, I can see that this was a pretty nasty problem to lob at a high-school class. First of all, you need to figure out that you should be looking at the position of \(Q\). If you don’t, you might be stuck spinning your wheels for a really long time. Then you need linear independence, which isn’t something we were ever formally taught. We kind of knew about it, but it would not have been obvious to us that we could make up ad-hoc coordinate systems like that. Interestingly enough, this idea is core to university-level linear algebra and beyond, so I can see why the authors wanted to use it somewhere.
So this one’s for you, Mrs. Perrett. Thanks for six great terms. And I did solve it on my own. It took me seven years, but I solved it.
(At least I think I did. Somebody should check my work.)