# Vectors, lines and laziness

What makes a mathematician a mathematician? Scientific studies say one thing above anything else: laziness ((Look up the reference yourself.)) 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?

### Vectors!

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.

### Back to the 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.

### Vector identities

Back in the day, you could rely 100% on the first question of the Fundamentals of Applied Maths ((“Where we emphasise the FUN!”)) 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!