Vectors Revision and Worksheets

What you need to know

A vector is something with both magnitude (size) and direction. On a diagram, they are denoted by an arrow, where the length of the arrow tells us the magnitude and the way the arrow is pointing tells us the direction.

When we add vectors, we add them end-to-end. For example, if you add two vectors $\mathbf{a}$ and $\mathbf{b}$ , then the result is the vector $\mathbf{a}+\mathbf{b}$ , which takes you from the start of $\mathbf{a}$ to the end of $\mathbf{b}$ (right).

The negative of a vector has the same magnitude of the original vector, it just goes in the exact opposite direction. When we subtract vectors, for example $\mathbf{a}-\mathbf{b}$ , we add on the negative of the vector that is being subtracted, i.e. we add the vector $-\mathbf{b}$ onto the vector $\mathbf{a}$ , as seen in the picture.

We can multiply a vector by a number. As with normal multiplication, it’s just the same as adding a number to itself multiple times, so the result of multiply $\mathbf{a}$ by 3 is

$3\mathbf{a}=\mathbf{a}+\mathbf{a}+\mathbf{a}$

Then, they are added end-to-end, just how we’ve already seen. Note: all vectors here are written in bold. When you’re writing this by hand, you should underline each letter that represents a vector.

Vectors are often split up into two parts – an $x$ part, which tells us how far the vector moves left or right, and a $y$ part, which tells us how far a vector moves up or down. When splitting up vectors like this, we express them as column vectors, where the top number is the $x$ part and the bottom number is the $y$ part.

For example, the vector $\mathbf{a}$ goes 3 spaces to the right and 2 spaces up, so would be expressed like $\begin{pmatrix}3\\2 \end{pmatrix}$ . If the vector goes left, the $x$ value is negative, and if it goes down, the $y$ value is negative.

To add/subtract column vectors, we add/subtract the $x$ and $y$ values separately. For example,

$\begin{pmatrix}-3\\4\end{pmatrix}+\begin{pmatrix}5\\2\end{pmatrix}=\begin{pmatrix}2\\6\end{pmatrix}$

To multiply a column vector by a number, we multiply both values in the vector by that number, e.g.

$5\times\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}10\\-15\end{pmatrix}$

It’s important to understand that the vectors you see diagrams of and vectors written in column form are just different ways of working with the same thing. Suppose you have two column vectors, which you then add together to get another vector. If you then drew those two column vectors on a diagram and added them end-to-end (like we saw above), the resulting vector would be precisely what you got when you added the two column vectors.

For the foundation course it is sufficient to have a good understanding how to represent vectors on a diagram and as a column, as well as knowing how to add/subtract/multiply them in both forms. However, for the higher course, there is a little more.

Firstly, a vector that goes from some point $A$ to another point $B$ may be denoted like $\overrightarrow{AB}$ . If a vector starts from the origin and goes to some point, say $A$ , it will be written with an O, like $\overrightarrow{OA}$ .

The way to add these vectors is explained by the picture on the right. As you can see, we still add them end-to-end. So, if you add the vector $\overrightarrow{AB}$ to the vector $\overrightarrow{BC}$ , then the result is a vector that starts at $A$ and ends at $C$ , thus is denoted by $\overrightarrow{AC}$ . The fact that it goes via $B$ doesn’t matter, what matters is where it starts and ends.

The easiest way to see a typical structure of a higher vector question is to see an example. Let’s go.

Example: In the diagram below, we have vectors $\overrightarrow{AB}=3\mathbf{a}$ and $\overrightarrow{AC}=4\mathbf{b}$ are shown. Point $D$ lies on the line $BC$ such that $BD:DC=1:3$ . Write the vector $\overrightarrow{AD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

There’s a lot going on here, but we’ll break it down. To find $\overrightarrow{AD}$ , we are going to add the vectors

$\overrightarrow{AB}+\overrightarrow{BD}$

Added end-to-end, they will give us a vector that starts at $A$ and ends at $B$ . For the first one, we don’t need to do anything – we know $\overrightarrow{AB}=3\mathbf{a}$ . The second part will require some work.

Firstly, recognise that if $BD:DC=1:3$ , then $D$ is $^{1}\!_{4}$ of the way along the line from $B$ to $C$ . Therefore, if we get the vector that goes from $B$ to $C$ and divide it by 4, we have $\overrightarrow{BD}$ . Now, to get from $B$ to $C$ using the vectors we’ve been given, we can go via $A$ . To get from $B$ to $A$ , we have to go backwards along $\overrightarrow{AB}$ , so we will take the negative of it. Therefore, we get

$\overrightarrow{BC}=-\overrightarrow{AB}+\overrightarrow{AC}=-3\mathbf{a}+4\mathbf{b}$

Dividing this by 4, we get

$\overrightarrow{BD}=\dfrac{1}{4}\overrightarrow{BC}=-\dfrac{3}{4}\mathbf{a}+\mathbf{b}$

Therefore, adding this to $\overrightarrow{AB}$ , we get the answer to be

$\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=3\mathbf{a}+\left(-\dfrac{3}{4}\mathbf{a}+\mathbf{b}\right)=\dfrac{9}{4}\mathbf{a}+\mathbf{b}$

Thus, we have our answer in terms of $\mathbf{a}$ and $\mathbf{b}$ .

Note: another common type of question is “prove one vector is parallel to another”. To do this, you work out both vectors in questions, and then if one is multiple of the other, they are parallel.

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Example Questions

1) Let $\mathbf{a}=\begin{pmatrix}3\\8\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}-7\\2\end{pmatrix}$ . Write $2\mathbf{a}+\mathbf{b}$ as a column vector.

Firstly, to multiply $\mathbf{a}$ by 2, we must multiply both of its components by 2:

$2\mathbf{a}=2\times\begin{pmatrix}3\\8\end{pmatrix}=\begin{pmatrix}6\\16\end{pmatrix}$

Then, to add this to $\mathbf{b}$ , we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be

$2\mathbf{a}+\mathbf{b}=\begin{pmatrix}6\\16\end{pmatrix}+\begin{pmatrix}-7\\2\end{pmatrix}=\begin{pmatrix}-1\\18\end{pmatrix}$

2) In the diagram below, we have vectors $\overrightarrow{AE}=3\mathbf{a}-2\mathbf{b}$ and $\overrightarrow{DC}=2\mathbf{a}+4\mathbf{b}$ . $E$ and $B$ are the midpoints of $AD$ and $AC$ respectively. Find an expression for $\overrightarrow{EB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ and state whether or not it is parallel to $\overrightarrow{DC}$ .

There are a lot of steps here, so take your time to read through it and make sure you understand.

We will find $\overrightarrow{EB}$ by doing

$\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}$

The first vector is straightforward, because we know $\overrightarrow{AE}$ , and that is just the same vector in the opposite direction. So, we get

$\overrightarrow{EA}=-\overrightarrow{AE}=-(3\mathbf{a}-2\mathbf{b})=-3\mathbf{a}+2\mathbf{b}$

Now we need $\overrightarrow{AB}$ . Since $B$ is the midpoint of $AC$ (given in the question), we must have that $\overrightarrow{AB}=\frac{1}{2} \overrightarrow{AC}$ . Therefore, looking at the diagram, we get that

$\overrightarrow{AB}=\dfrac{1}{2}\overrightarrow{AC}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{DC}\right)$

We’re given the second part of this, $\overrightarrow{DC}=2\mathbf{a}+4\mathbf{b}$ , and since $E$ is the midpoint of $AD$ , we can also work out the first part:

$\overrightarrow{AD}=2\overrightarrow{AE}=2(3\mathbf{a}-2\mathbf{b})=6\mathbf{a}-4\mathbf{b}$

Now, at long last, we have everything we need and can go back through our work, filling in the gaps. Now we have $\overrightarrow{AD}$ , we get that

$\overrightarrow{AB}=\dfrac{1}{2}\left(6\mathbf{a}-4\mathbf{b}+2\mathbf{a} +4\mathbf{b}\right)=\dfrac{1}{2}\left(8\mathbf{a}\right)=4\mathbf{a}$

Therefore, finally we have that

$\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}=-3\mathbf{a}+2\mathbf{b}+4\mathbf{a}=\mathbf{a}+2\mathbf{b}$

If $\overrightarrow{EB}$ and $\overrightarrow{DC}$ are parallel, then one must be a multiple of the other. Well, if we multiply $\overrightarrow{EB}$ by 2 then we get

$2\times\overrightarrow{EB}=2(\mathbf{a}+2\mathbf{b})=2\mathbf{a}+4\mathbf{b}=\overrightarrow{DC}$

Therefore, we’ve shown that $2\overrightarrow{EB}=\overrightarrow{DC}$ , and thus the two lines must be parallel.