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How do you use vectors to represent displacement, velocity and force, and find magnitude and direction?

Use vectors in mechanics: add and subtract vectors, multiply by a scalar, use component (i and j) form, and find the magnitude and direction of a vector.

A CCEA GCSE Further Mathematics answer on vectors in mechanics, covering component form, vector addition and subtraction, scalar multiples, and finding the magnitude and direction of displacement, velocity and force vectors in the Mechanics unit.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Component form
  3. Adding, subtracting and scaling
  4. Magnitude and direction
  5. Why this matters

What this dot point is asking

A vector has both size and direction, and CCEA GCSE Further Mathematics uses vectors throughout the Mechanics unit to represent displacement, velocity, acceleration and force. You must add and subtract vectors, multiply them by scalars, work in component (i\mathbf{i} and j\mathbf{j}) form, and find the magnitude and direction of a vector. These skills let you combine forces and describe motion in two dimensions cleanly.

Component form

Vectors are most easily handled in component form, splitting each into a part along i\mathbf{i} (horizontal) and a part along j\mathbf{j} (vertical). A displacement of 5i+3j5\mathbf{i} + 3\mathbf{j} means 55 units across and 33 units up. Writing every vector this way turns vector arithmetic into ordinary arithmetic on the separate components.

Adding, subtracting and scaling

Vector addition and subtraction work component by component, keeping i\mathbf{i} parts with i\mathbf{i} parts and j\mathbf{j} parts with j\mathbf{j} parts. Scalar multiplication multiplies both components by the number.

So (2i+3j)+(4ij)=6i+2j(2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} - \mathbf{j}) = 6\mathbf{i} + 2\mathbf{j}, and 3(2ij)=6i3j3(2\mathbf{i} - \mathbf{j}) = 6\mathbf{i} - 3\mathbf{j}. Adding force vectors gives the resultant force, the single force equivalent to several acting together, which is central to the forces topic.

Magnitude and direction

The magnitude of a vector is its length, found by applying Pythagoras to its components. The direction is the angle the vector makes, usually measured from the positive horizontal axis.

Why this matters

Vectors are the language of two-dimensional mechanics. Resolving and combining forces relies on component form, projectile motion splits velocity into horizontal and vertical components, and momentum is a vector too. The magnitude formula reuses Pythagoras from coordinate geometry, and the direction calculation reuses the trigonometry from the Pure unit, so this topic ties together skills you already have and applies them to motion and force.

Exam-style practice questions

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

CCEA Unit 2 (style)3 marksA particle has velocity v=(6i8j)\mathbf{v} = (6\mathbf{i} - 8\mathbf{j}) m/s. Find its speed.
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Speed is the magnitude of the velocity vector, found by Pythagoras on the components.

v=62+(8)2=36+64=100|\mathbf{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100}.

So the speed is 1010 m/s.

Marks are for using Pythagoras on the components and for the value. Squaring removes the sign, so the negative j\mathbf{j} component still contributes 6464.

CCEA Unit 2 (style)4 marksForces F1=(3i+2j)\mathbf{F_1} = (3\mathbf{i} + 2\mathbf{j}) N and F2=(i5j)\mathbf{F_2} = (\mathbf{i} - 5\mathbf{j}) N act on a particle. Find the resultant force and its magnitude.
Show worked answer →

Add the vectors component by component: R=(3+1)i+(25)j=4i3j\mathbf{R} = (3 + 1)\mathbf{i} + (2 - 5)\mathbf{j} = 4\mathbf{i} - 3\mathbf{j} N.

Magnitude: R=42+(3)2=16+9=25=5|\mathbf{R}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 N.

Marks are for the resultant in component form and for its magnitude. Adding the i\mathbf{i} parts to the j\mathbf{j} parts is the usual error; like components add to like.

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