Here are some vectors:

- \(\mathbf{a}=\begin{pmatrix}6\\1\end{pmatrix}\)
- \(\mathbf{b}\) has magnitude \(2\sqrt{2}\) and direction \(45^\circ\) anti-clockwise from the \(x\)-axis
- \(\mathbf{c}\) has magnitude \(4\) in the positive \(x\) direction
- \(\mathbf{d}=3\mathbf{i}-\mathbf{j}\)
- \(\mathbf{e}=7\mathbf{i}-3\mathbf{j}\)
- \(\mathbf{f}\) is \(2\) units in the negative \(x\) direction and \(3\) units in the positive \(y\) direction
- \(\mathbf{g}=\begin{pmatrix}-2\\4\end{pmatrix}\)

Select a starting point, \(M(2,5)\) or \(N(3,1)\), and two of the vectors from the list so that the sum of the vectors will get you from your start to the target, \(T(7,6)\).

How many different ways can you pair up vectors from the list to get to the target?

First, let’s write all of the vectors in the same format. We’ve chosen to use column vectors.

\[ \mathbf{a}=\begin{pmatrix}6\\1\end{pmatrix},\: \mathbf{b}=\begin{pmatrix}2\\2\end{pmatrix},\: \mathbf{c}=\begin{pmatrix}4\\0\end{pmatrix},\: \mathbf{d}=\begin{pmatrix}3\\-1\end{pmatrix},\: \mathbf{e}=\begin{pmatrix}7\\-3\end{pmatrix},\: \mathbf{f}=\begin{pmatrix}-2\\3\end{pmatrix},\: \mathbf{g}=\begin{pmatrix}-2\\4\end{pmatrix} \]

The vector from \(M\) to \(T\) is \(\overrightarrow{MT}=\begin{pmatrix}5\\1\end{pmatrix}\) and we can make this by adding, \[\mathbf{b}+\mathbf{d}=\begin{pmatrix}2+3\\2-1\end{pmatrix}=\begin{pmatrix}5\\1\end{pmatrix}.\]

Notice that we can add these two vectors the other way round, \(\mathbf{d}+\mathbf{b}\), with the same result. If we draw these two “routes” they look like this.

You can decide whether these two are different solutions or the same solution!

What shape have we constructed? Can you prove it?

We can also make \(\overrightarrow{MT}\) by adding \(\mathbf{e}\) and \(\mathbf{g}\), and we can make \(\overrightarrow{NT}\) by adding \(\mathbf{a}\) and \(\mathbf{g}\).

In all, we have three different pairs of vectors which will get us to the target, \(T\), and each pair can be combined in either of two orders.

You might also have noticed that starting at \(M\), if we apply \(\mathbf{c}\) first then \(\mathbf{b}\) it takes us through \(T\) but overshoots. So can you write \(\overrightarrow{MT}\) as a combination of *multiples* of \(\mathbf{c}\) and \(\mathbf{b}\)?

If we allow multiples (including negatives), is it also possible to write \(\overrightarrow{NT}\) as a combination of \(\mathbf{c}\) and \(\mathbf{b}\)? Which other points can you reach by adding multiples of \(\mathbf{c}\) and \(\mathbf{b}\)?

Can you add *three* of the vectors to get from one of the starts to the target?

\[\overrightarrow{NT}=\mathbf{b}+\mathbf{c}+\mathbf{f}\]

In how many different orders could we arrange these three vectors?