A Level Maths, A Level Vectors

Magnitude and Direction of Vectors – A Level Maths Revision

At A Level, Vectors start with different ways of representing 2D Vectors (check out that blog if you’re unsure with any of the ways vectors are written here) before moving onto using Pythagoras’ theorem and trigonometry to calculate the magnitude (or the length) and direction of vectors.

Magnitude and Direction of Vectors

If you’d like this blog on PowerPoint or PDF formats, click here. For more ways to practise magnitude and direction of vector questions, try either of these activities. If you’d like to practise some of the prior-knowledge required for this topic, try these multiple-choice questions.

Magnitude and Direction of Vectors

Vectors have both magnitude and direction, unlike scalars which only have magnitude. However, when vectors are presented in component form or as a column vector, the magnitude and direction aren’t always that clear.

You can use Pythagoras’ theorem to find the magnitude of a vector and trigonometry to find the angle and direction of the vector. This is known as magnitude-direction form.

Unless otherwise specified in the question, the direction of a vector is the angle between the vector and the positive horizontal unit vector, i. It is measured anticlockwise from i.

You can also be asked to give the direction of a vector as a bearing, which is the angle the vector makes with the positive vertical unit vector, j, measured in a clockwise direction.

Example Question 1

Find the magnitude and direction of the vector a = 4i + 2j.

First, sketch the vector with the horizontal and vertical components:

Example question 1 - sketch the vector with the horizontal and vertical components

To find the magnitude, use Pythagoras’ theorem. The i– and j-components are perpendicular, so this creates a right-angled triangle.

The magnitude of a is written as |a|.

\begin{aligned}|\textbf{a}|&=\sqrt{2^2+4^2}\\&=2\sqrt{5}\end{aligned}

When describing the direction of a vector, use the angle between the vector and the positive horizontal unit vector, i.

You know the length of the adjacent side is the i-component, 4, and the length of the opposite side is the j-component, 2. You can therefore use tan to find the angle:

\begin{aligned} &\tan{\theta}=\frac{2}{4}\\&\theta=\tan^{-1}{\left(\frac{2}{4}\right)}=26.6^{\circ} \text{ above }\textbf{i}\text{ (3s.f.)}\end{aligned}

You now have enough information to write the vector in magnitude-direction form – write the magnitude of the vector and then the direction of the vector in brackets.
So, the vector a in magnitude-component form is (2√5, 26.6°).

Example Question 2

Find the magnitude and bearing of the vector \binom{-3}{4} .

To find the magnitude, use Pythagoras’ theorem.

\sqrt{3^2+4^2} = 5

As you are asked for the bearing, you want the angle that the vector makes with the positive vertical unit vector, j, in a clockwise direction.

\begin{aligned} &a = \tan^{-1}{\left(\frac{4}{3}\right)}=53.13^{\circ}\\&b=270+53.13=323.1^{\circ} \end{aligned}

The bearing \binom{-3}{4} has magnitude 5 and a bearing of 323°.

Sometimes, you are given the vector in magnitude-direction form, and need to convert it into component form. This also uses trigonometry; this time, sine and cosine.

Example Question 3

Write the vector (8, 300°) in component form.

Start by drawing a sketch, taking care to mark the correct angle. The angle 300° is the angle anticlockwise from the positive i direction. Therefore, as angles around a point sum to 360°, the angle in the triangle is 60°.

Your sketch will also help work out if the components should be negative. In this case, both the i– and j-components are positive.

You can now find the i-component, 𝑎, and j-component, 𝑏, of the vector. Using trigonometry:

\begin{aligned} &a = 8\cos(60^{\circ}) = 4 \\ &b = 8\sin(60^{\circ}) = 4\sqrt{3}\end{aligned}

Therefore, the vector in component form is:

4\textbf{i} - 4\sqrt{3}\textbf{j}

Sometimes, you will be asked to find the unit vector in a specific direction – this is simply a vector of magnitude 1. The unit vector in the same direction as a vector a is written as â (this is said as “a hat”). To find â, change the magnitude of a to 1 unit by dividing a by |a|.

Example Question 4

Find the unit vector in the direction of the vector a = \binom{-4}{3}

First, find the magnitude of a:

\begin{aligned} |\textbf{a}| &= \sqrt{(-4)^2+(3)^2}\\&=5 \end{aligned}

Now, divide the vector by the magnitude:

\begin{aligned} \hat{\textbf{a}} &= \binom{-4}{3} \div 5 \\ &= \binom{\frac{-4}{5}}{\frac{3}{5}} \end{aligned}

The vector â is in the same direction as a but has a magnitude of 1.

When asked to find a vector, the answer should be in the same format as the question, where possible. If the question uses column vectors, your answer should be a column vector; if the question uses component form, so should your answer.

Practice Questions

1. Find the magnitude of each of these vectors. Give your answers in exact form.

a. 3i – 3j

b. -2i + 7j

c. \binom{5}{-1}

2. Find the direction of each vector.

a. \binom{8}{8}

b. 6i – 5j

c. –i + 8j

3. Given their magnitude and direction, write these vectors as column vectors. Give your answer correct to 3 significant figures where appropriate.

a. Magnitude 7 and direction 45°.

b. Magnitude \sqrt{3} and direction 340°.

c. Magnitude \frac{1}{2} and direction 100°.

4. For each of the vectors below, find the unit vector in the same direction. Give your answers in exact form.

a. a = 4i – 3j

b. b = 4i + 2j

c. c = \binom{1}{-2}

5. Vectors \overrightarrow{\text{PQ}} = 3i – 2j and \overrightarrow{\text{PR}} = -2i – 6j form two sides of a triangle PQR as shown below.

a. Find the angle QPR.
b. Find the area of triangle PQR.

6. The vector \overrightarrow{\text{AB}} = \binom{2}{4} . The vector \overrightarrow{\text{BC}} has magnitude 2√5 and is 10° below the horizontal.

a. Write the vector \overrightarrow{\text{BC}} as a column vector.

b. Show that triangle ABC is an isosceles triangle.

Test yourself then check the answers below

1. Find the magnitude of each of these vectors. Give your answers in exact form.

a. 3i – 3j

\boldsymbol{\sqrt{3^2+(-3)^2}=3\sqrt{2}}

b. -2i + 7j

\boldsymbol{\sqrt{(-2)^2+7^2}=\sqrt{53}}

c. \binom{5}{-1}

\boldsymbol{\sqrt{5^2+(-1)^2}=\sqrt{26}}

 

2. Find the direction of each vector.

a. \binom{8}{8}

\boldsymbol{\tan^{-1}\left(\frac{8}{8}\right)=45^{\circ}}

b. 6i – 5j

\boldsymbol{\begin{aligned}&\tan^{-1}\left(\frac{5}{6}\right)=39.8^{\circ}\\&360-39.8...=320^{\circ}\textbf{ (3s.f.)}\end{aligned}}

c. –i + 8j

\boldsymbol{\begin{aligned}&a=\tan^{-1}\left(\frac{8}{1}\right)=82.87...^{\circ}\\&b=180-82.87...=97.1^{\circ}\textbf{ (3s.f.)}\end{aligned}}

3. Given their magnitude and direction, write these vectors as column vectors. Give your answer correct to 3 significant figures where appropriate.

a. Magnitude 7 and direction 45°.

\boldsymbol{\begin{aligned}&\binom{7\cos(45)}{7\sin(45)}=\binom{\frac{7\sqrt{2}}{2}}{\frac{7\sqrt{2}}{2}}=\binom{4.95}{4.95}\textbf{ (3s.f.)}\end{aligned}}

b. Magnitude \sqrt{3} and direction 340°.

\boldsymbol{\begin{aligned}&360^{\circ}-340^{\circ}=20^{\circ}\\[0.5em]&\binom{\sqrt{3}\cos(20)}{-\sqrt{3}\sin(20)}=\binom{1.63}{-0.592}\textbf{ (3s.f.)}\end{aligned}}

c. Magnitude \frac{1}{2} and direction 100°.

\boldsymbol{\begin{aligned}&180^{\circ}-100^{\circ}=80^{\circ}\\[0.5em]&\binom{-\frac{1}{2}\cos(80)}{\frac{1}{2}\sin(80)}=\binom{-0.0868}{0.492}\textbf{ (3s.f.)}\end{aligned}}

4. For each of the vectors below, find the unit vector in the same direction. Give your answers in exact form.

a. a = 4i – 3j

\boldsymbol{\begin{aligned}\vert\textbf{\underline{a}}\vert&=\sqrt{4^2+(-3)^2}\\&=5\\ \hat{\textbf{\underline{a}}}&=\frac{4}{5}\textbf{\underline{i}}-\frac{3}{5}\textbf{\underline{j}}\end{aligned}}

b. b = 4i + 2j

\boldsymbol{\begin{aligned}\vert\textbf{\underline{b}}\vert&=\sqrt{4^2+2^2}\\&=2\sqrt{5}\\ \hat{\textbf{\underline{b}}}&=\frac{2\sqrt{5}}{5}\textbf{\underline{i}}+\frac{\sqrt{5}}{5}\textbf{\underline{j}}\end{aligned}}

 

c. c = \binom{1}{-2}

\boldsymbol{\begin{aligned}\vert\textbf{\underline{c}}\vert&=\sqrt{1^2+(-2)^2}\\&=\sqrt{5}\\ \hat{\textbf{\underline{c}}}&=\binom{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}}\end{aligned}}

5. Vectors \overrightarrow{\text{PQ}} = 3i – 2j and \overrightarrow{\text{PR}} = -2i – 6j form two sides of a triangle PQR as shown below.

a. Find the angle QPR.

\boldsymbol{\begin{aligned}&a=\tan^{-1}\left(\frac{2}{3}\right)=33.6\\&b=\tan^{-1}\left(\frac{6}{2}\right)=71.5\\&\textbf{QPR}=180-(33.6...+71.5...)=74.7^{\circ} \textbf{ (3s.f.)}\end{aligned}}

b. Find the area of triangle PQR.

\boldsymbol{\begin{aligned}\vert\overrightarrow{\textbf{PQ}}\vert&=\sqrt{3^2+(-2)^2}=\sqrt{13}\\ \vert\overrightarrow{\textbf{PR}}\vert&=\sqrt{(-2)^2+6^2}=2\sqrt{10}\\\textbf{Area}&=\frac{1}{2}\times\sqrt{13}\times 2\sqrt{10} \times \sin (74.7...) \\&=\textbf{11 square units}\end{aligned}}

6. The vector \overrightarrow{\text{AB}} = \binom{2}{4} . The vector \overrightarrow{\text{BC}} has magnitude 2√5 and is 10° below the horizontal.

a. Write the vector \overrightarrow{\text{BC}} as a column vector.

\boldsymbol{\begin{aligned}&\overrightarrow{\textbf{BC}} = \binom{2\sqrt{5}\cos(10^{\circ})}{2\sqrt{5}\sin(10^{\circ})}=\binom{4.40}{0.777} \textbf{ (3s.f.)}\end{aligned}}

b. Show that triangle ABC is an isosceles triangle.

\boldsymbol{\begin{aligned}&\overrightarrow{\textbf{AC}} = \binom{2+2\sqrt{5}\cos(10^{\circ})}{4+2\sqrt{5}\sin(10^{\circ})}=\binom{6.40}{4.777} \textbf{ (3s.f.)}\end{aligned}}

\boldsymbol{\begin{aligned}\vert\overrightarrow{\textbf{AB}}\vert&=\sqrt{2^2+4^2}=2\sqrt{5}\\\vert\overrightarrow{\textbf{BC}}\vert&=2\sqrt{5}\\\vert\overrightarrow{\textbf{AC}}\vert&=\sqrt{6.40...^2+4.77...^2}=7.99 \textbf{ (3s.f.)}\end{aligned}}

\boldsymbol{\begin{aligned}&\vert\overrightarrow{\textbf{AB}}\vert = \vert\overrightarrow{\textbf{BC}}\vert = 2\sqrt{5} \textbf{ and } \vert\overrightarrow{\textbf{AC}}\vert \neq \vert\overrightarrow{\textbf{AB}}\vert\textbf{. Therefore, ABC is isosceles.}\end{aligned}}

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