# Vector Problems

Here we will learn about more difficult vector problems, including vector routes involving midpoints, fractions and ratios of lengths. We will also look at parallel vectors.

There are also vector worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What are vector problems?

Vector problems use vectors to solve a variety of different types of problems. Vectors have both a magnitude and direction and can be used to show a movement. A quantity which has just magnitude (size) is called a scalar.

For example,

We can write vectors in several ways,

• Using an arrow
• Using boldface
• Underlined

\overrightarrow{AB} =\textbf{a}=\underline{a}

## Key facts for vector problems

In order to solve problems involving vectors it is helpful to use several key facts.

## How to solve vector problems

In order to solve vector problems:

1. Write any information you know onto the diagram.
2. Decide the route.
3. Write the vector.
4. Simplify your answer.

## Related lessons on vectors

Vector problems is part of our series of lessons to support revision on vectors. You may find it helpful to start with the main vectors lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

## Vector problem examples

### Example 1: parallel lines

The shape below is made from 8 equilateral triangles.

\overrightarrow{PQ} = \textbf{a},~ \overrightarrow{OQ} = \textbf{b}

Find the vector \overrightarrow{QT} .

1. Write any information you know onto the diagram.

Parallel vectors of the same magnitude are the same. These triangles are all identical therefore we can label the corresponding vectors a and b .

2Decide the route.

We need to find a route where we know the vectors.

3Write the vector.

\overrightarrow{QT}=-\textbf{b}-\textbf{b}+\textbf{a}-\textbf{b}

4Simplify your answer.

\overrightarrow{QT}=\textbf{a}-3\textbf{b}

### Example 2: extended line

\overrightarrow{AC}=\textbf{a}, ~\overrightarrow{BC}=\textbf{b}

The line AB is extended to the point D so that the length AD is three times the length AB .

Find the vector \overrightarrow{AD} .

Write any information you know onto the diagram.

Decide the route.

Write the vector.

Simplify your answer.

### Example 3: midpoint

\overrightarrow{BC} = \textbf{p}, ~\overrightarrow{BA}=\textbf{q}

D is the midpoint of AC . Find the vector \overrightarrow{BD} .

Write any information you know onto the diagram.

Decide the route.

Write the vector.

Simplify your answer.

### Example 4: fraction of a line

\begin{aligned} &\overrightarrow{AB} = 2\textbf{a}, ~ \overrightarrow{AC} =6\textbf{b}\\\\ &\text{BM }=\frac{1}{4}\text{ BA}, \text{ AN} = \frac{1}{3} \text{ AC} \end{aligned}

Find the vector \overrightarrow{MN} .

Write any information you know onto the diagram.

Decide the route.

Write the vector.

Simplify your answer.

### Example 5: ratio

\overrightarrow{FG} =\textbf{a}, ~ \overrightarrow{EH} =2\textbf{a}, ~ \overrightarrow{EF} =\textbf{b}

ED =2EH and the point J is such that GJ:JH = 2:1 .

Find the vector \overrightarrow{JD} .

Write any information you know onto the diagram.

Decide the route.

Write the vector.

Simplify your answer.

### Example 6: mix of information

\overrightarrow{AB} =6\textbf{a}, ~ \overrightarrow{BC} =10\textbf{b}

E is the point on AB such that \text{EB }=\frac{1}{3}\text{ AB} .

D is the point on BC such that CD:DB=1:4 .

F is the midpoint of ED .

Find the vector \overrightarrow{EF} .

Write any information you know onto the diagram.

Decide the route.

Write the vector.

Simplify your answer.

## How to show two vectors are parallel

Two vectors are parallel if one is a multiple of the other. This is because if one vector is a multiple of another, it is a bigger or smaller version of the other.

In order to show two vectors are parallel:

1. Work out each vector.
2. Show that one is a multiple of the other by factorising.

## Showing two vectors are parallel examples

### Example 7: two lines are parallel

Show that AB is parallel to CD .

Work out each vector.

Show that one is a multiple of the other by factorising.

### Example 8: two line segments form one straight line

ABCD is a parallelogram.

The line AB is extended to the point E such that \text{BE}=\frac{2}{3}\text{AB} .

The point F is on the line BC such that BF:FC=2:3 .

\overrightarrow{AB}=6\textbf{a}, \overrightarrow{CB}=5\textbf{b}

Show that DFE is a straight line.

Work out each vector.

Show that one is a multiple of the other by factorising.

### Common misconceptions

• Using the wrong sign

Remember to make the vector negative when going backwards along it.

• Mistakes with ratios

If two parts of a line are in the ratio 1:4, this means one part is ⅕ of a line and the other part ⅘ .

### Practice vector problem questions

1. ABCD is a parallelogram. The line AD is extended to the point E so that AE=3AD .

Find the vector  \overrightarrow{CE} .

-\textbf{p}+3\textbf{q}

-\textbf{p}+\textbf{q}

-\textbf{p}+2\textbf{q}

\textbf{p}+2\textbf{q}

\overrightarrow{CE}=-\textbf{p}+2\textbf{q}

2. \overrightarrow{PR}=4\textbf{a}, \overrightarrow{PQ}=3\textbf{b}

M is the midpoint of QR .

Find the vector \overrightarrow{RM} .

-2\textbf{a}+1.5\textbf{b}

-4\textbf{a}+3\textbf{b}

-2\textbf{a}+3\textbf{b}

2\textbf{a}-1.5\textbf{b}
\text{RM}=\frac{1}{2}\text{RQ}

\begin{aligned} &\overrightarrow{RQ}=-4\textbf{a}+3\textbf{b}\\\\ &\overrightarrow{RM}=\frac{1}{2}\overrightarrow{RQ}=\frac{1}{2}(-4\textbf{a}+3\textbf{b})=-2\textbf{a}+1.5\textbf{b} \end{aligned}

3. \overrightarrow{AB}=3\textbf{b}, \overrightarrow{BC}=2\textbf{a}

AD=2BC and the point E is on the line CD such that \text{ED}=\frac{1}{3}\text{CD} .

Find the vector \overrightarrow{AE} .

\frac{14}{3}\textbf{a}-\textbf{b}

\frac{10}{3}\textbf{a}+\textbf{b}

\textbf{b}+\frac{2}{3}\textbf{a}

\textbf{b}-\frac{2}{3}\textbf{a}

We know \text{DC}=\frac{1}{3}\text{DC} so we need to find the vector \overrightarrow{DC} .

\begin{aligned} &\overrightarrow{DC}=-4\textbf{a}+3\textbf{b}+2\textbf{a}=3\textbf{b}-2\textbf{a}\\\\ &\overrightarrow{DE}=\frac{1}{3}\overrightarrow{DC}=\frac{1}{3}(3\textbf{b}-2\textbf{a}) \end{aligned}

\begin{aligned} &\overrightarrow{AE}=4\textbf{a}+\frac{1}{3}(3\textbf{b}-2\textbf{a})\\\\ &\overrightarrow{AE}=4\textbf{a}+\textbf{b}-\frac{2}{3}\textbf{a}\\\\ &\overrightarrow{AE}=\frac{10}{3}\textbf{a}+\textbf{b} \end{aligned}

4. \overrightarrow{AC}=9\textbf{a}, \overrightarrow{AB}=8\textbf{b}

The point N is such that \text{AN}=\frac{1}{3}\text{AC} , and the point P is such that AP:PB=1:3 . M is the midpoint of BN . Find the vector \overrightarrow{PM} .

\frac{3}{2}\textbf{a}+2\textbf{b}

\frac{3}{2}\textbf{a}+\frac{4}{3}\textbf{b}

-\frac{3}{2}\textbf{a}-2\textbf{b}

\frac{3}{2}\textbf{a}-2\textbf{b}

We need to find a route from P to M . We can see \overrightarrow{PB}=6\textbf{b} .

We know that BM=\frac{1}{2}BN so we need to find the vector \overrightarrow{BN} .

\begin{aligned} &\overrightarrow{BN}=-8\textbf{b}+3\textbf{a}\\\\ &\overrightarrow{BM}=\frac{1}{2}\overrightarrow{BN}=\frac{1}{2}(-8\textbf{b}+3\textbf{a})\\\\ &\overrightarrow{PM}=6\textbf{b}+\frac{1}{2}(-8\textbf{b}+3\textbf{a})\\\\ &\overrightarrow{PM}=6\textbf{b}-4\textbf{b}+\frac{3}{2}\textbf{a}\\\\ &\overrightarrow{PM}=\frac{3}{2}\textbf{a}+2\textbf{b} \end{aligned}

5. \overrightarrow{AB}=2\textbf{a}+6\textbf{b}, \overrightarrow{AC}=3\textbf{a}+2\textbf{b}

D is the midpoint of BC and E is the point on AD such that AE:ED=1:3 .

Find the vector \overrightarrow{AE} .

\frac{5}{2}\textbf{a}+4\textbf{b}

\textbf{a}-4\textbf{b}

\frac{5}{8}\textbf{a}+\textbf{b}

-\frac{5}{4}\textbf{a}-\frac{23}{4}\textbf{b}

We need to begin by finding the vector \overrightarrow{AD} .

\begin{aligned} &\overrightarrow{AB}=2\textbf{a}+6\textbf{b}\\\\ &\overrightarrow{BC}=-2\textbf{a}-6\textbf{b}+3\textbf{a}+2\textbf{b}=\textbf{a}-4\textbf{b}\\\\ &\overrightarrow{BD}=\frac{1}{2}\overrightarrow{BC}=\frac{1}{2}\textbf{a}-2\textbf{b}\\\\ &\overrightarrow{AD}=2\textbf{a}+6\textbf{b}+\frac{1}{2}\textbf{a}-2\textbf{b}=\frac{5}{2}\textbf{a}+4\textbf{b}\\\\ &\overrightarrow{AE}=\frac{1}{4}(\frac{5}{2}\textbf{a}+4\textbf{b})=\frac{5}{8}\textbf{a}+\textbf{b} \end{aligned}

6. \overrightarrow{AB}=4\textbf{a}, \overrightarrow{AE}=8\textbf{b}
D is the midpoint of AE .

Which vector is parallel to \textbf{b}-\textbf{a} ?

\overrightarrow{AB}

\overrightarrow{AC}

\overrightarrow{CE}

\overrightarrow{BE}
\begin{aligned} &\overrightarrow{AB}=4\textbf{a}\\\\ &\overrightarrow{AC}=4\textbf{a}+4\textbf{b}\\\\ &\overrightarrow{CE}=4\textbf{b}-4\textbf{a}\\\\ &\overrightarrow{BE}=8\textbf{b}-4\textbf{a} \end{aligned}

4\textbf{b}-4\textbf{a}=4(\textbf{b}-\textbf{a}) so \overrightarrow{CE} is parallel to \textbf{b}-\textbf{a} .

### Vector problems GCSE questions

1. \overrightarrow{AB}=6\textbf{b},\overrightarrow{BC}=9\textbf{a},\overrightarrow{DF}=-3\textbf{a}+3\textbf{b}

F is the midpoint of AB . Show that AD is parallel to BC .

(2 marks)

Show answer
\overrightarrow{AD}=3\textbf{b}-(-3\textbf{a}+3\textbf{b})=3\textbf{a}

(1)

\overrightarrow{BC} =3\overrightarrow{AD} – they are multiples of each other therefore parallel.

(1)

2. \overrightarrow{AD}=12\textbf{a}, \overrightarrow{CD}=8\textbf{b}

AD=1.5BC.

E is the midpoint of AB .

EF:FC=1:3.

Find the vector \overrightarrow{BF} .

(4 marks)

Show answer
\overrightarrow{BC}=8\textbf{a}

(1)

\overrightarrow{EB}=\frac{1}{2}(12\textbf{a}-8\textbf{b}-8\textbf{a})=2\textbf{a}-4\textbf{b}

(1)

\overrightarrow{EC}=2\textbf{a}-4\textbf{b}+8\textbf{a}=10\textbf{a}-4\textbf{b}

(1)

\overrightarrow{EF}=\frac{1}{4}(10\textbf{a}-4\textbf{b})=\frac{5}{2}\textbf{a}-\textbf{b}

(1)

3. \overrightarrow{AB}=4\textbf{a}, \overrightarrow{BC}=2\textbf{b}

D is the midpoint of AB.

Point E is such that AE:EC=3:1 .

BF=2BC.

Determine whether DEF is a straight line.

(4 marks)

Show answer
\overrightarrow{CE}=\frac{1}{4}(-2\textbf{b}-4\textbf{a})=-\frac{1}{2}\textbf{b}-\textbf{a}

(1)

\overrightarrow{DE}=2\textbf{a}+2\textbf{b}-\frac{1}{2}\textbf{b}-\textbf{a}=\textbf{a}+\frac{3}{2}\textbf{b}

(1)

\overrightarrow{EF}=\frac{1}{2}\textbf{b}+\textbf{a}+2\textbf{b}=\textbf{a}+\frac{5}{2}\textbf{b}

(1)

DEF is not a straight line since \overrightarrow{DE} and \overrightarrow{EF} and are not multiples of each other and therefore are not parallel.

(1)

## Learning checklist

You have now learned how to:

• Solve complex problems involving vectors

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