Vector Addition & Components
Key idea: Add and subtract coplanar vectors and resolve vectors into perpendicular components using sine/cosine (A Level Physics).
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The core idea
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Learning objectives
- Resolve, add and subtract coplanar vectors.
1. Definitions (Must Know)
- Scalar: a quantity with magnitude only.
- Vector: a quantity with magnitude and direction.
- Magnitude: the size of a quantity (always non-negative).
- Resultant: the single vector that has the same effect as adding multiple vectors.
- Component: the part of a vector in a chosen direction (often horizontal/vertical).
- Coplanar vectors: vectors that lie in the same plane.
2. Key Ideas (What Earns Marks)
- In A Level, you must be able to:
- add and subtract coplanar vectors, and
- represent a vector as two perpendicular components.
- Addition: place vectors head-to-tail (or use the parallelogram method). The resultant goes from the start of the first to the end of the last.
- Subtraction: vector A- vector B = vector A + (- vector B) (reverse vector B).
- Resolving: if a vector vector A makes angle θ to the +x axis (measured anticlockwise),
- Aₓ = A cos θ
- A_y = A sin θ
- Resultant from components:
- Rₓ = ∑ Aₓ, R_y = ∑ A_y
- R = square root of (Rₓ² + R_y²)
- Direction: use atan2(R_y,Rₓ) when available. If you use tan⁻¹ (|R_y/Rₓ|), identify the quadrant from the component signs before stating the final direction.
Choose axes first (e.g. +x right, +y up). Left/down components are negative; the vector magnitude is never negative.
3. Detailed Explanations
A. Scalars vs vectors (don’t mix them)
| Scalar | Vector | |
|---|---|---|
| Has magnitude? | Yes | Yes |
| Has direction? | No | Yes |
| Examples | mass, time, energy | displacement, velocity, force |
You can only add/subtract vectors with vectors and scalars with scalars. If direction matters, treat it as a vector problem.
B. Graphical vector addition (quick)
Use this when the question explicitly asks for a graphical method:
- Choose a scale.
- Draw vectors head-to-tail.
- Draw the resultant.
For subtraction, reverse the vector you are subtracting, then add head-to-tail.
C. Component method (most reliable)
Use this when angles are given or when accuracy matters:
- Choose axes and define +x and +y.
- Resolve each vector into x and y components with correct signs.
- Add components to get Rₓ and R_y.
- Compute magnitude and direction of the resultant.
Sanity checks:
- The resultant magnitude must satisfy |A-B| ≤ R ≤ A + B for two vectors of magnitudes A and B.
- The magnitude of a component cannot exceed the original vector magnitude.
- The signs of Rₓ and R_y must agree with the quadrant you state.
Visual example: resolving into components
If a force has magnitude F at angle θ above +x, the point (Fₓ, F_y) is where the vector ends.
Mini-example (signs):
- If a force points left, its x-component is negative.
- If a velocity points down, its y-component is negative.
D. Finding the direction (quadrant check)
The value of tan⁻¹ (R_y/Rₓ) alone is not enough to guarantee the correct direction.
Workflow:
- Use the signs of Rₓ and R_y to decide the quadrant.
- Compute an acute angle (e.g. tan⁻¹ (|R_y/Rₓ|)).
- Adjust the angle to the correct quadrant.
4. Common Mistakes
- Swapping sin and cos because the angle is measured from the wrong axis.
- Forgetting negative signs for components (left/down).
- Using tan⁻¹ (R_y/Rₓ) but giving the wrong quadrant.
- Subtracting by “minus the magnitude” instead of reversing the vector direction.
5. Exam Tips
- Start with a labelled sketch (axes + angle definitions).
- Keep components in the same unit as the original quantity (e.g. N, m/s).
- If your resultant direction looks wrong, check the signs of Rₓ and R_y first.
- If the angle is measured from the +y axis, swap the trig:
- A_y = A cos θ, Aₓ = A sin θ (then apply signs).
- Next: use component resolution in Projectile Motion and uncertainty-aware results in Uncertainty & Error Propagation.
6. Worked Examples
Modelled example 1
Add two perpendicular vectors
Problem
Study the worked solution
Assign components
Method
Use Rₓ = 3 and R_y = 4.Reason
East and north are perpendicular positive axes.Working
vector R = (3,4) m.Find magnitude
Method
Obtain 5 m.Reason
Perpendicular components form a right triangle.Working
R = square root of (3² + 4²) = 5 mFind direction
Method
Obtain 53° north of east.Reason
The angle is measured from the east component.Working
θ = tan⁻¹ (4/3) = 53°
Guided practice 2
Resolve a force into components
Problem
Try this before viewing the solution
Hints
Hint 1: angle is measured from horizontal
View solution step by step
Resolve horizontally
Method
Use Fₓ = F cos θ.Reason
The x-component is adjacent to the stated angle.Working
Fₓ = 10 cos 30° = 8.7 NResolve vertically
Method
Use F_y = F sin θ.Reason
The y-component is opposite the angle.Working
F_y = 10 sin 30° = 5.0 N
Common misconception 3
Add two coplanar vectors using components
Learner claim
Try this before viewing the solution
View solution step by step
Resolve with signs
Method
Use B_y = -8 sin 20°.Reason
Vector B lies below the x-axis.Working
Rₓ = 12 cos 40° + 8 cos 20° = 16.7 N; R_y = 12 sin 40°-8 sin 20° = 4.98 N.Recombine
Method
Obtain 17.4 N at 16.6° above +x.Reason
Both resultant components are positive.Working
R = square root of (16.7² + 4.98²) = 17.4 N, θ = tan⁻¹ (4.98/16.7) = 16.6°
Examiner practice 4
Subtract vectors using components
Examination question
Try this before viewing the solution
View solution step by step
Reverse vector B
1 markMethod
Write vector A- vector B = vector A + (- vector B).Reason
Vector subtraction means adding the opposite vector.Working
- vector B points south.State components
1 markMethod
Use (Rₓ,R_y) = (10,-6).Reason
Take east and north as positive axes.Working
vector R = (10,-6) m s⁻¹.Find magnitude
1 markMethod
Obtain 11.7 m s⁻¹.Reason
Components are perpendicular.Working
R = square root of (10² + (-6)²) = 11.7 m s⁻¹Find direction
1 markMethod
State 31° south of east.Reason
The x-component is positive and y-component negative.Working
θ = tan⁻¹ (6/10) = 31° south of east.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark reversal, components, magnitude and direction.
Challenge 5
Relative velocity (aircraft + wind)
Independent transfer
Try this before viewing the solution
Hints
Hint 1: state the velocity relationship
View solution step by step
Add components
Method
Use (Rₓ,R_y) = (50,200) with east and north positive.Reason
Wind and airspeed are perpendicular velocity vectors.Working
vector v_ground = (50,200) m s⁻¹.Find ground speed
Method
Obtain 206 m s⁻¹.Reason
Speed is the resultant magnitude.Working
R = square root of (50² + 200²) = 206 m s⁻¹Find direction
Method
State 14° east of north.Reason
The reference direction is north, so use east component over north component.Working
θ = tan⁻¹ (50/200) = 14°
7. Mind Stretchers
Mind stretcher 1: Find the angle for a horizontal resultantExtension
Two forces act on a point:
- F₁ = 20 N at 30° above the +x axis
- F₂ = 15 N at angle θ below the +x axis
Find θ such that the resultant is horizontal.
Show Answer
For the resultant to be horizontal, vertical components must cancel:
20 sin 30° = 15 sin θ 20 × 0.5 = 15 sin θ sin θ = 10/15 = 2/3 θ ≈ 41.8°
Mind stretcher 2: Find the angle between two equal vectors (cosine rule)Extension
Two forces, each of magnitude 12 N, act with an angle θ between them. The resultant force has magnitude 20 N. Find θ.
Show Answer
Use the cosine rule for the magnitude of the sum: R² = A² + B² + 2AB cos θ
With A = B = 12 and R = 20: 20² = 12² + 12² + 2(12)(12) cos θ 400 = 288 + 288 cos θ cos θ = (400-288)/288 = 112/288 ≈ 0.389
θ ≈ cos⁻¹ (0.389) ≈ 67°
Mind stretcher 3: Optional (Enrichment)Extension
A. Equilibrium with two tensions (preview of Forces)
A weight of 10 N is supported symmetrically by two strings, each at 45° above the horizontal. Find the tension in each string.
Show Answer
By symmetry, the tensions are equal, T.
Horizontal components cancel.
Vertical components add to balance the weight:
2T sin 45° = 10 T = 10/(2 sin 45°) ≈ 7.07 N
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Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027