Vector Addition & Components

Key idea: Add and subtract coplanar vectors and resolve vectors into perpendicular components using sine/cosine (A Level Physics).

  • GCE A-Level H2 Physics 2027
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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.
Sign convention (components)

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)

ScalarVector
Has magnitude?YesYes
Has direction?NoYes
Examplesmass, time, energydisplacement, 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:

  1. Choose a scale.
  2. Draw vectors head-to-tail.
  3. 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:

  1. Choose axes and define +x and +y.
  2. Resolve each vector into x and y components with correct signs.
  3. Add components to get Rₓ and R_y.
  4. 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.

Resolving a vector into perpendicular componentsA 10 newton force points 30 degrees above the positive x-axis. Its horizontal component is 8.66 newtons and its vertical component is 5.00 newtons. Dashed projection lines form a right triangle, and arrows show that the components add head-to-tail to recover the original force.Perpendicular components+x+y30°Fₓ = 8.66 NFᵧ = 5.00 NF = 10.0 Nresultant of Fₓ and Fᵧ
For a force at angle θ from +x, the signed components are Fx = F cos θ and Fy = F sin θ. Here they add head-to-tail to recover the 10 N force.

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:

  1. Use the signs of Rₓ and R_y to decide the quadrant.
  2. Compute an acute angle (e.g. tan⁻¹ (|R_y/Rₓ|)).
  3. 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

Core

Problem

Add 3 m east and 4 m north, giving resultant magnitude and direction.
Study the worked solution
  1. Assign components

    Method

    Use Rₓ = 3 and R_y = 4.

    Reason

    East and north are perpendicular positive axes.

    Working

    vector R = (3,4) m.
  2. Find magnitude

    Method

    Obtain 5 m.

    Reason

    Perpendicular components form a right triangle.

    Working

    R = square root of (3² + 4²) = 5 m
  3. Find 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

About 4 min

Problem

Resolve a 10 N force at 30° above the horizontal into components.

Try this before viewing the solution

Hints

Hint 1: angle is measured from horizontal
The adjacent component uses cosine; the opposite component uses sine.
View solution step by step
  1. 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 N
  2. Resolve 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

Find and correct the mistake

Learner claim

vector A = 12 N at 40° above +x and vector B = 8 N at 20° below +x. A learner makes both y-components positive. Diagnose the sign and find vector A + vector B.

Try this before viewing the solution

Sign of Bᵧ

View solution step by step
  1. 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.
  2. 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

4 marks

Examination question

vector A = 10 m s⁻¹ east and vector B = 6 m s⁻¹ north. Find vector A- vector B in magnitude-direction form. [4 marks]

Try this before viewing the solution

View solution step by step
  1. Reverse vector B

    1 mark

    Method

    Write vector A- vector B = vector A + (- vector B).

    Reason

    Vector subtraction means adding the opposite vector.

    Working

    - vector B points south.
  2. State components

    1 mark

    Method

    Use (Rₓ,R_y) = (10,-6).

    Reason

    Take east and north as positive axes.

    Working

    vector R = (10,-6) m s⁻¹.
  3. Find magnitude

    1 mark

    Method

    Obtain 11.7 m s⁻¹.

    Reason

    Components are perpendicular.

    Working

    R = square root of (10² + (-6)²) = 11.7 m s⁻¹
  4. Find direction

    1 mark

    Method

    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.

Challenge 5

Relative velocity (aircraft + wind)

Minimal support

Independent transfer

An aircraft flies through the air at 200 m s⁻¹ north while wind is 50 m s⁻¹ east. Find its ground velocity.

Try this before viewing the solution

Hints

Hint 1: state the velocity relationship
Ground velocity equals air-relative velocity plus wind velocity.
View solution step by step
  1. 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.
  2. 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⁻¹
  3. Find 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

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027