What makes a mathematician a mathematician? Scientific studies say one thing above anything else: laziness1 We will go to extraordinary lengths to avoid doing any proper work.

For example, I had a situation: I had two points - call them $P$ and $Q$ - and a line with the equation $ax + by + c=0$; I needed to find the point where line $PQ$ crossed the line.

It’s not a difficult problem - I’d expect a good GCSE student to have a decent crack at it in a case where the numbers weren’t too tricky: find the equation of the line $PQ$ and solve the simultaneous equations that come out of it. But who has time to code all of that up?

I gave a MathsJam talk, and wrote a Chalkdust article, about the link between the equation of a line and vectors. It’s one of the neatest things I know.

There are two bits of set-up: firstly, a point with coordinates $(x,y)$ is treated as a vector, $(x,y,1)$; secondly, a line with equation $ax + by + c = 0$ is treated as the vector $(a,b,c)$.

A nice property: $ax + by + c = 0$ can be written as $(a,b,c) \cdot (x,y,1) = 0$ - so the vector corresponding to any point on the line is perpendicular to the vector representing the line.

A consequence: if you have two points on a line, taking the cross products of their vectors gives a vector corresponding to the line!

Try it! The ‘traditional’ way to find the line through $(3,5)$ and $(2,-1)$ would be to find the gradient (6) and use the correct line equation $y-y_1 = m(x-x_1)$ with either point: $y+1 = 6(x-2)$, or $0 = 6x - y - 13$. The vector way would be to calculate the cross product, $(3,5,1) \times (2,-1,1) = (6, -1, -13)$. It works!

Rather less obviously, if you take the cross product of the vectors representing two lines, you get… a multiple of the vector corresponding to the point where they cross. However, if you divide everything by the $z$ component to get it in the point-vector form, you get your point.

Knowing that, there’s no *way* I’m solving simultaneous equations for this problem.

So, I want to know where $PQ$ crosses a line with a given equation. This could hardly be set up any better for me! I’m going to say that point $P$ corresponds to vector $\vec{p}$ and similarly for point $Q$ and $\vec{q}$. Then the line $PQ$ corresponds to a vector $\vec{p} \times \vec{q}$.

If the line corresponds to $\vec{L}$, then the point where the two lines cross is a multiple of $(\vec{p} \times \vec{q})\times \vec{L}$.

It would be reasonable to stop there. But I’m a trained mathematician. There is more laziness to be done.

Back in the day, you could rely 100% on the first question of the Fundamentals of Applied Maths2 exam being “show that this vector identity holds”. It’d typically be calculus rather than algebra, and the *details* stayed in my mind just long enough to get a perfect 20 on the exam - but so thoroughly was I drilled in this that whenever I see several vector operators lined up like this I think “I shall consult a big list of vector identities to see if this can be simplified.”

And indeed it can: $(\vec{a} \times \vec{b}) \times \vec{c} \equiv (\vec{c} \cdot \vec{a}) \vec{b} - (\vec{c} \cdot \vec{b}) \vec{a}$.

So, after a solid half-hour of scrabbling around with projective geometry and vector identities, I’ve managed to reduce the problem to a one-liner! Splendid.

This is very closely related to the vector equation of a line in the form $\vec{r} = \lambda \vec {a} + (1-\lambda) \vec{b}$ - but I shall leave the details to the interested reader. I have some serious lazing to do!