# Vectors Questions, Worksheets and Revision

GCSE 6 - 7AQAEdexcelOCRWJECHigherAQA 2022Edexcel 2022OCR 2022WJEC 2022

## Vectors

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 multiply vectors and understand scalar multiples of vectors

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

Level 6-7 GCSE

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

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:

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

we add on the negative of the vector that is being subtracted.

Level 6-7 GCSE

## Multiplying vectors

Multiplying vectors by a Scalar – We can  multiply a vector by a number. For example, we can multiply $\mathbf{a}$ 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 multiply 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 $\mathbf{\textcolor{red}{a}}+\mathbf{\textcolor{blue}{b}}$

Level 6-7 GCSE

## Vector Notation

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

So from the diagram shown we can 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

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

Level 6-7 GCSE
Level 6-7 GCSE

## 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 $\overrightarrow{AB}=3\mathbf{a}$ but we need to find $\overrightarrow{BD}$

We know $BD:DC=1:3$, so the point $D$ 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 multiply this by $\dfrac{1}{4}$

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

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 $\mathbf{b}$.

Level 8-9GCSE

## Example Questions

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 $\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AC}$

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

The vector $\overrightarrow{EF}$ is the same as:

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

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 $\overrightarrow{CF} = \dfrac{3}{5}\times 2\mathbf{a}=\dfrac{6}{5}\mathbf{a}$

Hence,

$\overrightarrow{EF}= \dfrac{1}{2} \mathbf{b}+2\mathbf{a}+\dfrac{6}{5}\mathbf{a}=3.2\mathbf{a}+ \dfrac{1}{2} \mathbf{b}$

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:

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

Finally:

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

We will find $\overrightarrow{AC}$ by doing,

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

Thus

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

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:

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

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,

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

We have shown that $2\overrightarrow{EB}=\overrightarrow{DC}$, therefore, the two lines must be parallel.

Level 4-5GCSE

## Worksheet and Example Questions

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

Level 6-7 GCSENewOfficial MME

## You May Also Like...

### GCSE Maths Revision Cards

Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC.

£8.99

### GCSE Maths Revision Guide

The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!

From: £14.99

### Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99