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

This guide teaches section 1.10 Vectors (1.10a to 1.10h) of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. The specification places vectors in its Pure Mathematics content, which Papers 1, 2 and 3 all assess. Two-dimensional vectors are Stage 1 content, shared with AS. Three-dimensional vectors (1.10b) and kinematics with vectors (1.10h) are A Level only.

Use it with the vectors revision notes and the vectors practice questions. The course hub is at /boards/ocr/a-level/mathematics/, the printable checklist is at /checklists/ocr/a-level/mathematics/, and the free 10-minute diagnostics show you where to start. For the wider course see the pure mathematics overview.

What section 1.10 covers

Ref	You must be able to	Stage
1.10a	Use vectors in two dimensions, as xi + yj or as a column vector	Stage 1
1.10b	Use vectors in three dimensions, as xi + yj + zk or a column vector	A Level only
1.10c	Find magnitude and direction; convert between component and magnitude/direction form	Stage 1
1.10d	Add vectors diagrammatically and algebraically; multiply by scalars; know the geometry	Stage 1
1.10e	Understand and use position vectors	Stage 1
1.10f	Calculate the distance between two points from their position vectors	Stage 1
1.10g	Solve problems in pure mathematics and in context, including forces	Stage 1
1.10h	Solve problems in kinematics	A Level only

The specification extends 1.10c to 1.10g to three dimensions but excludes the direction of a vector in three dimensions. That is outside the specification, so a 3D vector is described by components or by a unit vector.

1.10a and 1.10b: notation

A scalar has size only (mass, time, speed, distance). A vector has size and direction (displacement, velocity, acceleration, force). Write vectors in bold in print and underlined by hand, a̲, so that you never confuse them with scalars. A vector from A to B is written →AB or a.

The same vector can be written in either form:

3i − 2j + 5k   =   ( 3)
                   (−2)
                   ( 5)

Here i, j and k each have length 1 and point along the x-, y- and z-axes. Equal vectors have equal components, so a vector does not depend on where it is drawn.

1.10c: magnitude, direction and unit vectors

The magnitude (modulus) of a vector is its length. Write |a| for the magnitude of a and |→AB| for the magnitude of →AB.

|xi + yj|      = √(x² + y²)
|xi + yj + zk| = √(x² + y² + z²)

A unit vector has magnitude 1. The unit vector in the direction of a is â = a/|a|.

In two dimensions the direction is the angle the vector makes with a line parallel to the positive x-axis, measured anticlockwise, in the interval [0°, 360°). Sketch the vector first, then use tan⁻¹ and adjust for the quadrant.

Example 1. Find the magnitude and direction of a = 4i − 9j.

|a| = √(16 + 81) = √97 ≈ 9.85
Sketch: x > 0, y < 0, so the fourth quadrant.
Acute angle = tan⁻¹(9/4) = 66.04...°
Direction = 360° − 66.04...° = 294.0° (1 d.p.)

Example 2. A vector has magnitude 26 and direction 215°. Write it in component form.

x = 26 cos 215° = −21.3
y = 26 sin 215° = −14.9
vector = −21.3i − 14.9j   (3 s.f.)

Both components are negative because 215° is in the third quadrant. The sign comes out correctly if you use the full angle in cos and sin.

Example 3 (A Level only). Find the unit vector in the direction of p = 4i + 4j − 7k.

|p| = √(16 + 16 + 49) = √81 = 9
p̂ = (4i + 4j − 7k)/9 = (4/9)i + (4/9)j − (7/9)k

A vector of magnitude 45 in the same direction is 45p̂ = 20i + 20j − 35k.

1.10d: adding vectors and scalar multiples

Algebraically, add or subtract matching components. Diagrammatically, place the tail of b at the head of a. The resultant a + b is the single arrow from the tail of a to the head of b, which gives a triangle of vectors. A closed loop of vectors has resultant 0.

Multiplying by a scalar k scales the length by |k|. If k is negative the direction reverses. So −a is a vector of the same length pointing the opposite way, and a − b means a + (−b).

Vectors a and b are parallel if b = ka for some non-zero scalar k. Equal vectors have the same magnitude and the same direction.

Example 4 (A Level only). The vector a = 3i − 2j + 5k is parallel to b = 6i + pj + qk. Find p and q.

6 = 2 × 3, so b = 2a
p = 2 × (−2) = −4
q = 2 × 5 = 10

Equating components. If a and b are not parallel, then ma + nb = ra + sb forces m = r and n = s. In component form, compare the i parts and the j parts to get two equations.

Example 5. Find m and n if m(3i − j) + n(i + 2j) = 10i + 6j.

i:  3m + n = 10
j: −m + 2n = 6
From the first, n = 10 − 3m.
−m + 20 − 6m = 6, so 7m = 14, m = 2, n = 4.

1.10e and 1.10f: position vectors and distance

The position vector of A is →OA, the journey from the origin O to A. If A is (2, −1, 3) then a = 2i − j + 3k.

A displacement vector between two points is “end minus start”:

→AB = →OB − →OA = b − a

The distance between A and B is |→AB|. In two dimensions, for the points ai + bj and ci + dj, it is √((c − a)² + (d − b)²). The straight-line version of this is in the straight lines and circles guide.

You will also meet the terms component vector (the part of a vector along one axis), resultant vector (the sum of vectors) and unit vector.

Example 6 (A Level only). A is (2, −1, 3) and B is (−2, 5, −1). Find →AB and the distance AB.

→AB = (−2 − 2)i + (5 − (−1))j + (−1 − 3)k = −4i + 6j − 4k
AB = √(16 + 36 + 16) = √68 = 2√17 ≈ 8.25

Dividing a line. When AP : PB = m : n, the point P has position vector p = a + m(b − a)/(m + n). Setting m = n gives the midpoint, (a + b)/2.

1.10g: problems in pure mathematics

Vectors prove geometric facts without coordinates. Pick two non-parallel vectors as a base, write every other vector in terms of them, and compare.

Example 7. ABCD is a trapezium with →AB = a, →AD = b and →DC = 2a. F is the point on AC with AF : FC = 1 : 2. Show that F lies on the diagonal BD and find BF : FD.

→AC = →AD + →DC = b + 2a
→AF = (1/3)→AC = (2/3)a + (1/3)b
→BD = →AD − →AB = b − a
→BF = →AF − →AB = (2/3)a + (1/3)b − a = (1/3)b − (1/3)a = (1/3)→BD

→BF is a multiple of →BD and both start at B, so B, F and D are collinear: F lies on BD. BF is one third of BD, so BF : FD = 1 : 2.

Finish with a sentence in words: the vectors are parallel and share a point, so the points are collinear.

1.10g: problems in context, including forces

A force is a vector, so forces acting at a point combine by vector addition to give a resultant. A particle is in equilibrium if that resultant is the zero vector. Newton’s second law, F = ma, also holds in vector form, so the resultant force and the acceleration point the same way.

Example 8. Forces F₁ = (6i + 2j) N and F₂ = (−9i + 7j) N act on a particle. A third force F₃ keeps it in equilibrium. Find F₃, its magnitude and its direction.

F₁ + F₂ = −3i + 9j
F₃ = −(F₁ + F₂) = 3i − 9j
|F₃| = √(9 + 81) = √90 = 3√10 ≈ 9.49 N
Fourth quadrant: tan⁻¹(9/3) = 71.57...°
Direction = 360° − 71.57...° = 288.4° (1 d.p.)

If F₃ were removed, the resultant would be −3i + 9j, which has the same magnitude, 9.49 N, in the opposite direction, 108.4°. For a particle of mass 2 kg this gives acceleration (−1.5i + 4.5j) m s⁻².

1.10h: kinematics (A Level only)

The constant acceleration formulae in section 3.02e hold in vector form, with the same letters but vector quantities:

s = ut + ½at²      v = u + at
s = vt − ½at²      s = ½(u + v)t

Here s is the displacement from the start. The position at time t is r = r₀ + s, where r₀ is the initial position. The velocity vector gives the direction of motion and the acceleration vector gives the direction of the resultant force. Speed is |v|, a scalar.

Example 9. A particle has position 4i + 6j m at t = 0, velocity (3i − 2j) m s⁻¹ and constant acceleration (−i + 0.5j) m s⁻². Find its velocity and position when t = 6, then its position when it is moving parallel to i.

t = 6:  v = (3i − 2j) + 6(−i + 0.5j) = −3i + j
        speed = √10 ≈ 3.16 m s⁻¹
        r = (4i + 6j) + 6(3i − 2j) + 18(−i + 0.5j)
          = (4 + 18 − 18)i + (6 − 12 + 9)j = 4i + 3j
Parallel to i: the j-component of v is zero.
        −2 + 0.5t = 0, so t = 4
        r = (4 + 12 − 8)i + (6 − 8 + 4)j = 8i + 2j

Common errors

• Forgetting the order in →AB = b − a. It is end minus start.
• Writing the direction as the acute angle from a calculator. A vector such as −3i + 7j has direction 113.2°, not 66.8°.
• Using √(x + y) or leaving out a square, or adding magnitudes: |a + b| is not |a| + |b|.
• Dividing a vector by its largest component instead of its magnitude to make a unit vector.
• Mixing scalars and vectors, for example writing a force as “6” when its direction matters, or giving speed as a vector.
• Proving collinearity with only one vector. You need two vectors with a common point, or equal ratios of components.
• Giving a direction for a 3D vector. The specification excludes it.

Next steps

Work through the revision notes, then the practice questions. The wider pure mathematics practice and exam preparation page cover mixed questions.

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.