Vector Geometric Proof

SPEC G25 (HIGHER) GEOMETRY & MEASURES

Use vector arithmetic to prove parallelism, collinearity, and ratio relationships. The structured 3-step strategy that handles every Higher-tier vector proof.

What you will learn

By the end of this lesson you will:

Apply the 3-step strategy for any vector proof: choose base vectors, express the required vectors, then compare

Use the midpoint and ratio formulas to express points along a line in terms of base vectors

Prove that two vectors are parallel by showing one is a scalar multiple of the other

Prove that three points are collinear by showing two vectors that share a common point are parallel

Write the concluding sentence that earns the final accuracy mark

Why this matters — exam intelligence

Vector proofs are a reliable 4–5 mark question at the back of every Higher paper. AQA examiner reports describe them as a grade 8–9 discriminator: students who set them up cleanly score full marks, students who don't typically score zero. Mastering the structure is high-leverage.

The Game You're Playing

A vector proof gives you a geometric fact to demonstrate — usually one of these three:

Type 1 — Parallel

Two lines (or vectors) point in the same direction. Prove by showing one vector is a scalar multiple of the other:XY = k PQ

Type 2 — Collinear

Three (or more) points lie on the same straight line. Prove by showing two vectors with a common endpoint are parallel:XY = k YZ

Type 3 — Ratio / Midpoint

A point divides a line in a stated ratio (e.g. 2:1) or is the midpoint. Express the point's position vector in terms of the base vectors and verify it matches the expected form.

All three reduce to one technique

Every type relies on the same engine: express the relevant vectors in terms of base vectors, then compare. The structure of your answer is the same; only the conclusion changes.

The Tools You Already Have

Vector proofs use only the arithmetic from Lesson 4.19. No new operations — just disciplined application:

Position vectors

If A has position vector a and B has position vector b, thenAB = b − a

Parallel test

Two vectors u and v are parallel iff one is a scalar multiple of the other:u = k v for some scalar k

Vector arithmetic identities you'll use repeatedly

Going via origin:XY = OY − OX

Going around a path:XZ = XY + YZ

Reversing direction:YX = −XY

Scaling: if M is the midpoint of XY, then XM = ½XY

The 3-Step Strategy

Every vector proof follows the same three steps:

1

Identify the base vectors. Usually given in the question as two non-parallel vectors, often labelled a and b. Every other vector you write must be expressed in terms of these.

2

Express the required vectors in terms of a and b . Use the path identities (going via origin, midpoint formula, ratio formula). Simplify each expression to the form pa + qb for scalars p, q.

3

Compare and conclude. Show one vector is a scalar multiple of another (or that two vectors are equal, or that a position vector matches the expected ratio form). Then write a concluding sentence stating the geometric result.

If you remember nothing else

"Express in a and b , then compare." That phrase is the entire strategy.

Step 1 — Picking the Base Vectors

The question almost always names them for you. Typical setup:

Standard exam phrasing

"In triangle OAB, OA = a and OB = b."

Two non-parallel vectors. Every point in the diagram has a position vector that can be written as some pa + qb.

Other vectors emerge naturally

Once a and b are fixed:

AB = b − a (destination minus start)

Position of A: a; position of B: b; position of O: 0

Any point M with OM = pa + qb sits inside or on the triangle for suitable p, q

Step 2 — Midpoint & Ratio Formulas

To express points on a line, two formulas dominate:

Midpoint: if M is the midpoint of XY, then XM = ½XY

Ratio: if P divides XY in ratio m:n, then XP = m / (m + n) XY

Worked applications

Midpoint of OA (where OA = a): if M is the midpoint, then OM = ½a.

Point P on AB with AP:PB = 2:3: AP = (2/5)AB = (2/5)(b − a).

To get the position vector of P from O:

OP = OA + AP = a + (2/5)(b − a) = (3/5)a + (2/5)b.

Useful identity

If M is the midpoint of AB, the position vector of M from O is the average of the position vectors of A and B:OM = ½(a + b)

01Worked Example — The Midpoint Theorem

In triangle OAB, OA = a and OB = b . M is the midpoint of OA, and N is the midpoint of OB.

Prove that MN is parallel to AB and half its length. [5 marks]

Express MN and AB in terms of base vectors a and b, then compare.

C — Calculate

Position vectors of M and N (using the midpoint formula):

OM = ½a and ON = ½b

Apply MN = ON − OM:

MN = ½b − ½a = ½(b − a)

And AB directly:

AB = OB − OA = b − a

Compare:

MN = ½AB

E — Evaluate (the conclusion sentence)

Since MN is a scalar multiple of AB , MN is parallel to AB.

Since the scalar is ½, MN is half the length of AB. ∎

This is the textbook midpoint theorem — one of the most-tested vector-proof setups. Notice how every step lives in the a, b world; no specific numbers ever appear.

The Conclusion Sentence

The single most-flagged error on AQA examiner reports for this topic: students compute the right vector relationship and stop, never stating the geometric conclusion. This routinely costs the final accuracy mark.

The trap

Writing "MN = ½AB" alone is not a complete proof — it's a numerical equality. The marker needs to see you state what that equality means geometrically.

A complete conclusion includes

Why parallel? — "because one is a scalar multiple of the other"

What's the relationship? — the actual ratio (1/2, 2/3, etc.)

For collinearity: add "and they share point [name]"

Incomplete

"MN = ½AB."

Stops at the equation. Final A1 lost.

Complete

"Since MN = ½AB, MN is a scalar multiple of AB, so MN is parallel to AB. Since the scalar is ½, MN is half the length of AB."

All marks earned.

Proving Two Vectors Are Parallel

The structure of every "prove parallel" question:

A

Express the two vectors of interest, XY and PQ, in terms of the base vectors a and b.

B

Show one is a scalar multiple of the other: XY = k PQ for some specific scalar k.

C

Conclude: "Since XY is a scalar multiple of PQ, XY is parallel to PQ. The length of XY is |k| times the length of PQ."

When does this method fail?

If after simplifying you can't get one vector as a scalar multiple of the other, then they are not parallel — and the question must be asking you to prove a different relationship (e.g. ratio).

02Worked Example — Parallel via Ratio

In triangle OAB, OA = a and OB = b . P is the point on OA such that OP : PA = 2 : 1, and Q is the point on OB such that OQ : QB = 2 : 1.

Prove that PQ is parallel to AB and find the ratio of their lengths. [5 marks]

OP : PA = 2 : 1 means P divides OA so that OP is two-thirds of the whole. Same logic for Q on OB. Express PQ in terms of a, b and compare with AB.

C — Calculate

Position vectors of P and Q from O:

OP = (2/3)a and OQ = (2/3)b

Apply PQ = OQ − OP:

PQ = (2/3)b − (2/3)a = (2/3)(b − a)

And AB = b − a, so:

PQ = (2/3) AB

E — Evaluate

Since PQ is a scalar multiple of AB , PQ is parallel to AB. The length of PQ is two-thirds of the length of AB. ∎

This is the same structural pattern as the midpoint theorem but with k = 2/3 instead of 1/2. The proof technique generalises to any ratio.

Proving Three Points Are Collinear

Three points X, Y, Z lie on a straight line iff there is a scalar k such that:

XY = k YZ

(or equivalently XZ is a scalar multiple of XY)

Why this works

If XY and YZ are parallel, the lines XY and YZ point in the same (or opposite) direction. They share the point Y. So Z lies on the same line as X and Y — i.e. all three are collinear.

Common mistake — examiner-flagged trap

"DE + EF = DF therefore D, E, F collinear." This is not valid. The path equation DE + EF = DF is true for any three points D, E, F — it just says "walking D to E to F gets you to F". It does NOT prove they're collinear. You must show one vector is a scalar multiple of another.

A complete "collinear" conclusion

"Since XY = k YZ, the vectors are parallel. They share the point Y. Therefore X, Y and Z are collinear."

03Worked Example — Collinearity

Three points P, Q, R have position vectors:

OP = a , OQ = 2 a + b , OR = 4 a + 3 b

Prove that P, Q, R are collinear. [4 marks]

Find PQ and QR in terms of a and b. Show one is a scalar multiple of the other. They share the point Q, so collinearity follows.

C — Calculate

Use PQ = OQ − OP:

PQ = (2a + b) − a = a + b

Use QR = OR − OQ:

QR = (4a + 3b) − (2a + b) = 2a + 2b = 2(a + b)

Compare:

QR = 2 PQ

E — Evaluate

Since QR is a scalar multiple of PQ , the vectors PQ and QR are parallel. They share the common point Q. Therefore P, Q and R are collinear. ∎

Notice the THREE elements of the conclusion: scalar multiple → parallel → common point → collinear. Skipping "they share Q" loses a mark.

General — Not Specific

One trap that loses entire questions: using specific numbers to "prove" a general result.

Why specific examples don't prove anything

"Take a = (1, 0) and b = (0, 1). Then MN = (−0.5, 0.5) and AB = (−1, 1). So MN = ½AB." — This shows the relationship for ONE choice of a and b, not for all of them.

A correct proof is general

Keep a and b as arbitrary, unspecified vectors throughout. Every step of arithmetic should still work no matter which two non-parallel vectors they happen to represent. The conclusion holds for all triangles OAB, not just one example.

How to spot the trap

If at any point in your working you see numerical components — (1, 0), (3, −2), 5 — you've drifted into a specific example. The proof should look like ½a, (2/3)(b − a), and so on, all the way to the conclusion.

Examiner Alerts

Alert 1 — Stopping at the equation

Computing "XY = k PQ" without writing the geometric conclusion is the most common error AQA examiners flag. What to write instead: finish with a sentence stating "parallel because scalar multiple" (and "collinear because they share a point" for collinearity).

Alert 2 — Path equation as proof of collinearity

Writing "DE + EF = DF" is true for any three points and proves nothing on its own (a recurrent examiner-flagged error). What to write instead: show DE = k EF as a scalar multiple; the path equation alone earns zero marks toward collinearity.

Alert 3 — Wrong AB direction

Writing AB = a − b instead of b − a flips the sign on every subsequent vector. The proof can still be salvaged with care, but typically loses an M-mark and an A-mark. What to write instead: write the rule out at the top of your working: "AB = b − a (destination minus start)" before substituting.

Common Errors

Students often writeCorrect versionWhy it matters

" DE + EF = DF , so D, E, F collinear."Show DE = k EF as a scalar multiple.The path equation is true for any three points.

Computes MN = ½ AB and stops.Add "parallel because scalar multiple; half length because k = ½."Conclusion sentence earns the final A1.

Substitutes a = (3, 1) and works numerically.Keep a and b general throughout.One example doesn't prove a general result.

AB = a − bAB = b − aDestination minus start. Flipped sign cascades through the proof.

P divides AB in ratio 2:3, so AP = (2/3) AB .AP = (2/5) AB — the denominator is the sum of the parts.2:3 splits AB into 5 parts; AP gets 2 of them.

For collinear: shows XY parallel to PQ with no shared point.Find two parallel vectors that share a point.Without a common point, parallel vectors lie on different lines.

Practice

Three points A, B, C have position vectors:

OA = 2 a + b , OB = 5 a + 7 b , OC = 8 a + 13 b

(a) Express AB and BC in terms of a and b . [2 marks]

(b) Hence prove that A, B and C are collinear, and state the position of B on the line. [3 marks]

(a) Apply AB = OB − OA and BC = OC − OB.

(b) Compare to find the scalar relationship; note the shared point B for collinearity; comment on the size relationship.

C — Calculate

(a)

AB = (5a + 7b) − (2a + b) = 3a + 6b = 3(a + 2b)

BC = (8a + 13b) − (5a + 7b) = 3a + 6b = 3(a + 2b)

(b)

AB = BC

Since AB = BC , the vectors are equal — so they are parallel (k = 1) and have the same length. They share the point B. Therefore A, B and C are collinear, and B is the midpoint of AC. ∎

Lesson checkpoint — you should now be able to:

Apply the 3-step strategy: identify base vectors, express required vectors, compare

Use midpoint and ratio formulas to express points on a line

Prove parallelism by showing one vector is a scalar multiple of another

Prove collinearity by combining a scalar multiple with a shared point

Write the concluding sentence that earns the final accuracy mark

End of Module 4 — Geometry & Measures

That completes the geometry and measures module. Vector proofs are the final — and arguably the most powerful — technique in the AQA Higher geometry toolkit.

The Higher-tier geometry hierarchy you've built

4.10 — Circle theorems : 4–5 mark proof questions on circle properties

4.15 / 4.16 — Pythagoras & SOH CAH TOA : foundation for all length/angle work

4.17 — 3D Pythagoras & trig [H]: identify the right triangle inside the solid

4.18 — Sine & cosine rules [H]: any triangle, not just right-angled

4.19 — Vector arithmetic : column form, addition, scalar multiplication

4.20 — Vector proofs [H]: the back-of-paper grade-9 question

What's next

Module 5 covers probability — a fresh topic with its own structure but a familiar requirement: clear notation and explicit logical steps.

Practice priority for vector proofs

Of all the topics you've learned, vector proofs reward spaced practice the most. The structure is mechanical once internalised, but only repetition makes it automatic. Aim for at least one past-paper vector-proof question per revision session.