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A-Level Physics · Mechanics

Adding Vectors by Components

This is the A-Level companion to the GCSE lesson on adding vectors in two dimensions. There, we found the resultant of two vectors by scale drawing and, where they were perpendicular, by Pythagoras. Here we develop the general method that works for any vectors at any angle: resolving each vector into components.

Pythagoras alone only helps when the vectors are perpendicular. For every other situation, physicists use a wonderfully general trick: break every vector into perpendicular parts.

Any vector V can be split into an x-component, Vx (how far it points in the east–west direction), and a y-component, Vy (how far it points in the north–south direction). The two components, joined tip-to-tail, take you from the tail of V to its tip — together they form the vector. Splitting a vector into its components like this is called resolving the vector.

Interactive: Components of a Vector

Use the + and buttons to change Vx and Vy separately. The solid arrow is the vector V; the dotted arrows are its two components along the axes.

Vx Vy V θ
sin θ = Vy / V
cos θ = Vx / V
tan θ = Vy / Vx
V² = Vx² + Vy²
Vx:
Vy:

Two ways to specify a vector

There are two equally correct ways to specify a vector:

By its components — state Vx and Vy.

By its magnitude and angle — state its length V and the angle θ it makes with a known direction.

Both contain exactly the same information. In physics we usually state a vector by its magnitude and angle — "300 km/h at 45° north of east" means more to a pilot than a pair of components — but we usually calculate with components, because components along the same axis add together just like the 1D vectors of the previous lesson.

Converting between the two

From magnitude and angle to components (θ measured from the positive x-axis):

Vx=Vcosθ
Vy=Vsinθ

From components to magnitude and angle:

V=Vx2+Vy2
θ=tan1(Vy/Vx)

One caution: the tan⁻¹ button on a calculator always returns an angle between −90° and +90°, so on its own it cannot tell you which quadrant your vector is in. Always sketch the vector first — the sketch tells you the quadrant; the calculation gives you the precise angle.

Worked Example: A Hiker's Displacement

A hiker walks 8.0 km east across a valley, then turns and walks 5.0 km in a direction 60.0° north of west. What is her displacement from the start?

0 2 4 6 8 2 4 East → ↑ North 8.0 km 5.0 km 60° (a)

Step 1 — Resolve each vector into components. Choose east as the positive x-direction and north as the positive y-direction.

Vector 1 points due east, so it has only an x-component:

D1x=+8.0 km,D1y=0

Vector 2 points west and north, so its x-component is negative:

D2x=(5.0 km)(cos60.0°)=(5.0 km)(0.500)=2.5 km
D2y=+(5.0 km)(sin60.0°)=+(5.0 km)(0.866)=+4.33 km
0 2 4 6 8 2 4 East → ↑ North D2x D2y 5.0 km 60° (b)

Step 2 — Add the components along each axis separately — this is exactly the 1D addition from the previous lesson:

Rx=D1x+D2x=8.0+(2.5)=+5.5 km
Ry=D1y+D2y=0+4.33=+4.33 km

Step 3 — Rebuild the resultant from its components:

R=Rx2+Ry2=30.25+18.75=49.0=7.0 km
tanθ=Ry/Rx=4.33/5.5=0.787θ=38.2°
0 2 4 6 8 2 4 East → ↑ North 8.0 km 5.0 km R = 7.0 km θ = 38.2° (c)

The hiker's displacement is 7.0 km at 38.2° north of east.

Notice what the components did: they turned one 2D problem into two 1D problems, one along each axis. That is the whole method — and it works for any number of vectors at any angles.

Exercise: Resultant Vectors

Answer the questions below, then press Enter or click Check. Give magnitudes as whole numbers and angles in degrees to 1 decimal place. The three circles show your progress — a circle turns green when you answer that type of question correctly. Click "New question" at any time for different numbers.

Magnitude: km Angle: °
Guided Exercise: A Drone's Velocity Components

A drone flies at a steady m/s in a direction ° west of north. (a) Find the northerly and westerly components of its velocity. (b) How far north and how far west has the drone travelled after minutes?

x y north east

Work through the steps below — each answer is checked before you move on. Sketch the vector first: the angle is measured from north, not from the x-axis!

The angle is measured from north, so north is the adjacent side of the triangle: vN = v cos θ. Calculate the north component, to 1 decimal place.

m/s

West is the opposite side: vW = v sin θ. Calculate the west component, to 1 decimal place.

m/s

Distance = speed × time. Convert minutes to seconds, then find how far north the drone has travelled: dN = vN × t. Give your answer in metres.

m

Now the same for the westerly distance: dW = vW × t.

m
Exercise: Force Components — Now On Your Own

A gardener pulls a garden roller with a force of N along a handle that makes an angle of ° above the horizontal. Find the horizontal and vertical components of the force, each to 1 decimal place.

No steps this time — but this problem works exactly like the drone question above. Sketch the right-angled triangle, decide which component is adjacent to the given angle, then use cos and sin.

Horizontal: N Vertical: N

That completes the groundwork on vectors. If you would like more practice on the mathematics itself, look out for our mathematics resources on vectors — but what you have learned here is enough to carry you through GCSE and A-Level Physics.

You will meet these ideas again and again: in projectile motion, where velocity is resolved into horizontal and vertical components; in forces and equilibrium, where resultant forces decide whether and how objects accelerate; in momentum, where 2D collisions are solved component by component; in moments and torque, where only the perpendicular component of a force does the turning; and in circular motion and in electric and magnetic fields, where direction is everything. Master components now, and every one of those topics becomes easier.

Where this fits in your exam specification
Exam boardTopic
AQA A-Level Physics3.4.1.1 Scalars and vectors (resolving)
Edexcel A-Level PhysicsTopic 2 — Mechanics
OCR A A-Level PhysicsModule 3 — Forces and Motion

Studying a different exam board? The method is the same — only the topic name differs.