Vectors and resultantsEdexcel International A Level Maths: Revision notes

Section 1

Vectors, magnitude and direction

A vector has both magnitude and direction; a scalar has magnitude only. In mechanics, force, velocity, acceleration and displacement are vectors. In component form, $\mathbf{a}=x\mathbf{i}+y\mathbf{j}$ , where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors. The magnitude is $|\mathbf{a}|=\sqrt{x^2+y^2}$ and the direction is found from $\tan\theta=\dfrac{y}{x}$ , with $\theta$ the angle from the $\mathbf{i}$ direction. Example: $\mathbf{a}=5\mathbf{i}+3\mathbf{j}$ has $|\mathbf{a}|=\sqrt{34}=5.83$ and $\theta=31.0^\circ$ above $\mathbf{i}$ .

Key termsvectorscalarunit vector

Exam tip

Sketch the vector first to see which quadrant it is in; $\tan^{-1}$ on a calculator only gives an acute angle for positive ratios.

Section 2

Adding vectors: the resultant

The resultant is the single vector with the same effect as several vectors acting together. Add components: $(a\mathbf{i}+b\mathbf{j})+(c\mathbf{i}+d\mathbf{j})=(a+c)\mathbf{i}+(b+d)\mathbf{j}$ . In a vector diagram, place the vectors nose to tail; the resultant joins the start to the finish (the triangle law). For $\mathbf{F}_1=7\mathbf{i}-3\mathbf{j}$ and $\mathbf{F}_2=-2\mathbf{i}+6\mathbf{j}$ , the resultant is $5\mathbf{i}+3\mathbf{j}$ . A particle is in equilibrium when the resultant is $\mathbf{0}$ , so each component sums to zero.

Key termsresultantequilibrium

Common mistake

Subtracting a vector by accident. Add all the $\mathbf{i}$ components together and all the $\mathbf{j}$ components together, keeping their signs.

Section 3

Resolving a vector into components

To resolve a vector of magnitude $F$ at angle $\theta$ to the $\mathbf{i}$ direction, use trigonometry in a right-angled triangle: the component along $\mathbf{i}$ is $F\cos\theta$ and along $\mathbf{j}$ is $F\sin\theta$ , so $\mathbf{F}=(F\cos\theta)\mathbf{i}+(F\sin\theta)\mathbf{j}$ . Example: $F=24$ N at $40^\circ$ gives $(18.4\mathbf{i}+15.4\mathbf{j})$ N. If the angle is measured from the $\mathbf{j}$ direction, the roles of sine and cosine swap.

Key termsresolvecomponent

Common mistake

Using $\cos$ for the component away from the angle. The component adjacent to the angle uses cosine; the component opposite uses sine.

Section 4

Vectors from magnitude and direction

A vector in the direction of $x\mathbf{i}+y\mathbf{j}$ with magnitude $M$ is $M\dfrac{x\mathbf{i}+y\mathbf{j}}{\sqrt{x^2+y^2}}$ . Example: a 26 N force in the direction $5\mathbf{i}+12\mathbf{j}$ is $26\times\dfrac{5\mathbf{i}+12\mathbf{j}}{13}=10\mathbf{i}+24\mathbf{j}$ . A 3, 4, 5 or 5, 12, 13 triangle gives exact components, so look out for them.

Key termsunit vector direction

Section 5

Solving problems with resultants

To find an unknown force: write the required resultant, subtract the known vectors and read off the components. For three forces with resultant $20\mathbf{i}$ , $\mathbf{F}_3=20\mathbf{i}-(\mathbf{F}_1+\mathbf{F}_2)$ . Then find its magnitude with Pythagoras and its direction with $\tan\theta$ , saying clearly which line the angle is measured from (for example 'below the direction of $\mathbf{i}$ '). Keep exact values or at least 4 significant figures until the end, and round to 3 s.f.

Key termsdirection

Exam tip

Always name the line the angle is measured from, and say above or below it.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Vectors and resultants

Two forces act on a particle: $\mathbf{F}_1=(7\mathbf{i}-3\mathbf{j})$ N and $\mathbf{F}_2=(-2\mathbf{i}+6\mathbf{j})$ N, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors in the horizontal plane, in the directions east and north.
Find the angle between the resultant force and the direction east, giving your answer in degrees to 1 decimal place.2 marks
A force $\mathbf{F}$ of magnitude 24 N acts on a particle in a direction $40^\circ$ above the direction of the unit vector $\mathbf{i}$ , where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors in a vertical plane, horizontal and vertically upwards respectively.
A second force $\mathbf{G}=(-10\mathbf{i}-6\mathbf{j})$ N also acts on the particle. Find the magnitude of the resultant of $\mathbf{F}$ and $\mathbf{G}$ .2 marks
Three forces $\mathbf{P}=(3\mathbf{i}+5\mathbf{j})$ N, $\mathbf{Q}=(-7\mathbf{i}+2\mathbf{j})$ N and $\mathbf{R}=(p\mathbf{i}+q\mathbf{j})$ N act on a particle, where $p$ and $q$ are constants. The particle is in equilibrium.
Find the values of $p$ and $q$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).