Vector Geometry

This lesson covers geometric proofs using vectors — proving collinearity, parallelism, and finding intersection points — as required by the Edexcel A-Level Mathematics specification (9MA0). Vector geometry questions are among the most challenging and rewarding at A-Level.

Proving Collinearity

Three points A, B, C are collinear if they lie on the same straight line. To prove collinearity using vectors:

Method: Show that AB = k AC (or AB = k BC) for some scalar k.

If one displacement vector is a scalar multiple of another, and they share a common point, the three points lie on the same line.

Worked Example

Show that P(2, 1, -1), Q(4, 5, 3) and R(5, 7, 5) are collinear.

PQ = (4 - 2, 5 - 1, 3 - (-1)) = (2, 4, 4) PR = (5 - 2, 7 - 1, 5 - (-1)) = (3, 6, 6)

Is PR = k × PQ? Check: 3/2 = 6/4 = 6/4 = 1.5

Yes, PR = 1.5 PQ, so P, Q, R are collinear.

Furthermore, Q divides PR in the ratio 2 : 1 (since PQ = (2/3) PR, Q is two-thirds of the way from P to R).

Proving Parallelism

Two line segments are parallel if their direction vectors are scalar multiples of each other.

AB is parallel to CD if AB = k CD for some scalar k.

Note: Parallel lines have the same direction but need not share a point (unlike collinear points).

Worked Example

Show that AB is parallel to CD, where A(1, 0, 2), B(3, 4, 6), C(0, 1, -1), D(1, 3, 1).

AB = (2, 4, 4) and CD = (1, 2, 2)

AB = 2 × CD, so AB is parallel to CD.

Proving Lines Are NOT Parallel

If AB ≠ k CD for any scalar k, the lines are not parallel.

Example: a = (1, 2, 3) and b = (2, 3, 4). Since 1/2 ≠ 2/3, the vectors are not parallel.

Vector Proofs in Geometry

Many A-Level questions give you a shape (triangle, parallelogram, etc.) with some points defined in terms of vectors, and ask you to prove geometric properties.

Common Strategies

Express all relevant vectors in terms of the given vectors (usually a, b, or p, q, etc.).
Show two vectors are parallel (one is a scalar multiple of the other) to prove line segments are parallel.
Show two vectors are equal to prove line segments are equal in length and direction.
Show a vector equals k times another to find ratios.

Worked Example: Proving a Parallelogram

OABC is a quadrilateral where OA = a and OC = c. B is the point such that OB = a + c. Prove that OABC is a parallelogram.

We need to show that opposite sides are parallel and equal.

OA = a and CB = OB - OC = (a + c) - c = a

Since OA = CB (same vector), OA is parallel to CB and equal in length.

OC = c and AB = OB - OA = (a + c) - a = c

Since OC = AB, OC is parallel to AB and equal in length.

Both pairs of opposite sides are parallel and equal, so OABC is a parallelogram.

Worked Example: Midpoint Theorem

In triangle OAB, M is the midpoint of OA and N is the midpoint of OB. Prove that MN is parallel to AB and MN = (1/2)AB.

Let OA = a and OB = b.

OM = (1/2)a (M is the midpoint of OA) ON = (1/2)b (N is the midpoint of OB)

MN = ON - OM = (1/2)b - (1/2)a = (1/2)(b - a) = (1/2)AB

Since MN = (1/2)AB, MN is parallel to AB and half its length. ∎

Worked Example: Intersection of Medians

In triangle OAB, let OA = a and OB = b. The midpoint of AB is M, and the midpoint of OB is N. Find the position vectors of the points where OM and AN intersect.

Point on OM: OM = (1/2)(a + b). A general point on OM: t × (1/2)(a + b) = (t/2)a + (t/2)b for parameter t.

Point on AN: AN = ON - OA = (1/2)b - a. A general point on AN: a + s((1/2)b - a) = (1 - s)a + (s/2)b for parameter s.

At the intersection: (t/2)a + (t/2)b = (1 - s)a + (s/2)b

Comparing coefficients:

a: t/2 = 1 - s
b: t/2 = s/2 → t = s

From the first equation: s/2 = 1 - s → s/2 + s = 1 → 3s/2 = 1 → s = 2/3

So t = 2/3. The intersection point has position vector: (2/3)(1/2)(a + b) = (1/3)a + (1/3)b = (1/3)(a + b)

This is the centroid of the triangle.

Finding Where a Line Meets an Axis or Plane

Example: A line passes through A(1, 2, 3) in the direction d = (2, -1, 4). Find where the line crosses the xz-plane (y = 0).

Parametric form: r = (1, 2, 3) + t(2, -1, 4) = (1 + 2t, 2 - t, 3 + 4t)

For the xz-plane, y = 0: 2 - t = 0 → t = 2

Point: (1 + 4, 0, 3 + 8) = (5, 0, 11)

Common Mistakes

Wrong direction for displacement vectors — Always use finish minus start: AB = b - a.
Confusing parallel with equal — Parallel means same direction (possibly different magnitude). Equal means same direction AND same magnitude.
Incomplete collinearity proof — You must show vectors are parallel AND share a common point.
Arithmetic errors in component calculations — Work systematically and check each component.

Summary

Collinear points: Show AB = k AC for some scalar k (the displacement vectors are parallel and share a common point).
Parallel lines: Show their direction vectors are scalar multiples.
Geometric proofs involve expressing all vectors in terms of given base vectors and showing relationships (parallelism, equality, ratios).
The midpoint theorem, centroid, and parallelogram properties can all be proved using vectors.
Use parametric forms to find where lines meet axes or planes.

A-Level Deep Dive: Vector Geometry

Spec mapping

Edexcel 9MA0-02 specification section 12 — Vectors covers vectors to solve problems in pure mathematics and in context, including geometric and mechanical applications (refer to the official specification document for exact wording). This sub-strand is examined in Paper 2 — Pure Mathematics, but the techniques recur in Paper 3 — Statistics and Mechanics (force diagrams, displacement, relative velocity), giving vectors a synoptic reach across the whole qualification. The Edexcel formula booklet provides the magnitude formula $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ and the scalar product, but the central geometric ideas — collinearity, midpoints, ratios of division, parallel and concurrent lines — must be reconstructed in every question from first principles. Section 12 sits alongside section 11 (Vectors as a separate strand at AS) for two-dimensional work and extends to three dimensions only at A2.

Worked example with full mark scheme

Question (8 marks): $ABCD$ is a parallelogram with $\overrightarrow{AB} = \mathbf{a}$ and $\overrightarrow{AD} = \mathbf{b}$ . The diagonals $AC$ and $BD$ intersect at $M$ . Prove, using vectors, that $M$ is the midpoint of both diagonals (so the diagonals bisect each other).

Solution with mark scheme:

Step 1 — set up position vectors with $A$ as origin.

Take $A$ as origin. Then $\overrightarrow{AB} = \mathbf{a}$ , $\overrightarrow{AD} = \mathbf{b}$ . Since $ABCD$ is a parallelogram, $\overrightarrow{DC} = \overrightarrow{AB} = \mathbf{a}$ , so $\overrightarrow{AC} = \overrightarrow{AD} + \overrightarrow{DC} = \mathbf{a} + \mathbf{b}$ .

M1 — correct use of the parallelogram property to obtain $\overrightarrow{AC}$ . A common slip is writing $\overrightarrow{AC} = \mathbf{a} - \mathbf{b}$ , confusing $AC$ with $DB$ .

A1 — $\overrightarrow{AC} = \mathbf{a} + \mathbf{b}$ stated cleanly.

Step 2 — parameterise both diagonals.

A general point on $AC$ has position vector $\lambda(\mathbf{a} + \mathbf{b})$ for some scalar $\lambda \in [0,1]$ .

A general point on $BD$ starts at $B$ (position $\mathbf{a}$ ) and moves toward $D$ (position $\mathbf{b}$ ), so has position vector $\mathbf{a} + \mu(\mathbf{b} - \mathbf{a}) = (1 - \mu)\mathbf{a} + \mu \mathbf{b}$ for $\mu \in [0,1]$ .

M1 — parameterising both diagonals using two distinct scalar parameters. Using one parameter for both is the classic mistake — it begs the question.

A1 — both parameterisations correct.

Step 3 — equate components at $M$ .

At $M$ , the two parameterisations give the same position vector:

$\lambda(\mathbf{a} + \mathbf{b}) = (1 - \mu)\mathbf{a} + \mu \mathbf{b}$

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel (otherwise $ABCD$ would be degenerate), they form a basis: their coefficients must match independently.

M1 — invoking linear independence of $\mathbf{a}$ and $\mathbf{b}$ to equate coefficients. This is the crucial reasoning step; without an explicit statement that $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, the deduction is unjustified.

Equating coefficients of $\mathbf{a}$ : $\lambda = 1 - \mu$ . Equating coefficients of $\mathbf{b}$ : $\lambda = \mu$ .

A1 — both equations correct.

Step 4 — solve and conclude.

From the two equations: $\mu = 1 - \mu$ , so $\mu = \tfrac{1}{2}$ , and hence $\lambda = \tfrac{1}{2}$ .

A1 — both parameter values found.

Since $\lambda = \tfrac{1}{2}$ , point $M$ is halfway along $AC$ , so $M$ is the midpoint of $AC$ . Since $\mu = \tfrac{1}{2}$ , point $M$ is halfway along $BD$ , so $M$ is the midpoint of $BD$ . Therefore the diagonals of the parallelogram bisect each other.

A1 — explicit conclusion linking parameter values to the midpoint claim, with the geometric statement quoted back from the question.

Total: 8 marks (M3 A5). The split rewards the geometric set-up and the linear-independence reasoning equally.

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Relative to an origin $O$ , the points $P$ , $Q$ , $R$ have position vectors $\mathbf{p}$ , $\mathbf{q}$ , $\mathbf{r}$ respectively. The point $S$ lies on $PQ$ such that $PS:SQ = 2:1$ , and the point $T$ is the midpoint of $QR$ .

(a) Express $\overrightarrow{OS}$ and $\overrightarrow{OT}$ in terms of $\mathbf{p}$ , $\mathbf{q}$ , $\mathbf{r}$ . (3)

(b) Hence find $\overrightarrow{ST}$ and express it as a scalar multiple of $\left(-\tfrac{2}{3}\mathbf{p} - \tfrac{1}{3}\mathbf{q} + \mathbf{r}\right)$ , stating the constant of proportionality. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — applying the section formula $\overrightarrow{OS} = \mathbf{p} + \tfrac{2}{3}(\mathbf{q} - \mathbf{p})$ .
A1 (AO1.1b) — simplifying to $\overrightarrow{OS} = \tfrac{1}{3}\mathbf{p} + \tfrac{2}{3}\mathbf{q}$ .
A1 (AO1.1b) — $\overrightarrow{OT} = \tfrac{1}{2}(\mathbf{q} + \mathbf{r})$ .

(b)

M1 (AO2.1) — forming $\overrightarrow{ST} = \overrightarrow{OT} - \overrightarrow{OS} = \tfrac{1}{2}\mathbf{q} + \tfrac{1}{2}\mathbf{r} - \tfrac{1}{3}\mathbf{p} - \tfrac{2}{3}\mathbf{q} = -\tfrac{1}{3}\mathbf{p} - \tfrac{1}{6}\mathbf{q} + \tfrac{1}{2}\mathbf{r}$ .
M1 (AO2.4) — extracting a common scalar factor of $\tfrac{1}{2}$ : $\overrightarrow{ST} = \tfrac{1}{2}\left(-\tfrac{2}{3}\mathbf{p} - \tfrac{1}{3}\mathbf{q} + \mathbf{r}\right)$ .
A1 (AO2.4) — stating the constant of proportionality $\tfrac{1}{2}$ , matching the printed target form.

Total: 6 marks split AO1 = 3, AO2 = 3. This is a balanced AO question — Edexcel uses ratio-of-division and midpoint problems to test both procedural fluency and the reasoning step of recognising scalar multiples.

Synoptic links

Connects to:

Section 11 — Position vectors and basic vector algebra: every geometric proof in section 12 starts by choosing an origin and writing each named point as a position vector. The arithmetic of $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ is the bedrock — without it, no geometric claim can be encoded as an equation between vectors.
Section 11 — Vector lines and equations of lines: parameterising a diagonal as $\mathbf{r}(\lambda) = \lambda(\mathbf{a} + \mathbf{b})$ is exactly the vector equation of a line, $\mathbf{r} = \mathbf{r}_0 + \lambda \mathbf{d}$ . Intersection problems for diagonals reduce to solving simultaneous vector-line equations, the same technique used for skew/intersecting lines in 3D.
Section 12 — Scalar product for perpendicular tests: while pure ratio-and-midpoint problems use only linear independence, harder section-12 questions ask whether two diagonals (e.g. of a rhombus) are perpendicular. Then $\mathbf{u} \cdot \mathbf{v} = 0$ becomes the test of choice, and the algebra moves from "matching coefficients" to "expanding scalar products".
Section 7 — Coordinate geometry in the $(x,y)$ -plane: every theorem provable by vector methods has a coordinate counterpart. The midpoint-of-diagonals result above can be proved by placing $A = (0,0)$ , $B = (a_1, a_2)$ , $D = (b_1, b_2)$ and computing the average of opposite vertices. Vectors give the cleaner, basis-free proof.
Affine geometry (beyond A-Level): the proofs in this section are affine — they use only addition and scalar multiplication of vectors, never length or angle. This is why they generalise instantly to any dimension and any "parallelogram-like" structure in linear algebra. Recognising the affine character of a proof is the bridge from school geometry to undergraduate linear algebra.

Mark-scheme literacy

Vector-geometry questions on 9MA0-02 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	40–50%	Writing position vectors, applying the section formula, computing $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ , parameterising lines
AO2 (reasoning / interpretation)	35–45%	Invoking linear independence to equate coefficients, recognising parallel/concurrent structure, justifying ratios and midpoint claims
AO3 (problem-solving)	10–20%	Constructing a multi-step proof, choosing which point to take as origin, deciding whether to parameterise or to combine vector equations

Vector Geometry

Vector Geometry

Proving Collinearity

Worked Example

Proving Parallelism

Worked Example

Proving Lines Are NOT Parallel

Vector Proofs in Geometry

Common Strategies

Worked Example: Proving a Parallelogram

Worked Example: Midpoint Theorem

Worked Example: Intersection of Medians

Finding Where a Line Meets an Axis or Plane

Common Mistakes

Summary

A-Level Deep Dive: Vector Geometry

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Synoptic links

Mark-scheme literacy

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