Dear Uncle Colin,

I’ve got three points: $A$, with a position vector of $(2\bi + 4\bj)$, $B$, with a position vector of $(6\bi + 8\bj)$ and $C$, with a position vector of $(k\bi + 25\bj)$, and they all lie on the same straight line.

I have to find $k$, and I don’t know where to start!

-- Points In Collinear Kerfuffle

Hello, PICK!

A good place to start would be to think about what a straight line is, as far as vector geometry goes: you can think of it all of the points ‘in the same direction’ from a given point – or as any multiple of a specific direction vector added to a reference point.

In this case, if you took your reference point as $A$ and your direction vector as $\vec{AB}$, an equation of your line would be $$\\mathbf{r} = (2\\bi + 4\\bj) + \\lambda(4\\bi + 4\\bj)$$.

Now, you know that $\mathbf{r} = (k\bi + 25\bj)$ lies on this line, so $(k\bi + 25\bj) = (2\bi + 4\bj) + \lambda(4\bi + 4\bj)$ for some values of $k$ and $\lambda$.

We can split this out into two equations: in $\bi$, $k = 2 + 4\lambda$; in $\bj$, $25 = 4 + 4\lambda$.

The second equation gives $\lambda = \frac{21}{4}$, so $k = 23$.

This is the vector way of looking at it, PICK. There is a possibly simpler way, which is to look at the position vectors of $A$, $B$ and $C$ as points in the Cartesian plane.

What’s the equation of the line through (2,4) and (6,8)? It’s $y = x + 2$. So when $y=25$, like it does at point $C$, then $x = 23$, which is your value of $k$.

Hope that helps!

-- Uncle Colin

* Edited 2016-09-28 to fix broken LaTeX. Thanks to @christianp and @dragondodo for pointing it out.