IGCSE Maths · Transformations and Vectors · Core & Extended

Vectors: Notation, Arithmetic and Position Vectors

A vector describes a movement — a distance and a direction — and once you can add, subtract, and scale them, you can describe positions and prove geometric relationships without measuring anything. Notation, magnitude, arithmetic, and position vectors are shared by both tiers. Using vectors to prove parallel or collinear relationships is Extended-only.

1. Vector Notation

A vector is written as a column: the top number is the horizontal movement, the bottom is the vertical movement. Vectors are often labelled with a bold lowercase letter (a, b) in print, or an arrow over two capital letters (like AB) to show a specific directed line segment from A to B.

a = (x, y)  — written as a column, x on top, y below
Reading a directed line segment A B The vector AB (from A to B) moves 6 units right and 5 units up: AB = (6, 5).
Try it — Vector Notation
  1. Point A is at (1,2) and point B is at (7,9). Write the vector AB as a column vector.
  2. The vector CD is (−3, 4), and C is at (5,1). Find the coordinates of D.
Show answers
  1. (6,7)
  2. D = C + vector = (5−3, 1+4) = (2,5)

2. Magnitude of a Vector

The magnitude (written |a|) is the vector’s length — found using Pythagoras’ theorem on its horizontal and vertical parts.

|a| = √(x² + y²)
Worked example Find the magnitude of a = (3, 4).
|a| = √(3² + 4²) = √25 = 5
Try it — Magnitude
  1. Find the magnitude of (6, 8).
  2. Find the magnitude of (5, 12).
  3. A vector has magnitude 13 and horizontal component 5. Find its vertical component.
Show answers
  1. √(36+64) = 10
  2. √(25+144) = 13
  3. 13²=5²+y² → y²=169−25=144 → y=12

3. Adding and Subtracting Vectors

Add or subtract the top numbers together, and separately the bottom numbers.

Worked example a = (2, 3), b = (5, −1). Find a + b and a − b.
a + b = (2+5, 3−1) = (7, 2)
a − b = (2−5, 3−(−1)) = (−3, 4)
Try it — Adding and Subtracting
  1. a=(4,1), b=(−2,6). Find a+b.
  2. a=(7,−3), b=(2,5). Find a−b.
  3. a=(3,4), b=(−1,−2), c=(5,0). Find a+b−c.
Show answers
  1. (2,7)
  2. (5,−8)
  3. (3−1−5, 4−2−0) = (−3,2)

4. Multiplying a Vector by a Scalar

Multiplying by a plain number (a scalar) scales both parts of the vector by that number — it changes the vector’s length (and flips its direction if the scalar is negative), but not its underlying direction otherwise.

Worked example a = (2, 3). Find 3a and −2a.
3a = (6, 9)
−2a = (−4, −6)
Try it — Scalar Multiplication
  1. a=(4,−5). Find 2a.
  2. b=(−3,6). Find −4b.
  3. a=(2,3), b=(1,−1). Find 2a+3b.
Show answers
  1. (8,−10)
  2. (12,−24)
  3. 2(2,3)+3(1,−1) = (4,6)+(3,−3) = (7,3)

5. Position Vectors

A position vector describes a point’s location relative to a fixed origin O. If A and B have position vectors a and b, the vector from A to B is found by subtracting: AB = b − a (destination minus start).

Worked example Point A has position vector a = (2, 3). Point B has position vector b = (7, 1). Find the vector AB.
AB = b − a = (7−2, 1−3) = (5, −2)
Try it — Position Vectors
  1. A has position vector (1,5), B has position vector (6,2). Find AB.
  2. A has position vector (4,−1), B has position vector (−2,3). Find BA (careful with the order).
Show answers
  1. (6−1,2−5) = (5,−3)
  2. BA = a−b = (4−(−2),−1−3) = (6,−4)

6. Vector Geometry — Proving Relationships Extended only

Vectors can prove that lines are parallel (one vector is a scalar multiple of the other) or that points are collinear (lie on the same straight line) — all through pure algebra, no measuring or protractors required.

Worked example — the midpoint theorem O C D A B In triangle OAB, OA = a and OB = b. C is the midpoint of OA, and D is the midpoint of OB. Show that CD is parallel to AB, and find CD in terms of AB.
OC = ½a (C is the midpoint of OA)    OD = ½b (D is the midpoint of OB)
CD = OD − OC = ½b − ½a = ½(b − a)
AB = OB − OA = b − a
So CD = ½AB — since CD is a scalar multiple of AB, the two lines are parallel, and CD is exactly half the length of AB.
Worked example — a point dividing a line in a given ratio In triangle OAB, OA = a and OB = b. P is the point on AB such that AP:PB = 2:1. Find OP in terms of a and b.
Since AP:PB = 2:1, P is 2/3 of the way from A to B.
OP = OA + (2/3)AB = a + (2/3)(b − a) = a + (2/3)b − (2/3)a = (1/3)a + (2/3)b
Answer: OP = ⅓a + ⅔b
Try it — Vector Geometry Extended
  1. In triangle OAB, OA=a, OB=b. M is the midpoint of AB. Show that OM = ½(a+b).
  2. In triangle OAB, OA=a, OB=b. Q is the point on AB such that AQ:QB = 1:2. Find OQ in terms of a and b.
  3. Points X, Y, Z have position vectors x, y, z. If z − y = 2(y − x), explain what this tells you about the three points.
Show answers
  1. OM = OA + ½AB = a + ½(b−a) = a + ½b − ½a = ½a + ½b = ½(a+b)
  2. OQ = a + (1/3)(b−a) = (2/3)a + (1/3)b
  3. Since YZ = 2(XY), YZ is a scalar multiple of XY, meaning the two vectors are parallel — and since they share the point Y, this means X, Y, and Z all lie on the same straight line (they’re collinear).
Common mistakes to watch for
  • Getting the order wrong in AB = b − a — it’s always “end point minus start point,” not the other way round.
  • Forgetting to square both components before adding, when finding a magnitude — a common slip carried over from the distance formula.
  • In vector geometry, concluding two lines are parallel just because they “look” similar — you must show algebraically that one vector is a scalar multiple of the other.
  • Forgetting that two vectors being parallel is not enough on its own to prove collinearity — they must also share a common point.
  • Mixing up a ratio like “AP:PB = 2:1” with the fraction of the vector needed — this means P is 2/3 of the way along AB from A, not 1/2 or 2/1.

7. Quick-Fire Challenge Round

True or False?
  1. The vector AB is found by calculating (position of B) − (position of A).
  2. Multiplying a vector by a negative scalar reverses its direction.
  3. Two vectors are parallel only if they are equal.
  4. Showing two vectors share a common point is enough on its own to prove three points are collinear.
Show answers
  1. True.
  2. True.
  3. False — vectors are parallel if one is any scalar multiple of the other, not just when they’re identical. Ext.
  4. False — you also need to show the vectors are parallel (scalar multiples of each other); a shared point alone doesn’t guarantee collinearity. Ext.
Spot the Error
  1. “A is at (3,5), B is at (8,1). Vector AB = (3−8, 5−1) = (−5,4).”
  2. “Magnitude of (5,12): |v| = 5+12 = 17.”
  3. “CD = ½AB, so CD and AB are parallel and CD is twice as long as AB.”
Show answers
  1. AB = (end)−(start) = B−A = (8−3, 1−5) = (5,−4), not A−B.
  2. Magnitude needs Pythagoras, not simple addition: |v|=√(25+144)=√169=13.
  3. If CD = ½AB, then CD is half the length of AB, not twice as long.

8. Mixed Practice — Bringing It All Together

Core & Extended
  1. Find the magnitude of (9,12).
  2. a=(3,−2), b=(−4,5). Find a+b and a−b.
  3. a=(2,1). Find −3a.
  4. A has position vector (2,6), B has position vector (9,2). Find AB.
Show answers
  1. √(81+144) = 15
  2. a+b=(−1,3). a−b=(7,−7)
  3. (−6,−3)
  4. (7,−4)
Extended-level mixed set Extended only
  1. In triangle OAB, OA=a, OB=b. R is the midpoint of AB. Find OR in terms of a and b.
  2. In triangle OAB, OA=a, OB=b. S is the point on AB with AS:SB=3:1. Find OS in terms of a and b.
Show answers
  1. OR = ½(a+b)
  2. OS = a + (3/4)(b−a) = ¼a + ¾b

Practice at Home

A large set to close out Topic 7 — take it steadily.

Part A — Notation and magnitude
  1. A is at (2,3), B is at (9,7). Write vector AB as a column vector.
  2. Find the magnitude of (8,15).
  3. A vector has magnitude 25 and horizontal component 7. Find its vertical component.
Part B — Arithmetic and scalar multiplication
  1. a=(5,2), b=(−3,4). Find a+b and a−b.
  2. a=(4,−1). Find 5a and −3a.
  3. a=(1,2), b=(3,−1). Find 4a−2b.
Part C — Position vectors
  1. A has position vector (5,1), B has position vector (2,8). Find AB.
  2. P has position vector (−3,4), Q has position vector (6,−2). Find PQ and QP.
Part D — Vector geometry Extended
  1. In triangle OAB, OA=a, OB=b. C is the midpoint of OB. Find AC in terms of a and b.
  2. In triangle OAB, OA=a, OB=b. T is on AB with AT:TB=1:3. Find OT in terms of a and b.
  3. Points L, M, N have position vectors l, m, n. If n−m = 3(m−l), explain what this shows about the three points.
Part E — Applied / challenge
  1. OABCD is not needed — instead: in triangle OAB, OA=a, OB=b. E is the midpoint of OA, F is the midpoint of OB. Show EF is parallel to AB, then find the ratio of the area of triangle OEF to the area of triangle OAB (hint: think about the length scale factor and how area scales).
  2. A ship’s position vector (in km, relative to a harbour) is (12,9) at noon. It travels at a constant velocity vector of (−3,4) km/h. Find its position vector at 2pm, and its distance from the harbour at that time.
  3. Prove that the diagonals of a parallelogram OABC (with OA=a, OC=c, and B = A+C in position) bisect each other, using position vectors.

Key Vocabulary Recap

Magnitude — the length of a vector, |a|
Scalar — an ordinary number used to scale a vector’s size
Position vector — a vector describing a point’s location from a fixed origin
Collinear — points lying on the same straight line Ext.
Parallel vectors — vectors that are scalar multiples of each other Ext.

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section C7.2 (Core subject content) and section E7.2 (Extended subject content) — Vectors in two dimensions. Vector notation, magnitude, addition/subtraction, scalar multiplication, and position vectors are shared by both tiers. Using vector geometry to prove parallel or collinear relationships is Extended-only, marked above. This closes out Topic 7 — Transformations and Vectors.

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