Scalar vs vector quantities · Representing vectors with arrows · Adding perpendicular vectors using Pythagoras and trigonometry
Distinguish between quantities that have magnitude only and those that have both magnitude and direction.
Correctly identify speed, velocity, distance, displacement, mass, weight and force as scalar or vector.
Use scaled arrows to represent the magnitude and direction of a vector quantity.
Use Pythagoras' theorem to find the magnitude of a resultant vector from two perpendicular components.
Apply tan, sin and cos to calculate the angle of the resultant vector relative to a reference direction.
Describe resultant velocity or force with correct magnitude, unit and direction in real-world scenarios.
This distinction is crucial in physics. Two cars travelling at 30 m/s in opposite directions have the same speed (scalar) but very different velocities (vector). If they collide, their directions matter enormously!
| Scalar Quantities | Vector Quantities |
|---|---|
| Distance (m) | Displacement (m) |
| Speed (m/s) | Velocity (m/s) |
| Mass (kg) | Force / Weight (N) |
| Energy (J) | Acceleration (m/s²) |
| Temperature (°C or K) | Momentum (kg m/s) |
| Time (s) | Electric field strength (N/C) |
Notice the pairs: distance ↔ displacement, speed ↔ velocity. Each pair describes similar physical ideas, but one is a scalar and the other is a vector. For example, if you walk 5 m north then 5 m south, your total distance is 10 m but your displacement is 0 m (back where you started).
Because vectors have direction, we represent them using arrows. The convention is:
For example, a force of 30 N acting to the right might be drawn as a 3 cm arrow pointing right (scale: 1 cm = 10 N). A force of 15 N acting upward would be drawn as a 1.5 cm arrow pointing up.
To add two vectors graphically, place the tail of the second vector at the tip of the first. The resultant vector is drawn from the tail of the first to the tip of the second.
When two perpendicular vectors (at 90° to each other) are added, the resultant is the hypotenuse of the right-angled triangle they form. Pythagoras' theorem gives the magnitude:
| Symbol | Meaning | Unit |
|---|---|---|
| R | Magnitude of the resultant vector | N, m/s, m, etc. |
| a | First component vector (magnitude) | same unit as R |
| b | Second component vector (magnitude), perpendicular to a | same unit as R |
This works for any vector quantity — forces, velocities, displacements — as long as the two components are perpendicular (90° apart).
After finding the magnitude with Pythagoras, we find the direction using trigonometry. For a right-angled triangle with the resultant as hypotenuse and angle θ measured from one of the component vectors:
The terms "opposite" and "adjacent" refer to the sides relative to angle θ:
You can also use sin and cos if needed:
A complete vector answer must always include:
Understanding vectors is essential throughout physics and engineering. Here are key contexts you'll meet in GCSE questions:
A swimmer aims directly across a river, but the river current pushes them downstream. Their actual velocity is the resultant of their swimming velocity (perpendicular to the bank) and the current velocity (parallel to the bank). These two are perpendicular, so Pythagoras applies.
A hanging mass experiences weight (downward) and tension (upward). If two ropes pull at right angles, the resultant tension can be found using vectors.
A plane flying north in a crosswind from the west has a resultant velocity found by combining its airspeed (north) and the wind speed (east). Both perpendicular — perfect for Pythagoras.
East displacement: a = 40 m | South displacement: b = 30 m
These are perpendicular (90° apart), so we can use Pythagoras' theorem.
R = √(a² + b²)
R = √(40² + 30²)
R = √(1600 + 900)
R = √2500
R = 50 m
We want the angle θ below the east direction (i.e. south of east).
tan θ = opposite ÷ adjacent = 30 ÷ 40 = 0.75
θ = tan⁻¹(0.75) = 36.9°
North component: v_N = 200 m/s | East component: v_E = 50 m/s
North and east are perpendicular, so Pythagoras applies.
R = √(200² + 50²)
R = √(40000 + 2500)
R = √42500
R = 206.2 m/s
tan θ = opposite ÷ adjacent = 50 ÷ 200 = 0.25
θ = tan⁻¹(0.25) = 14.0°
Horizontal (right) and vertical (up) are at 90° to each other. ✓
R = √(12² + 9²)
R = √(144 + 81)
R = √225
R = 15 N
tan θ = opposite ÷ adjacent = 9 ÷ 12 = 0.75
θ = tan⁻¹(0.75) = 36.9°
North: 6 km | East: 8 km — these are perpendicular. ✓
R = √(6² + 8²) = √(36 + 64) = √100 = 10 km
tan θ = 8 ÷ 6 = 1.333...
θ = tan⁻¹(1.333) = 53.1°
The angle is 53.1° clockwise from north = bearing of 053°.
Question 1: Which of the following is a vector quantity?
Question 2: Which of the following correctly pairs a scalar with its vector equivalent?
Question 3: Two forces act at right angles on an object: 5 N east and 12 N north. What is the magnitude of the resultant force?
Question 4: A force of 20 N acts horizontally and a force of 20 N acts vertically upward. Calculate the magnitude of the resultant force. Give your answer to 1 decimal place in newtons.
Question 5: A swimmer swims at 3 m/s perpendicular to a river bank. The river flows at 4 m/s parallel to the bank. What is the angle of the swimmer's resultant velocity relative to the direction they are swimming? Give your answer to 1 decimal place in degrees.
Challenge 1: A remote-controlled drone moves 24 m due east, then 10 m due north, then 7 m due south. Calculate the magnitude and direction (bearing to the nearest degree) of the drone's resultant displacement from its starting position.
Challenge 2: A yacht sails with a velocity of 6.0 m/s due east relative to the water. A tidal current flows at 2.5 m/s due south. (a) Calculate the magnitude of the yacht's resultant velocity relative to the seabed. (b) Calculate the angle the resultant makes south of east. (c) Explain why stating only the speed "6.5 m/s" would be an incomplete description of the yacht's motion.
Challenge 3: Two children push a box. Child A pushes with 40 N due north. Child B pushes with 30 N due east. A frictional force of 10 N acts due south on the box. (a) Find the net northward force. (b) Calculate the resultant of all three forces, giving magnitude and direction.
Challenge 4 (Extended): A plane travels at 300 m/s due north. A crosswind blows at 40 m/s due west. The pilot wants to compensate by adjusting heading so that the resultant velocity is exactly due north. (a) Calculate what the actual resultant speed would be if no correction were made. (b) Describe qualitatively what heading adjustment the pilot would need to make to maintain a due-north course despite the westward wind. (c) Using Pythagoras, calculate the speed through the air the pilot would need to maintain if the resultant ground speed must remain 300 m/s due north despite the 40 m/s westward wind.