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Solving Geometric Problems

Vectors turn geometry into algebra. Rather than chasing angles round a diagram, you name two vectors, write everything else in terms of them, and let the algebra deliver the result. Proving the midpoints of a quadrilateral form a parallelogram takes about four lines this way — and works for $every$ quadrilateral at once.

35 min Vectors

The big picture

What makes vector proof powerful is that it is general by construction. A coordinate proof fixes particular numbers and then has to argue the result was not a fluke; a vector proof never commits to specific values, so the conclusion holds for every configuration from the outset. The method is always the same three moves: pick a base pair of vectors, express every vector in the figure in terms of them, then read the geometry off the algebra — a scalar multiple means parallel, equal vectors mean a parallelogram, and a shared point plus parallel means collinear. Recognising which conclusion you are aiming at tells you what algebraic form to work towards.

What you'll be able to do

Express vectors in a figure in terms of a chosen base pair
Prove lines are parallel or points collinear using scalar multiples
Use ratios along a line to locate intersection points
Structure a vector proof so each step is justified

Choosing a base and expressing everything in it

Start by naming two non-parallel vectors from a convenient point — often two sides of the figure meeting at a vertex, called $a$ and $b$ . Every other vector in the diagram can then be written in terms of these two.

The two must be non-parallel. If $b$ were a multiple of $a$ , both would point along one line and could not describe a two-dimensional figure between them.

To express a new vector, walk a route through the figure using vectors you already know, adding head-to-tail. Any route gives the same answer — which is itself a useful check.

1Route:

O \to A \to M

, so

O M = O A + A M

A B = b - a

, and

M

is halfway, so

A M = \frac{1}{2} (b - a)

3So

O M = a + \frac{1}{2} b - \frac{1}{2} a

Answer

O M = \frac{1}{2} (a + b)

Tip — Write the route before the algebra: " $O$ to $A$ , then $A$ to $M$ ". Most errors in vector proof are wrong routes, not wrong arithmetic.

What the algebra tells you about the geometry

Each geometric conclusion has an algebraic signature, and knowing them tells you what to aim for.

$Parallel$ : one vector is a scalar multiple of the other. $Equal in length and direction$ : the vectors are identical, which for opposite sides means a parallelogram. $Collinear$ : two displacements are scalar multiples and share a point. $Ratio along a line$ : the multiple itself is the fraction of the way along.

So a proof is finished when the algebra reaches the matching form — and the final line should say which signature was found and what it means.

P Q = k R S \Rightarrow P Q ∥ R S

Scalar multiple ⇒ parallel.

P Q = S R \Rightarrow P QR S a parallelogram

Equal opposite sides.

O P = \frac{1}{2} a

and

O Q = \frac{1}{2} b

P Q = O Q - O P = \frac{1}{2} b - \frac{1}{2} a

A B = b - a

, so

P Q = \frac{1}{2} A B

4A scalar multiple, so parallel; the scalar is

\frac{1}{2}

, so half the length.

Answer

P Q = \frac{1}{2} A B

— parallel and half the length.

Why this works

That result is the midpoint theorem, proved in three lines and for every triangle simultaneously. A congruent-triangles proof of the same fact takes considerably longer and needs a diagram to point at.

Finding an intersection with two unknowns

When two lines cross at an unknown point, give each line its own unknown fraction — say $λ$ along one and $μ$ along the other — and write the intersection point both ways.

Setting the two expressions equal gives one vector equation. Because $a$ and $b$ are non-parallel, their coefficients must match separately, so one vector equation becomes two scalar equations — enough to find both unknowns.

That independence is the engine of the whole method. It works only because the base pair is non-parallel, which is why the earlier condition mattered.

1Along

A M

O G = a + λ (\frac{1}{2} b - a) = (1 - λ) a + \frac{λ}{2} b

2Along

B N

O G = b + μ (\frac{1}{2} a - b) = \frac{μ}{2} a + (1 - μ) b

3Match

a

1 - λ = \frac{μ}{2}

. Match

b

\frac{λ}{2} = 1 - μ

4From the second,

μ = 1 - \frac{λ}{2}

. Substituting:

1 - λ = \frac{1}{2} - \frac{λ}{4}

, so

\frac{1}{2} = \frac{3 λ}{4}

and

λ = \frac{2}{3}

Answer

O G = \frac{1}{3} a + \frac{1}{3} b

Tip — Comparing coefficients is only valid because $a$ and $b$ are non-parallel. Say so in the proof — it is a credited justification, not a technicality.

Writing the proof so it reads as one

A vector proof is a written argument, and marks are awarded for the reasoning as much as the algebra. Four things belong in it: what the base vectors are, the route used for each expression, the algebraic relationship found, and the geometric conclusion drawn from it.

The last of those is the one most often left out. " $P Q = \frac{1}{2} A B$ " is a fact about vectors; "so $P Q$ is parallel to $A B$ and half its length" is the answer to the question that was asked.

1Let

O A B C

be a parallelogram with

O A = a

and

O C = c

; then

O B = a + c

2Midpoint of diagonal

O B

\frac{1}{2} (a + c)

3Midpoint of diagonal

A C

a + \frac{1}{2} (c - a) = \frac{1}{2} (a + c)

4The two midpoints have the same position vector, so they are the same point.

AnswerBoth diagonals have the same midpoint, so they bisect each other.

Why this works

Notice the proof never assumed anything about the parallelogram’s angles or side lengths — $a$ and $c$ stayed general throughout. That is why one argument covers every parallelogram, including squares and rhombuses.

Think like an examiner

Define your base vectors explicitly at the start. An unlabelled $a$ costs marks even if the algebra is right.
End with the geometric conclusion in words, not just the vector relationship.
When comparing coefficients, state that the base vectors are non-parallel — that is what licenses the comparison.
Show the route for each vector (" $O M = O A + A M$ "). Method marks live in those lines.

Common misconceptions

Choosing two parallel vectors as the base pair.

They must be non-parallel, or they cannot express every vector in the plane and coefficients cannot be compared.

Stopping at

P Q = 2 R S

Add the conclusion: therefore

P Q

is parallel to

R S

and twice its length.

Using one unknown for two different lines in an intersection problem.

Each line needs its own parameter,

λ

and

μ

, or you have assumed the answer.

Taking

A M = \frac{1}{2} b

when

M

is the midpoint of

A B

A M = \frac{1}{2} A B = \frac{1}{2} (b - a)

— half the

displacement

, not half a position vector.

Proof signatures

P Q = k R S

Parallel, with ratio

k

P Q = S R

Parallelogram.

A B ∥ B C

With shared

B

: collinear.

α a + β b = γ a + δ b \Rightarrow α = γ, β = δ

Comparing coefficients (non-parallel base).

Stretch yourself

In triangle $O A B$ , $O A = a$ and $O B = b$ . $P$ divides $O A$ in the ratio $2 : 1$ and $Q$ divides $O B$ in the ratio $2 : 1$ . Prove that $P Q$ is parallel to $A B$ , and state the ratio of their lengths.

Hint — Write $O P$ and $O Q$ as fractions of $a$ and $b$ , then find $P Q$ .

Questions students ask

How do I choose which vectors to call

a

and

b

Pick two non-parallel vectors from a point where several lines meet — usually two sides of the figure sharing a vertex. That minimises the number of routes you have to construct.

Why must the base vectors be non-parallel?

Two parallel vectors span only a line, so they cannot describe every point of a plane figure, and comparing coefficients becomes invalid because the same vector could be written many ways.

Is a vector proof better than a coordinate proof?

For general results, yes — vectors stay general, so the proof covers every case at once. Coordinates fix particular numbers and then need an argument that the choice was not special.

What if my route differs from the mark scheme’s?

Any valid route gives the same vector, so a different route earns full marks provided each step is justified. Vector addition is route-independent.

Key takeaways

Pick two non-parallel base vectors and express every other vector in terms of them.
Build each expression by walking a route through the figure, adding head-to-tail.
A scalar multiple means parallel; equal vectors mean a parallelogram; parallel plus a shared point means collinear.
For intersections, give each line its own parameter and compare coefficients.
Finish with the geometric conclusion stated in words.

How this fits the course

Build on

Vectors in Three Dimensions

Leads to

Vectors in Three Dimensions

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Real past-paper questions on Solving Geometric Problems, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

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Solving Geometric Problems

Choosing a base and expressing everything in it

What the algebra tells you about the geometry

Finding an intersection with two unknowns

Writing the proof so it reads as one

Practise this topic

Related topics

Solving Geometric Problems

Choosing a base and expressing everything in it

What the algebra tells you about the geometry

Finding an intersection with two unknowns

Writing the proof so it reads as one

Practise this topic

Related topics