2D Vectors – A Level Maths Revision

2D Vectors

You’ll have met 2D vectors at GCSE – A Level builds on that knowledge; first, with different ways of representing 2D vectors in similar-looking problems, before moving onto calculating the magnitude and direction of vectors, position vectors and vectors in 3D.

2D Vectors – Contents

Representing Vectors

There are two types of quantity in mathematics: scalar quantities and vector quantities. A scalar just has a magnitude, such as distance or mass. A vector has both magnitude and direction, such as displacement or weight.

To understand the difference, think about two different walks. The first starts and finishes at home. The distance is how far you walked, but the displacement is zero, because you start and finish in the same place. For the second, you start at home and walk to the shops. This distance is still how far you walked, but the displacement is now the (straight-line) distance from your home to the shops.

Vectors can be represented in a few different ways. They can be shown with a lower-case letter in bold or underlined: a, or a. They can also be represented by their end-points: the vector from P to Q is:

\overrightarrow{\text{PQ}}

Here, the arrow represents the direction.

There are also different ways to give information about the size of a vector. You should already be familiar with column vectors:

\binom{a}{b}

Here, the top number, π‘Ž, is how many units the vector travels in the π‘₯-direction and the bottom number, 𝑏, is how many units it travels in the 𝑦-direction.

There are two special cases, the vectors i and j:

\textbf{i}=\binom{1}{0} \text{ and } \textbf{j}=\binom{0}{1}

i and j are known as unit vectors. They are exactly 1 unit long in the π‘₯- and 𝑦-direction, respectively. Unit vectors are another way to show the size of vectors:

\binom{a}{b}=a\textbf{i}+b\textbf{j}

For example:

\binom{2}{-3}=2\textbf{i}-3\textbf{j}

This is also known as component form. The top number, or coefficient of i, is referred to as the i-component and the bottom number, or coefficient of j, as the j-component.

Vector Addition

Vector addition is incredibly useful and is as simple as adding the i-components and the j-components. The outcome when adding two vectors is called the resultant. For example:

\binom{1}{-3}+\binom{2}{4}=\binom{1\,+\,2}{-3\,+\,4}=\binom{3}{1}

If you are given the vectors in component form, it’s very similar. Let’s write the vectors above in the form 𝑝i + π‘žj:

\begin{aligned} &\binom{1}{-3}=\textbf{i}-3\textbf{j}\\&\binom{2}{4}=2\textbf{i}+4\textbf{j}\end{aligned}

Adding these would give:

(\textbf{i}-3\textbf{j})+(2\textbf{i}+4\textbf{j})=(1+2)\textbf{i}+(-3+4)\textbf{j}=3\textbf{i}+\textbf{j}


Example Question 1

Given that \overrightarrow{AB} = 3\textbf{i}-2\textbf{j}\text{, } \overrightarrow{AC}=-\textbf{i}+5\textbf{j} and \overrightarrow{CD}=2\textbf{i}, find the vector \overrightarrow{BD} .

With a question like this, it’s always sensible to draw a quick diagram to help you understand the problem better.

It doesn’t need to be accurate, but the points A, B, C and D should be in roughly the right place. For example, C is above and to the left of A, while B is below and to the right of A. You are looking for the vector \overrightarrow{BD} , so you need to start at B and travel to D.

The first stage of this is to go from B to A. This is the vector \overrightarrow{BA} , which is equal to -\overrightarrow{AB} . After that, simply travel along
\overrightarrow{AC} then \overrightarrow{CD} . Writing this as an addition gives:

\begin{aligned} \overrightarrow{BD} &=-\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{CD}\\&=-(3\textbf{i}-2\textbf{j})+(-\textbf{i}+5\textbf{j})+2\textbf{i}\\&=(-3-1+2)\textbf{i}+(2+5)\textbf{j}\\&=-2\textbf{i}+7\textbf{j}\end{aligned}


Scalar Multiplication

As well as adding and subtracting vectors, you can also multiply them by a scalar. For example:

2\times \binom{5}{-4}=\binom{10}{-8}

or:

2(5\textbf{i}-4\textbf{j})=10\textbf{i}-8\textbf{j}

The result when multiplying by a positive scalar is a vector in the same direction but of a different magnitude. This means the starting vector and the resultant will be parallel. This also means that you can prove that two vectors are parallel by showing that one is a scalar multiple of the other.

A scalar multiplier is usually written using Ξ». This is the lower-case Greek letter, lambda, which is written as an upside-down y.


Example Question 2

Show that the vectors \binom{9}{-3} and \binom{-6}{2} are parallel.

If the vectors are parallel, then:

\begin{aligned}&\binom{9}{-3}=\lambda \binom{-6}{2}\\&\binom{9}{-3}=\binom{-6\lambda}{2\lambda}\end{aligned}

You need a value of Ξ» such that 9 = -6Ξ» and -3 = 2Ξ». Solving these equations gives:

\begin{aligned}&9=-6\lambda\\&\lambda=-\frac{9}{6}\\&\lambda=-\frac{3}{2} \end{aligned}

\begin{aligned} &-3=2\lambda \\ &\lambda = -\frac{3}{2}\end{aligned}

So, the value of Ξ» is -\frac{3}{2} and \binom{9}{-3} and \binom{-6}{2} are parallel.

You could have written \binom{-6}{2} = \lambda\binom{9}{-3} which would give a different value of Ξ». This is fine – the two vectors are parallel as long as we get the same value of Ξ» for the i– and j-components.


Example Question 3

The vectors 4i – 2j and 7i + 𝑛j are parallel. Find the value of 𝑛.

Since the vectors are parallel, 7i + 𝑛j = Ξ»(4i – 2j). Compare the i-components:

\begin{aligned}&7=4\lambda\\&\lambda=\frac{7}{4}\end{aligned}

Substituting this into the j-components gives:

\begin{aligned} &n=-2\lambda\\&n=-2\times\frac{7}{4}\\&n=-\frac{7}{2} \end{aligned}


Geometrical Problems

These skills can all be applied to problem-solving and modelling problems. It is really important you are confident with these skills.


Example Question 4

In the diagram below, M is a point on the line segment QR such that \overrightarrow{QM}:\overrightarrow{MR} = 1:3. Vector \overrightarrow{PQ} = 3i + 5j and \overrightarrow{PR} = 4i – 2j. Find the vector \overrightarrow{QM}

Start by finding the vector \overrightarrow{QR}:

\begin{aligned} \overrightarrow{QR} &= - \overrightarrow{PQ}+\overrightarrow{PR} \\ &=-(3\textbf{i}+5\textbf{j})+(4\textbf{i}-2\textbf{j})\\&=\textbf{i}-7\textbf{j}\end{aligned}

Since \overrightarrow{QM}:\overrightarrow{MR} = 1:3, \overrightarrow{QM} is \frac{1}{4} of \overrightarrow{QR} :

\begin{aligned} \overrightarrow{QM} &= \frac{1}{4}\overrightarrow{QR} \\ &=\frac{1}{4}(\textbf{i}-7\textbf{j}) \\ &= \frac{1}{4}\textbf{i} - \frac{7}{4}\textbf{j}\end{aligned}

Practice Questions

1. For the vectors p = 4i, q = \frac{1}{2} i + 3j, r = i – 2j and s = 2i –\frac{3}{4}j, write each of the following in the form π‘Ži + 𝑏j.

a. p + q

b. 2rq

c. 4q – 3r + s

2. In the diagram below, M is the midpoint of QR and \overrightarrow{\text{PQ}} = 3i + \frac{3}{2}j, \overrightarrow{\text{QR}} = 5i – 2j and \overrightarrow{\text{SP}} = -4i + \frac{5}{2}j.

a. Find the vector \overrightarrow{\text{PM}} in the form π‘Ži + 𝑏j.

b. Show that \overrightarrow{\text{PQ}} and \overrightarrow{\text{SR}} are parallel.

3. Given that a = \binom{p}{2p}, b = \binom{2q}{-q}, c = \binom{5}{2} and a + 5b = c, find the values of 𝑝 and π‘ž.

4. In the diagram below, M is the midpoint of BC and AN:ND = 3:2. Given that \overrightarrow{\text{AB}} = i + 4j, \overrightarrow{\text{AC}} = 7i + 6j and \overrightarrow{\text{DC}} = -3i + j, find the vector \overrightarrow{\text{MN}}.

2D Vectors - Practice Question 4

5. The resultant of a = \frac{3}{2}i + j and b = π‘ši – 2𝑛j is equal to the vector c = \frac{1}{2}j. Find the value of π‘š and 𝑛.

Answers

1. For the vectors p = 4i, q = \frac{1}{2} i + 3j, r = i – 2j and s = 2i –\frac{3}{4}j, write each of the following in the form π‘Ži + 𝑏j.

a. p + q

\boldsymbol{(4\textbf{\underline{i}})+(\frac{1}{2}\textbf{\underline{i}}+3\textbf{\underline{j}})=\frac{9}{2}\textbf{\underline{i}}+3\textbf{\underline{j}}}

b. 2rq

\boldsymbol{2(\textbf{\underline{i}}-2\textbf{\underline{j}})-(\frac{1}{2}\textbf{\underline{i}}+3\textbf{\underline{j}})=\frac{3}{2}\textbf{\underline{i}} - 7 \textbf{\underline{j}}}

c. 4q – 3r + s

\boldsymbol{4(\frac{1}{2}\textbf{\underline{i}} + 3\textbf{\underline{j}})-3(\textbf{\underline{i}}-2\textbf{\underline{j}})+(2\textbf{\underline{i}}-\frac{3}{4}\textbf{\underline{j}})=\textbf{\underline{i}}+\frac{69}{4}\textbf{\underline{j}}}

2. In the diagram, M is the midpoint of QR and \overrightarrow{\text{PQ}} = 3i + \frac{3}{2}j, \overrightarrow{\text{QR}} = 5i – 2j and \overrightarrow{\text{SP}} = -4i + \frac{5}{2}j.

Β 

a. Find the vector \overrightarrow{\text{PM}} in the form π‘Ži + 𝑏j.

\boldsymbol{\begin{aligned}\overrightarrow{\textbf{QM}}&=\frac{1}{2}\overrightarrow{\textbf{QR}}\\[0.5em]&=\frac{5}{2}\textbf{\underline{i}}-\textbf{\underline{j}}\end{aligned}}

\boldsymbol{\begin{aligned}\overrightarrow{\textbf{PM}}&=\overrightarrow{\textbf{PQ}}+\overrightarrow{\textbf{QM}}\\[0.5em]&=(3\textbf{\underline{i}}+\frac{3}{2}\textbf{\underline{j}})+(\frac{5}{2}\textbf{\underline{i}}-\textbf{\underline{j}})\\[0.5em]&=\frac{11}{2}\textbf{\underline{i}}+\frac{1}{2}\textbf{\underline{j}}\end{aligned}}

b. Show that \overrightarrow{\text{PQ}} and \overrightarrow{\text{SR}} are parallel.

\boldsymbol{\begin{aligned}\overrightarrow{\textbf{SR}}&=\overrightarrow{\textbf{SP}}+\overrightarrow{\textbf{PQ}}+\overrightarrow{\textbf{QR}}\\&=(-4\textbf{\underline{i}}+\frac{5}{2}\textbf{\underline{j}})+(3\textbf{\underline{i}}+\frac{3}{2}\textbf{\underline{j}})+(5\textbf{\underline{i}}-2\textbf{\underline{j}})\\&=4\textbf{\underline{i}}+2\textbf{\underline{j}}\end{aligned}}

\boldsymbol{\begin{aligned}&\overrightarrow{\textbf{SR}}=\lambda\overrightarrow{\textbf{PQ}}\\&4\textbf{\underline{i}}+2\textbf{\underline{j}}=\lambda(3\textbf{\underline{i}}+\frac{3}{2}\textbf{\underline{j}})\end{aligned}}

\boldsymbol{\begin{aligned}\\\textbf{for \textbf{\underline{i}} component: }4&=3\lambda\\\lambda&=\frac{4}{3}\end{aligned}}

\boldsymbol{\begin{aligned}\\\textbf{for \textbf{\underline{j}} component: }2&=\frac{3}{2}\lambda\\[0.5em]\lambda&=\frac{4}{3}\end{aligned}}

\boldsymbol{\overrightarrow{\textbf{SR}}=\frac{4}{3}\overrightarrow{\textbf{PQ}}\textbf{ so }\overrightarrow{\textbf{PQ}}\textbf{ and } \overrightarrow{\textbf{SR}} \textbf{ are parallel.}}

Β 

3. Given that a = \binom{p}{2p}, b = \binom{2q}{-q}, c = \binom{5}{2} and a + 5b = c, find the values of 𝑝 and π‘ž.

Β 

\boldsymbol{\begin{aligned}&\textbf{\underline{a}}+5\textbf{\underline{b}} = \textbf{\underline{c}}\\&\binom{p}{2p}+5\binom{2q}{-q}=\binom{5}{2}\\[0.5em]&\binom{p\,+\,10q}{2p\,-\,5q}=\binom{5}{2}\end{aligned}}

This gives 2 simultaneous equations:

\boldsymbol{\begin{aligned} &p + 10q=5 \\ &2p-5q=2\end{aligned}}

Β 

Solving these gives:

\boldsymbol{\begin{aligned} &p = \frac{9}{5} \\ &q=\frac{8}{25}\end{aligned}}

4. In the diagram, M is the midpoint of BC and AN:ND = 3:2. Given that \overrightarrow{\text{AB}} = i + 4j, \overrightarrow{\text{AC}} = 7i + 6j and \overrightarrow{\text{DC}} = -3i + j, find the vector \overrightarrow{\text{MN}}.

\boldsymbol{\begin{aligned}\overrightarrow{\textbf{BC}}&=-\overrightarrow{\textbf{AB}}+\overrightarrow{\textbf{AC}} \\ &= -(\textbf{\underline{i}}+4\textbf{\underline{j}})+(7\textbf{\underline{i}}+6\textbf{\underline{j}})\\&=6\textbf{\underline{i}}+2\textbf{\underline{j}}\\\overrightarrow{\textbf{BM}}&=\frac{1}{2}\overrightarrow{\textbf{BC}}\\&=3\textbf{\underline{i}}+\textbf{\underline{j}}\end{aligned}}

\boldsymbol{\begin{aligned}\overrightarrow{\textbf{AD}}&=\overrightarrow{\textbf{AC}}-\overrightarrow{\textbf{DC}} \\ &= (7 \textbf{\underline{i}}+6 \textbf{\underline{j}})-(-3\textbf{\underline{i}}+\textbf{\underline{j}})\\&=10\textbf{\underline{i}}+5\textbf{\underline{j}}\\\overrightarrow{\textbf{AN}}&=\frac{3}{5}\overrightarrow{\textbf{AD}}\\&=6\textbf{\underline{i}}+3\textbf{\underline{j}}\end{aligned}}

\boldsymbol{\begin{aligned}\overrightarrow{\textbf{MN}}&=-\overrightarrow{\textbf{BM}}-\overrightarrow{\textbf{AB}} +\overrightarrow{\textbf{AN}}\\ &= -(3\textbf{\underline{i}}+\textbf{\underline{j}})-(\textbf{\underline{i}}+4\textbf{\underline{j}})+(6\textbf{\underline{i}}+3\textbf{\underline{j}})\\&=2\textbf{\underline{i}}-2\textbf{\underline{j}}\end{aligned}}

5. The resultant of a = \frac{3}{2}i + j and b = π‘ši – 2𝑛j is equal to the vector c = \frac{1}{2}j. Find the value of π‘š and 𝑛.

\boldsymbol{\begin{aligned} \textbf{\underline{a}}+\textbf{\underline{b}} & = (\frac{3}{2}\textbf{\underline{i}} + \textbf{\underline{j}}) + (m\textbf{\underline{i}} - 2n\textbf{\underline{j}}) \\ &= (\frac{3}{2} + m)\textbf{\underline{i}}+(1-2n)\textbf{\underline{j}}\end{aligned}}

Β 

\boldsymbol{\frac{3}{2} + m = 0 \textbf{ so } m = -\frac{3}{2}}

\boldsymbol{\begin{aligned} &1 - 2n =\frac{1}{2}\\ &2n =\frac{1}{2} \\ &n = \frac{1}{4} \end{aligned}}


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