OCR A Level Mathematics A: Vectors in two and three dimensions (H240) – Revision Notes

These notes condense section 1.10 Vectors (1.10a to 1.10h) of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026). For the reasoning and longer worked examples, go to the study guide. Test yourself afterwards with the practice questions.

Two-dimensional work is Stage 1 (shared with AS). Three-dimensional vectors (1.10b) and kinematics with vectors (1.10h) are A Level only. The specification excludes the direction of a vector in three dimensions. Course hub: /boards/ocr/a-level/mathematics/. Printable checklist: /checklists/ocr/a-level/mathematics/. The free 10-minute diagnostics will show which parts need most work.

Key definitions

Term	Meaning
Scalar	Size only: mass, time, speed, distance
Vector	Size and direction: displacement, velocity, acceleration, force
Magnitude (modulus)	The length of a vector, written |a| or |→AB|
Unit vector	A vector of magnitude 1; i, j, k are along the axes
Position vector	→OA: the displacement from the origin O to A
Displacement vector	The vector from one point to another, →AB
Resultant vector	The sum of two or more vectors
Component vector	The part of a vector along one axis
Parallel vectors	b = ka for some non-zero scalar k
Equal vectors	Same magnitude and same direction

Formulas

Result	Formula
Length of a 2D vector	|xi + yj| = √(x² + y²)
Length of a 3D vector (A Level only)	|xi + yj + zk| = √(x² + y² + z²)
Direction, 2D	Angle θ from the positive x-axis, 0° ≤ θ < 360°, found from tan⁻¹(y/x) and the quadrant
Components from magnitude r and direction θ	x = r cos θ, y = r sin θ
Unit vector	â = a/|a|
Displacement	→AB = b − a
Distance	AB = |b − a| = √((c − a)² + (d − b)²) in 2D
Point dividing AB in ratio m : n	p = a + m(b − a)/(m + n)
Midpoint of AB	(a + b)/2
Resultant force	F = F₁ + F₂ + … ; equilibrium when F = 0
Newton’s second law, vector form	F = ma
Constant acceleration, vectors (A Level only)	s = ut + ½at² and v = u + at; also s = vt − ½at² and s = ½(u + v)t

Method boxes

Magnitude and direction of xi + yj

1. Sketch the vector and name the quadrant.
2. Magnitude = √(x² + y²).
3. Acute angle α = tan⁻¹(|y|/|x|).
4. First quadrant: θ = α. Second: θ = 180° − α. Third: θ = 180° + α. Fourth: θ = 360° − α.

Components from magnitude and direction

1. Take x = r cos θ and y = r sin θ with the full angle θ.
2. The signs then come out correctly in every quadrant.

Parallel vectors

1. Write one vector as a multiple of the other: b = ka.
2. Find k from the component you know, then use it on the others.

Proving collinearity

1. Express two vectors with a common point, such as →AB and →AC, in terms of the same base vectors.
2. Show one is a scalar multiple of the other.
3. Write the conclusion in words: parallel and sharing a point, so collinear.

Position and ratio problems

1. Choose two non-parallel base vectors, often a and b from the origin.
2. Reach every point by a route of known vectors, using →AB = b − a for any step you have to reverse.
3. Equate coefficients of a and b when two routes reach the same point.

Forces

1. Add the forces component by component.
2. Equilibrium: the sum is 0. Unknown force = minus the sum of the known ones.
3. Find magnitude and direction last, from the final components.

Kinematics (A Level only)

1. Write the known quantities as vectors.
2. Use v = u + at for velocity and r = r₀ + ut + ½at² for position.
3. “Moving parallel to i” means the j-component of v is 0. “Due east of the origin” means the j-component of r is 0 (with i east and j north).

Small worked reminders

Direction. For −5i + 2j: the second quadrant, α = tan⁻¹(2/5) = 21.8°, so θ = 180° − 21.8° = 158.2°.

Ratio point. A = (1, 2), B = (7, 8). The point P with AP : PB = 1 : 2 has p = a + (b − a)/3 = (i + 2j) + (2i + 2j) = 3i + 4j.

Equal ratios. (k, 12) is parallel to (5, −3) when k/5 = 12/(−3), so k = −20.

Must-know distinctions

• Speed or velocity. Speed is |v|, a scalar. Velocity is a vector.
• Position or displacement. Position is measured from O. Displacement is a change of position, end minus start.
• Parallel or equal. Parallel vectors can have different lengths and can point in opposite directions. Equal vectors cannot.
• Magnitude or direction. A 2D direction is an angle from the positive x-axis. In 3D you give components or a unit vector, never a direction angle.
• Vector or its magnitude. The vector 3i − 4j is not equal to 5. Its magnitude is 5, and the two are never interchangeable in an answer.
• Component form or magnitude/direction form. Use components to add vectors. Use magnitude and direction to describe the result.
• Sum of magnitudes or magnitude of a sum. |a + b| = |a| + |b| only when the vectors point the same way.

More worked reminders

Three dimensions (A Level only). P is (3, −6, 2) and Q is (−1, 0, 5). Then →PQ = −4i + 6j + 3k and PQ = √(16 + 36 + 9) = √61 ≈ 7.81. The unit vector from P towards Q is (−4i + 6j + 3k)/√61.

Resultant force with direction. Forces (7i − 3j) N and (−2i + 8j) N have resultant 5i + 5j, with magnitude √50 = 5√2 ≈ 7.07 N. It lies in the first quadrant at tan⁻¹(5/5) = 45° to the positive x-axis.

Equating coefficients. If a and b are not parallel and (p + 2)a + 3b = 5a + (q − 1)b, then p + 2 = 5 and 3 = q − 1, so p = 3 and q = 4.

Isosceles triangle. A, B, C have position vectors 2i, 5i + 4j and 2i + 8j. Then →AB = 3i + 4j and →BC = −3i + 4j, so AB = BC = 5 and the triangle is isosceles.

Kinematics (A Level only). A particle starts at 2i m with velocity (4i + j) m s⁻¹ and acceleration −2i m s⁻². At t = 3, r = 2i + 3(4i + j) + 4.5(−2i) = 5i + 3j.

Checklist before you leave a question

• Have you drawn a sketch, even a rough one, for any direction or geometry question?
• Is every vector in bold, underlined or written with an arrow, so that scalars and vectors stay separate?
• Does your final answer say what was asked: a vector, a magnitude, an angle, or a distance?
• Is an angle in the right interval and the right quadrant?
• Have you stated the units, and kept exact values until the end?

Quick self-test

1. Find |−3i + 7j| to 3 s.f.
2. Find the direction of −3i + 7j, in [0°, 360°), to 1 d.p.
3. (A Level only) Find the unit vector in the direction of 12i − 4j + 3k.
4. (A Level only) If a = (3, −1, 4) and b = (−2, 5, 0) as column vectors, find 2a − 3b.
5. For what value of k is (k, 9) parallel to (4, −6)?
6. A is (1, 2) and B is (7, −6). Find →AB and the distance AB.
7. Write the vector of magnitude 12 and direction 300° in exact component form.
8. Forces (2i + 5j) N, (ai + bj) N and (−7i + j) N keep a particle in equilibrium. Find a and b.
9. (A Level only) Find the distance from the origin to P(3, −6, 2).
10. A particle has velocity (2i + 3j) m s⁻¹ and constant acceleration (i − j) m s⁻². Find its velocity after 4 s.

Answers

1. √58 ≈ 7.62
2. Second quadrant: 180° − tan⁻¹(7/3) = 180° − 66.8° = 113.2°
3. |v| = √(144 + 16 + 9) = 13, so (12i − 4j + 3k)/13
4. (6, −2, 8) − (−6, 15, 0) = (12, −17, 8)
5. k/4 = 9/(−6), so k = −6
6. →AB = 6i − 8j; AB = √(36 + 64) = 10
7. x = 12 cos 300° = 6, y = 12 sin 300° = −6√3, so 6i − 6√3 j
8. Sum of i parts: 2 + a − 7 = 0, so a = 5; sum of j parts: 5 + b + 1 = 0, so b = −6
9. √(9 + 36 + 4) = √49 = 7
10. v = (2i + 3j) + 4(i − j) = 6i − j m s⁻¹

Where marks are usually lost

• Reporting the calculator’s acute angle instead of the direction in [0°, 360°). Sketch first.
• Subtracting the wrong way round: →AB is b − a, and →BA is a − b.
• Dropping a negative sign in a squared term, for example (−4)² written as −16.
• Rounding too early. Keep exact surds (√68 = 2√17) until the last line, then round to 3 s.f. if the question asks.
• Finishing a collinearity or parallelogram proof with no conclusion in words.
• Reading “show that” as “state”. Show every line of the working that reaches the given result.
• Forgetting the direction of a force or velocity when the question asks for “magnitude and direction”.
• Treating speed as a vector, or finding velocity when the question asks for speed.
• Using r = ut + ½at² and forgetting to add the initial position r₀.
• Giving units late or not at all: newtons for force, m s⁻¹ for velocity, m s⁻² for acceleration.

Official syllabus

OCR, A Level in Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018, section 1.10 (Vectors), with the constant acceleration formulae from section 3.02e.