Vectors

This lesson covers Vectors as required by the A-Level Mathematics Pure 1 specification. Vectors are quantities that have both magnitude (size) and direction, distinguishing them from scalars which have magnitude only. Vectors are used extensively in both pure mathematics and mechanics.

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Scalars and Vectors

Type	Definition	Examples
Scalar	A quantity with magnitude only	Speed, mass, temperature, distance
Vector	A quantity with magnitude and direction	Velocity, force, displacement, acceleration

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Vector Notation

Vectors can be written in several ways:

• As a column vector: components arranged vertically.
• As i-j notation: ai + bj where i is the unit vector in the x-direction and j is the unit vector in the y-direction.
• As a position vector: the vector from the origin O to a point A, written as OA.
• A bold letter: a, or an underlined letter: a̲.

Example: The vector from (0, 0) to (3, 4) can be written as 3i + 4j.

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Vector Operations

Addition and Subtraction

Vectors are added (or subtracted) component-wise:

(a₁, b₁) + (a₂, b₂) = (a₁ + a₂, b₁ + b₂)
(a₁, b₁) − (a₂, b₂) = (a₁ − a₂, b₁ − b₂)

Example: If a = 2i + 3j and b = −i + 5j, then:

a + b = (2 + (−1))i + (3 + 5)j = i + 8j
a − b = (2 − (−1))i + (3 − 5)j = 3i − 2j

Scalar Multiplication

Multiplying a vector by a scalar k multiplies each component:

k × (a, b) = (ka, kb)

Example: If v = 3i − 2j, then 2v = 6i − 4j.

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Magnitude of a Vector

The magnitude (or modulus) of vector v = ai + bj is:

|v| = √(a² + b²)

Example: Find the magnitude of v = (−5, 12).

|v| = √(25 + 144) = √169 = 13

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Unit Vectors

A unit vector has magnitude 1. To find the unit vector in the direction of v:

v̂ = v / |v|

Example: Find the unit vector in the direction of (3, 4).

|v| = √(9 + 16) = 5
v̂ = (3/5, 4/5) = (0.6, 0.8)

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Position Vectors and Displacement

The position vector of a point P is the vector from the origin O to P, written OP or p.

The vector from point A (with position vector a) to point B (with position vector b) is:

AB = b − a

Example: If A = (1, 3) and B = (4, 7), find the vector AB.

AB = b − a = (4 − 1)i + (7 − 3)j = 3i + 4j

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Parallel Vectors

Two vectors are parallel if one is a scalar multiple of the other:

a is parallel to b  ⟺  a = kb for some scalar k

Example: (4, −6) is parallel to (−2, 3) because (4, −6) = −2(−2, 3).

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Collinear Points

Three points A, B, C are collinear (lie on the same straight line) if AB = k × AC for some scalar k.

Example: A = (1, 2), B = (4, 8), C = (7, 14). Show that A, B, C are collinear.

AB = (3, 6), AC = (6, 12)
AC = 2 × AB

Since AC is a scalar multiple of AB, the points are collinear.

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Geometric Proofs Using Vectors

Example: In triangle OAB, OA = a and OB = b. M is the midpoint of AB. Show that OM = ½(a + b).

AB = b − a
AM = ½AB = ½(b − a)
OM = OA + AM = a + ½(b − a) = a + ½b − ½a = ½a + ½b = ½(a + b) ∎

Example: In triangle OAB, P divides OA in ratio 2:1 and Q divides OB in ratio 2:1. Show that PQ is parallel to AB.

OP = (2/3)a, OQ = (2/3)b
PQ = OQ − OP = (2/3)b − (2/3)a = (2/3)(b − a) = (2/3)AB

Since PQ is a scalar multiple of AB, PQ is parallel to AB. ∎

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Distance Between Two Points

The distance between points with position vectors a and b is |b − a|.

Example: Find the distance between P(−3, 4) and Q(5, −2).

PQ = (8, −6)
|PQ| = √(64 + 36) = √100 = 10

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Summary

• Vectors have both magnitude and direction; scalars have magnitude only.
• Vectors can be written as column vectors or in i-j notation.
• Add/subtract vectors component-wise; scalar multiplication multiplies each component.
• The magnitude of ai + bj is √(a² + b²).
• A unit vector has magnitude 1: divide by the magnitude.
• The vector from A to B is AB = b − a (position vectors).
• Parallel vectors are scalar multiples of each other.
• Use vectors to prove collinearity and other geometric properties.

Exam Tip: When working with vectors, always clearly define your notation. State position vectors and show your working for each component. When proving geometric properties, clearly state what you are showing (e.g., "Since AB = 2BC, the points A, B, C are collinear"). Remember that AB = b − a (destination minus start). The magnitude of a vector is always positive.

A-Level Deep Dive: Vectors

Spec mapping

AQA 7357 specification, Pure Mathematics section I — Vectors. The single-year (AS-level) component covers 2D vectors only, while the full A-Level extends to 3D and to applications in geometry. Vectors are a synoptic backbone of the A-Level: they reappear in Pure 2 (scalar product, vector equations of straight lines), Mechanics (force diagrams, resolution into components, equilibrium), and physics teaching of motion and fields. The AQA formula booklet provides the magnitude formulae for 2D and 3D but does not list the position-vector ratio formula or the parallelogram/triangle laws — these must be memorised. AQA examiners routinely fuse vectors with coordinate geometry, requiring candidates to translate between vector arrows and Cartesian coordinates fluently.

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Worked example with full mark scheme

Question (8 marks): $ABCD$ABCD is a parallelogram with $\vec{AB} = \mathbf{p}$AB=p and $\vec{AD} = \mathbf{q}$AD=q. The diagonals $AC$AC and $BD$BD meet at $M$M. Prove that $M$M is the midpoint of both diagonals (i.e. the diagonals bisect each other).

Solution with mark scheme:

Step 1 — express the diagonals in terms of $\mathbf{p}$p and $\mathbf{q}$q.

Since $ABCD$ABCD is a parallelogram, $\vec{BC} = \vec{AD} = \mathbf{q}$BC=AD=q, so $\vec{AC} = \vec{AB} + \vec{BC} = \mathbf{p} + \mathbf{q}$AC=AB+BC=p+q. Also $\vec{BD} = \vec{AD} - \vec{AB} = \mathbf{q} - \mathbf{p}$BD=AD−AB=q−p.

M1 — using the triangle/parallelogram law to write the diagonals as combinations of $\mathbf{p}$p and $\mathbf{q}$q. A1 — both diagonal expressions correct.

Step 2 — parametrise points on each diagonal.

Let $M$M divide $AC$AC in ratio $\lambda : (1-\lambda)$λ:(1−λ), so $\vec{AM} = \lambda(\mathbf{p} + \mathbf{q})$AM=λ(p+q). Let $M$M also divide $BD$BD in ratio $\mu : (1-\mu)$μ:(1−μ) from $B$B, so $\vec{AM} = \vec{AB} + \mu \vec{BD} = \mathbf{p} + \mu(\mathbf{q} - \mathbf{p}) = (1 - \mu)\mathbf{p} + \mu \mathbf{q}$AM=AB+μBD=p+μ(q−p)=(1−μ)p+μq.

M1 — introducing two parameters and writing both expressions for $\vec{AM}$AM.

Step 3 — equate coefficients (since $\mathbf{p}$p and $\mathbf{q}$q are non-parallel, hence linearly independent).

$\lambda \mathbf{p} + \lambda \mathbf{q} = (1 - \mu)\mathbf{p} + \mu \mathbf{q}$λp+λq=(1−μ)p+μq

Matching $\mathbf{p}$p and $\mathbf{q}$q coefficients:

$\lambda = 1 - \mu, \qquad \lambda = \mu$λ=1−μ,λ=μ

M1 — equating coefficients with explicit appeal to the linear independence of $\mathbf{p}$p and $\mathbf{q}$q. A1 — both equations correct.

Step 4 — solve and conclude.

Substituting $\lambda = \mu$λ=μ into $\lambda = 1 - \mu$λ=1−μ gives $2\mu = 1$2μ=1, so $\mu = \lambda = \tfrac{1}{2}$μ=λ=21​.

A1 — correct values.

E1 — concluding statement: since $\lambda = \mu = \tfrac{1}{2}$λ=μ=21​, $M$M is the midpoint of both $AC$AC and $BD$BD, so the diagonals bisect each other, as required.

Total: 8 marks (M3 A3 E1 plus implicit setup). The non-parallelism justification is the AO2-flavoured step that examiners actively look for.

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): Points $A$A, $B$B, $C$C have position vectors $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$a=2i−j+3k, $\mathbf{b} = 4\mathbf{i} + \mathbf{j} - \mathbf{k}$b=4i+j−k and $\mathbf{c} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}$c=−i+5j+2k.

(a) Find $|\vec{AB}|$∣AB∣. (2)

(b) Find a unit vector parallel to $\vec{AC}$AC. (2)

(c) Determine whether $\vec{AB}$AB and $\vec{AC}$AC are parallel, justifying your answer. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — $\vec{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$AB=b−a=2i+2j−4k, then $|\vec{AB}| = \sqrt{4 + 4 + 16}$∣AB∣=4+4+16​.
• A1 (AO1.1b) — $|\vec{AB}| = \sqrt{24} = 2\sqrt{6}$∣AB∣=24​=26​.

(b)

• M1 (AO1.1b) — $\vec{AC} = -3\mathbf{i} + 6\mathbf{j} - \mathbf{k}$AC=−3i+6j−k with magnitude $\sqrt{9 + 36 + 1} = \sqrt{46}$9+36+1​=46​.
• A1 (AO1.1b) — unit vector $\dfrac{1}{\sqrt{46}}(-3\mathbf{i} + 6\mathbf{j} - \mathbf{k})$46​1​(−3i+6j−k).

(c)

• M1 (AO2.4) — testing whether $\vec{AC} = k \vec{AB}$AC=kAB for some scalar $k$k: comparing first components, $-3 = 2k$−3=2k gives $k = -3/2$k=−3/2; checking the $\mathbf{j}$j component, $2k = -3 \neq 6$2k=−3=6.
• A1 (AO2.4) — concluding the vectors are not parallel because the ratio of components is not constant.

Total: 6 marks split AO1 = 4, AO2 = 2. The justification in (c) — explicitly identifying inconsistency between component ratios — is the AO2 reasoning step. A bare "they are not parallel" with no test scores zero on (c).

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Synoptic links

1. Pure 2 — scalar product: the dot product $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta$a⋅b=a1​b1​+a2​b2​+a3​b3​=∣a∣∣b∣cosθ extends magnitude and direction into a single algebraic operation. Perpendicularity ($\mathbf{a} \cdot \mathbf{b} = 0$a⋅b=0) and projection both rest on the foundations laid in Pure 1.
2. Pure 2 — vector equations of lines: the line through $\mathbf{a}$a parallel to $\mathbf{d}$d is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$r=a+λd. Recognising this as a position vector plus a scalar multiple of a direction vector is exactly the mental model built here.
3. Mechanics — resolving forces: writing a force as $\mathbf{F} = F\cos\theta \, \mathbf{i} + F\sin\theta \, \mathbf{j}$F=Fcosθi+Fsinθj is component decomposition, and adding forces in equilibrium ($\sum \mathbf{F} = \mathbf{0}$∑F=0) is vector addition.
4. Coordinate geometry — distance formula: $|\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$∣AB∣=(x2​−x1​)2+(y2​−y1​)2​ is just $|\mathbf{b} - \mathbf{a}|$∣b−a∣ written in coordinates. The vector formulation generalises cleanly to 3D where Pythagoras would otherwise be reapplied awkwardly.
5. Physics — motion and fields: velocity, acceleration, electric and magnetic fields are all vector quantities. Position vectors trace trajectories; relative velocity is vector subtraction. The conceptual transfer between mathematics and physics happens through this section.

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Mark-scheme literacy

Vectors questions on AQA 7357 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Computing magnitudes, finding unit vectors, adding/subtracting position vectors, expressing displacements in component form
AO2 (reasoning / interpretation)	25–35%	Justifying parallelism via scalar multiples, appealing to linear independence when equating coefficients, interpreting geometric statements vectorially
AO3 (problem-solving)	5–15%	Multi-step geometry proofs (parallelogram diagonals, ratio-of-division problems), open-ended "show that …" questions

Examiner-rewarded phrasing: "since $\mathbf{p}$p and $\mathbf{q}$q are non-parallel, we may equate coefficients"; "the unit vector parallel to $\mathbf{v}$v is $\mathbf{v}/|\mathbf{v}|$v/∣v∣"; "the vectors are parallel iff one is a non-zero scalar multiple of the other". Phrases that lose marks: claiming two vectors are parallel without identifying the scalar; writing a magnitude as $\sqrt{24}$24​ when $2\sqrt{6}$26​ is the simplified form; confusing position vectors (from origin) with displacement vectors (between two points).