Vectors | Pearson Edexcel GCSE Maths Notes

Visual model

Vectors describe movement

A

B

\vec{AB}

\text{right}

\text{up}

horizontal move plus vertical move

Follow arrows in order.

Reverse direction means negative vector.

Scale vectors before adding them.

Gold-standard guide

26 mins

What you will learn

Describe movement and solve geometric vector problems.

Use a clear step-by-step method for vectors.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier geometry and measures skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Vector AB = OB − OA (start at A, go back to O, then to B)

Step 1

Find AB by computing OB − OA

AB = OB − OA = (−1, 4) − (3, −1) = (−1 − 3, 4 − (−1)) = (−4, 5)

Step 2

Find the magnitude of AB

|AB| = sqrt((−4)² + 5²) = $\sqrt{16 + 25}$ = $\sqrt{41}$ ≈ 6.40

Watch out

Students compute AB = OA − OB (wrong direction)

Column vector

move right/left first, then up/down.

Vector addition

add matching components.

Worked example

OA = (3, −1) and OB = (−1, 4). Find the vector AB and its magnitude.

Find AB by computing OB − OA: AB = OB − OA = (−1, 4) − (3, −1) = (−1 − 3, 4 − (−1)) = (−4, 5).

Find the magnitude of AB: |AB| = sqrt((−4)² + 5²) = $\sqrt{16 + 25}$ = $\sqrt{41}$ ≈ 6.40.

Final answer

AB = (−4, 5); |AB| = $\sqrt{41}$ ≈ 6.40

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

OA = (3, −1) and OB = (−1, 4). Find the vector AB and its magnitude.

3 marks4 minsvectors-worked

Show solution

Worked solution

1.Find AB by computing OB − OA: AB = OB − OA = (−1, 4) − (3, −1) = (−1 − 3, 4 − (−1)) = (−4, 5).
2.Find the magnitude of AB: |AB| = sqrt((−4)² + 5²) = $\sqrt{16 + 25}$ = $\sqrt{41}$ ≈ 6.40.

Final answer

AB = (−4, 5); |AB| = $\sqrt{41}$ ≈ 6.40

Mark points

M1: find ab by computing ob − oa
M1: find the magnitude of ab
A1: AB = (−4, 5); |AB| = $\sqrt{41}$ ≈ 6.40

Watch out

Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

Diagnosticrecall

OA = (2, 5) and OB = (6, 1). Find the midpoint M of AB as a column vector.

1 mark2 minsvectors-q1

Show solution

Worked solution

1.Spot the skill: Vector AB = OB − OA (start at A, go back to O, then to B).
2.Use the find ab by computing ob − oa stage first, then find the magnitude of ab.
3.Keep the final answer visible: (4, 3).

Final answer

(4, 3)

Mark points

M1: use the correct vector ab = ob − oa (start at a, go back to o, then to b). magnitude |v| = $\sqrt{x^{2} + y^{2}}$ .column vectors: top is horizontal (positive = right), bottom is vertical (positive = up).scalar × vector: multiply each component.
A1: (4, 3)

Watch out

Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

Easyprocedure

Vector a = (3, 4). Find |a| and 2a.

2 marks3 minsvectors-q2

Show solution

Worked solution

1.Spot the skill: Vector AB = OB − OA (start at A, go back to O, then to B).
2.Use the find the magnitude of ab stage first, then find ab by computing ob − oa.
3.Keep the final answer visible: |a| = 5; 2a = (6, 8).

Final answer

|a| = 5; 2a = (6, 8)

Mark points

M1: use the correct vector ab = ob − oa (start at a, go back to o, then to b). magnitude |v| = $\sqrt{x^{2} + y^{2}}$ .column vectors: top is horizontal (positive = right), bottom is vertical (positive = up).scalar × vector: multiply each component.
A1: |a| = 5; 2a = (6, 8)

Watch out

Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

Mediumreasoning

If p = (1, −2) and q = (−3, 5), find 3p − 2q.

3 marks4 minsvectors-q3

Show solution

Worked solution

1.Spot the skill: Vector AB = OB − OA (start at A, go back to O, then to B).
2.Use the find ab by computing ob − oa stage first, then find the magnitude of ab.
3.Keep the final answer visible: (9, −16).

Final answer

(9, −16)

Mark points

M1: use the correct vector ab = ob − oa (start at a, go back to o, then to b). magnitude |v| = $\sqrt{x^{2} + y^{2}}$ .column vectors: top is horizontal (positive = right), bottom is vertical (positive = up).scalar × vector: multiply each component.
A1: (9, −16)

Watch out

Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

Hardproblem solving

Show that vectors (2, 6) and (1, 3) are parallel.

3 marks5 minsvectors-q4

Show solution

Worked solution

1.Spot the skill: Vector AB = OB − OA (start at A, go back to O, then to B).
2.Use the find the magnitude of ab stage first, then find ab by computing ob − oa.
3.Keep the final answer visible: (2, 6) = 2(1, 3), so one is a scalar multiple of the other.

Final answer

(2, 6) = 2(1, 3), so one is a scalar multiple of the other

Mark points

M1: use the correct vector ab = ob − oa (start at a, go back to o, then to b). magnitude |v| = $\sqrt{x^{2} + y^{2}}$ .column vectors: top is horizontal (positive = right), bottom is vertical (positive = up).scalar × vector: multiply each component.
A1: (2, 6) = 2(1, 3), so one is a scalar multiple of the other

Watch out

Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

Exam-stylemulti-step

OABC is a parallelogram with OA = a and OC = c. Find the vector from B to the midpoint of OC.

4 marks6 minsvectors-q5

Show solution

Worked solution

1.Spot the skill: Vector AB = OB − OA (start at A, go back to O, then to B).
2.Use the find ab by computing ob − oa stage first, then find the magnitude of ab.
3.Keep the final answer visible: BC = −a; mid-OC is c/2; vector from B = c/2 − (a + c) = −a − c/2.

Final answer

BC = −a; mid-OC is c/2; vector from B = c/2 − (a + c) = −a − c/2

Mark points

M1: use the correct vector ab = ob − oa (start at a, go back to o, then to b). magnitude |v| = $\sqrt{x^{2} + y^{2}}$ .column vectors: top is horizontal (positive = right), bottom is vertical (positive = up).scalar × vector: multiply each component.
A1: BC = −a; mid-OC is c/2; vector from B = c/2 − (a + c) = −a − c/2

Watch out

Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

Grade 9 stretchproblem solving

In triangle OAB, OA = a and OB = b. Point P lies on OA with OP = $\frac{2}{3}$ a, and Q lies on OB with OQ = $\frac{2}{3}$ b. Prove that PQ is parallel to AB and state the length scale factor from triangle OAB to triangle OPQ.

3a

-2b

scale first, then combine

4 marks7 minsvectors-g9

Show solution

Worked solution

1.Write AB using the route A to O to B.
2.Write PQ using the route P to O to Q.
3.Compare the two vectors and use the scalar-multiple test for parallel lines.

Final answer

AB = b - a and PQ = $\frac{2}{3}$ (b - a) = $\frac{2}{3}$ AB, so PQ is parallel to AB; the length scale factor is $\frac{2}{3}$

Mark points

M1: obtain AB = b - a
M1: obtain PQ = $\frac{2}{3}$ b - $\frac{2}{3}$ a
A1: show PQ = $\frac{2}{3}$ AB
C1: conclude parallel and give scale factor $\frac{2}{3}$

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

16 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Vectors - 2 marksOA = (2, 5) and OB = (6, 1). Find the midpoint M of AB as a column vector.Mark answer

Answer

(4, 3)

2Angles, lines and polygons - 2 marksThe exterior angle of a regular polygon is 24°. How many sides?Mark answer

Answer

3Properties of shapes - 2 marksName all 2D shapes with equal diagonals that bisect each other.Mark answer

Answer

Rectangle, square

4Perimeter, area and volume - 3 marksFind the perimeter of a rectangle with length 13 cm and width 8 cm.Mark answer

Answer

42 cm

Mastery check

I can explain the method for vectors.
I can show clear working without skipping key steps.
I can avoid this mistake: Students compute AB = OA − OB (wrong direction).AB goes FROM A TO B, so subtract A's position from B's position: AB = b − a.Drawing a small arrow diagram of the journey prevents direction errors.

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