What are resolving a vector into components?

Any vector V at angle θ to a chosen axis splits into two perpendicular components:

Three coplanar forces act at a point: 5.0\ N east, 4.0\ N north, and 3.0\ N west. Find the resultant. [3 marks]

SingaporePhysicsSyllabus dot point

How do we distinguish scalars from vectors, and how do we add and resolve vectors to solve physical problems?

Distinguish scalar and vector quantities, add coplanar vectors, and resolve a vector into perpendicular components

A focused answer to the H2 Physics Measurement learning outcome on scalars and vectors. The distinction, vector addition by the parallelogram and triangle rules, and resolving a vector into perpendicular components.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to classify quantities as scalar or vector, to add coplanar vectors using the triangle or parallelogram rule (or by components), and to resolve a single vector into two perpendicular components. Resolution is the most-used skill in all of mechanics: projectile motion, inclined planes, equilibrium and circular motion all depend on it.

The answer

Scalars and vectors

A scalar has magnitude only. A vector has both magnitude and direction.

Scalars	Vectors
distance, speed	displacement, velocity
mass, time	acceleration, force
energy, temperature	momentum, field strength

The key consequence is that vectors must be combined using their directions, not by simple arithmetic addition.

Adding two vectors

Two vectors are added head-to-tail (the triangle rule) or as two sides of a parallelogram (the parallelogram rule). The resultant is the single vector from the start of the first to the end of the second.

For two perpendicular vectors $A$ and $B$ , the resultant has magnitude:

R = \sqrt{A^2 + B^2}

at angle $\theta$ above $A$ where $\tan\theta = \dfrac{B}{A}$ .

For two vectors at a general angle, either draw a scale diagram or use components.

Resolving a vector into components

Any vector $V$ at angle $\theta$ to a chosen axis splits into two perpendicular components:

V_x = V\cos\theta, \qquad V_y = V\sin\theta

The component along the direction you measure $\theta$ from carries the cosine; the perpendicular component carries the sine. Getting this association right is essential, because it depends entirely on where the angle is measured from.

Adding many vectors by components

For three or more coplanar vectors, resolve each into $x$ and $y$ components, sum each direction separately, then recombine:

R_x = \sum V_x, \qquad R_y = \sum V_y, \qquad R = \sqrt{R_x^2 + R_y^2}

This is the systematic method that never fails, even for awkward angles.

Worked example

A hiker walks $3.0\ \text{km}$ due east, then $4.0\ \text{km}$ in a direction $60^\circ$ north of east. Find the magnitude and direction of the resultant displacement.

Step 1: Resolve each leg into east (x) and north (y) components

Leg 1: $x_1 = 3.0\ \text{km}$ , $y_1 = 0$ .

Leg 2: $x_2 = 4.0\cos 60^\circ = 2.0\ \text{km}$ , $y_2 = 4.0\sin 60^\circ = 3.46\ \text{km}$ .

Step 2: Sum the components

$R_x = 3.0 + 2.0 = 5.0\ \text{km}$ .

$R_y = 0 + 3.46 = 3.46\ \text{km}$ .

Step 3: Recombine into a resultant

R = \sqrt{5.0^2 + 3.46^2} = \sqrt{25 + 12.0} = \sqrt{37.0} = 6.08\ \text{km}

Step 4: Find the direction

$\tan\phi = \dfrac{R_y}{R_x} = \dfrac{3.46}{5.0} = 0.692$ , so $\phi = 34.7^\circ$ north of east.

The resultant displacement is $6.1\ \text{km}$ at $35^\circ$ north of east.

Examples in context

Example 1. Projectile launch velocity. A ball is launched at $20\ \text{m s}^{-1}$ at $30^\circ$ above the horizontal. Its horizontal component is $20\cos 30^\circ = 17.3\ \text{m s}^{-1}$ and its vertical component is $20\sin 30^\circ = 10\ \text{m s}^{-1}$ . These two independent components are analysed separately in projectile problems, which is why resolution is the entry point to all projectile motion.

Example 2. Equilibrium of three forces. A street lamp hangs from two cables. Resolving the cable tensions into horizontal and vertical components and setting each sum to zero ( $\sum F_x = 0$ , $\sum F_y = 0$ ) lets you solve for both tensions. The component method turns a geometry problem into two simultaneous equations.

Try this

Q1. State the difference between a scalar and a vector, giving one example of each. [2 marks]

Cue. Scalar has magnitude only (e.g. mass); vector has magnitude and direction (e.g. force).

Q2. A force of $24\ \text{N}$ acts at $40^\circ$ to the horizontal. Find its horizontal and vertical components. [2 marks]

Cue. $F_x = 24\cos 40^\circ = 18.4\ \text{N}$ , $F_y = 24\sin 40^\circ = 15.4\ \text{N}$ .

Q3. Three coplanar forces act at a point: $5.0\ \text{N}$ east, $4.0\ \text{N}$ north, and $3.0\ \text{N}$ west. Find the resultant. [3 marks]

Cue. $R_x = 5.0 - 3.0 = 2.0\ \text{N}$ , $R_y = 4.0\ \text{N}$ ; $R = \sqrt{2.0^2 + 4.0^2} = 4.47\ \text{N}$ at $\tan^{-1}(4.0/2.0) = 63.4^\circ$ north of east.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA box rests on a slope inclined at

30^\circ

to the horizontal. Its weight is

50\ \text{N}

. Resolve the weight into components parallel and perpendicular to the slope, and state which component tends to make the box slide.

Show worked answer →

Resolve the weight relative to the slope surface, not the horizontal.

Component along the slope (down the incline): $W_\parallel = W \sin\theta = 50 \sin 30^\circ = 25\ \text{N}$ .

Component perpendicular to the slope (into the surface): $W_\perp = W \cos\theta = 50 \cos 30^\circ = 43.3\ \text{N}$ .

The parallel component, $25\ \text{N}$ acting down the slope, is the one that tends to make the box slide.

Markers reward correct identification of which trig function applies to which component (the $\sin\theta$ component is along the slope here because $\theta$ is measured from the horizontal), correct values, and a clear statement that the parallel component causes sliding.

Original3 marksTwo forces of

6.0\ \text{N}

and

8.0\ \text{N}

act at a single point at right angles to each other. Determine the magnitude and direction of the resultant force.

Show worked answer →

Because the forces are perpendicular, use Pythagoras and the tangent ratio.

Magnitude: $R = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \text{N}$ .

Direction: the angle $\theta$ from the $8.0\ \text{N}$ force satisfies $\tan\theta = \dfrac{6.0}{8.0} = 0.75$ , so $\theta = 36.9^\circ$ .

The resultant is $10\ \text{N}$ at $36.9^\circ$ to the $8.0\ \text{N}$ force.

Markers reward the Pythagorean magnitude, the correct trig ratio for the angle, and a fully specified direction (angle relative to a stated reference force).