Ask Uncle Colin: A Vector Line

Dear Uncle Colin,

I’ve got three points: $A$, with a position vector of $(2\mathbf{i}+4\mathbf{j})$, $B$, with a position vector of $(6\mathbf{i}+8\mathbf{j})$ and $C$, with a position vector of $(k\mathbf{i}+25\mathbf{j})$, 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 $\overrightarrow{AB}$, an equation of your line would be $$\begin{gathered} mathbfr=(2 \\ bi+4 \\ bj)+ \\ lambda(4 \\ bi+4 \\ bj) \end{gathered}$$.

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

We can split this out into two equations: in $\mathbf{i}$, $k=2+4\lambda$; in $\mathbf{j}$, $25=4+4\lambda$.

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.