## What you need to know

A **vector **is something with both **magnitude**and **direction**. On diagrams they are denoted by an arrow, where the length tells us the magnitude and the arrow tells us direction.

– You will need to **add and subtract vectors**

– You will also need to multiple and understand **scalar multiples of vectors **

Make sure you are happy with the following topics before continuing:

Adding and subtracting vectors

When we **add** vectors:

\textcolor{red}{\mathbf{a}+\mathbf{b}}

This 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:

\textcolor{limegreen}{\mathbf{a}-\mathbf{b}},

Adding on the negative of the vector that is being subtracted.

## Multiplying vectors

**Multiplying vectors by a Scalar **– We can **multiply**a vector by a number. we can multiple \mathbf{a} by 3:

3\mathbf{\textcolor{red}{a}}=\mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{red}{a}}

This means the the vectors are added end to end.

We can **multiple more complicated vectors **

3\mathbf{(\textcolor{red}{a}+\textcolor{blue}{b})}=\mathbf{3\textcolor{red}{a}}+\mathbf{3\textcolor{blue}{b}}

**Note: **all vectors here are written in bold. When you’re writing this by hand, you should underline each letter that represents a vector.

**Scalar multiples **– Scalar multiples are all parallel to each other:

\mathbf{3\textcolor{red}{a}}+\mathbf{3\textcolor{blue}{b}} is parallel to \mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{blue}{b}}

## Vector Notation

A vector from point X to point Y will be shown as \overrightarrow{XY}.

So from the diagram shown we cans see that:

\overrightarrow{XY} = \mathbf{a}

\overrightarrow{YZ} = \mathbf{b}

\overrightarrow{ZX} = \mathbf{c}

If we go against the arrow the vector becomes negative.

\overrightarrow{YX} = \mathbf{-a}

We can also combine vectors, meaning

\overrightarrow{XZ} = \mathbf{a + b} = \mathbf{-c}

**Example: Involving ratios**

In the diagram below, we have vectors

\overrightarrow{AB}=3\mathbf{a} and \overrightarrow{AC}=4\mathbf{b} .

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}.

**[3 marks]**

To find \overrightarrow{AD}, we are going to add the vectors

\overrightarrow{AB}+\overrightarrow{BD}

We know \overrightarrow{AB}=3\mathbf{a} , we need to find \overrightarrow{BD}

We know BD:DC=1:3, so D is \dfrac{1}{4} of the way along \overrightarrow{BC}

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

Now we need to multiple this be \dfrac{1}{4}

\overrightarrow{BD}=\dfrac{1}{4}(3\mathbf{a}+4\mathbf{b})=-\dfrac{3}{4}\mathbf{a}+\mathbf{b}

Now we need to add \overrightarrow{AB} to the answer:

\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}.

### Example Questions

**Question 1: **The diagram shows a scalene triangle, ABC, with vectors \overrightarrow{AB} =3\mathbf{a} and \overrightarrow{BC}=2\mathbf{b}. The point M is the midpoint of AC.

Find an expression for the vector \overrightarrow{AM} in terms of \mathbf{a} and \mathbf{b}.

First we can find the vector \overrightarrow{AC}

\overrightarrow{AC} =\overrightarrow{AB} +\overrightarrow{BC} =3\mathbf{a} +2\mathbf{b}

Since M is the midpoint of AC we know that \overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AC} , thus

\overrightarrow{AM}=\dfrac{1}{2}(3\mathbf{a} +2\mathbf{b}) = \dfrac{3}{2}\mathbf{a} +\mathbf{b}

**Question 2:** The diagram shows a rhombus, ABCD, where the opposite sides are parallel. Given that \overrightarrow{CF} is an extension of \overrightarrow{DC}, such that DC : CF = 5 : 3 . The point E is the midpoint of AD.

Find an expression for the vector \overrightarrow{EF} in terms of \mathbf{a} and \mathbf{b}.

We will find \overrightarrow{EF} by doing,

\overrightarrow{EF}=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CF}

First we can find the vector \overrightarrow{ED} as \dfrac{1}{2}\mathbf{b} .

Then we also know \overrightarrow{DC} = \overrightarrow{AB} hence \overrightarrow{DC}=2\mathbf{a} .

The final section \overrightarrow{CF} = \dfrac{3}{8}\times 2\mathbf{a}=\dfrac{3}{4}\mathbf{a}

Hence,

\overrightarrow{EF}=\mathbf{b}+2\mathbf{a}+\dfrac{3}{4}\mathbf{a}=2.75\mathbf{a}+\mathbf{b}

**Question 3: **The diagram shows a scalene triangle, ABC, with vectors \overrightarrow{AB} =\mathbf{a} and \overrightarrow{BC}=2\mathbf{b}. The point D is positioned on the line BC such that BD : DC = 3 : 2 . The point E is positioned on the line AD such that AE : ED = 1 : 2 .

Find an expression for the vector \overrightarrow{AE} in terms of \mathbf{a} and \mathbf{b}.

First we can find the vector \overrightarrow{BC}

\overrightarrow{BC} =-\overrightarrow{AB} +\overrightarrow{AC} =-\mathbf{a} +2\mathbf{b}

Then we can find,

\overrightarrow{BD} =\dfrac{3}{5}\overrightarrow{BC} =-\dfrac{3}{5}\mathbf{a} +\dfrac{6}{5}\mathbf{b}

Next,

\overrightarrow{AD} =\overrightarrow{AB}+\overrightarrow{BD} =\mathbf{a} +(-\dfrac{3}{5}\mathbf{a} +\dfrac{6}{5}\mathbf{b})=\dfrac{2}{5}\mathbf{a}+\dfrac{6}{5}\mathbf{b}

Finally,

\overrightarrow{AE}=\dfrac{1}{3}\overrightarrow{AD}=\dfrac{1}{3}(\dfrac{2}{5}\mathbf{a}+\dfrac{6}{5}\mathbf{b})= \dfrac{2}{15}\mathbf{a}+\dfrac{2}{5}\mathbf{b}

**Question 4: **The diagram shows a regular hexagon, ABCDEF, centre O, with with vectors \overrightarrow{OB} =\mathbf{a} and \overrightarrow{DE}=\mathbf{b}.

Find an expression for the vector \overrightarrow{AC} in terms of \mathbf{a} and \mathbf{b}.

We will find \overrightarrow{AC} by doing,

\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{OB}+\overrightarrow{OC}

thus,

\overrightarrow{AC} =\mathbf{b}-\mathbf{a}+\mathbf{b}= 2\mathbf{b}-\mathbf{a}

**Question 5:** The diagram below shows the vectors \overrightarrow{AE}=3\mathbf{a}-2\mathbf{b} and \overrightarrow{DC}=2\mathbf{a}+4\mathbf{b}. The points 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}.

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, we have everything we need and can go back through our work, filling in the gaps we find:

\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,

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.

### Worksheets and Exam Questions

#### (NEW) Vectors Exam Style Questions - MME

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