Vectors in two dimensions for IGCSE Mathematics - Additional: Methods and Worked Examples

Why vectors feel different from the rest of the syllabus

You're used to working with numbers that sit neatly on a number line. Vectors break that habit. A vector carries two pieces of information at once: how far and in which direction. That double payload is exactly what makes vectors powerful, but it's also why they can feel unfamiliar the first time you meet them in your IGCSE Additional Mathematics course.

The good news? The arithmetic itself is straightforward. If you can add, subtract, and multiply ordinary numbers, you can do all of those things with vectors. The trick is keeping track of what each operation means geometrically. Once that clicks, vectors stop being abstract notation and start being a tool you actually want to reach for.

Notation: the three ways you'll see vectors written

Cambridge uses several notations, sometimes within the same question. Recognising all of them quickly saves you time.

The arrow notation tells you the direction of travel. The vector from A to B is written with an arrow pointing right above the letters AB. Reversing the journey gives the negative: the vector from B to A equals the negative of the vector from A to B. This sign convention matters enormously in proof questions, so build the habit early.

Position vectors

Every point in a coordinate plane can be described by its position vector: the vector from the origin O to that point. If point P has coordinates (5, -1), then the position vector of P is the column vector (5 over -1). You'll often see this written as OP with an arrow above.

Why does this matter? Because once every point has a position vector, you can describe any journey between two points using subtraction:

• The vector from A to B equals the position vector of B minus the position vector of A
• In symbols: AB = b - a, where a and b are the position vectors of A and B

This single idea, that you subtract the start from the finish, underpins almost every vector calculation in the IGCSE Additional Mathematics exam.

Worked example 1: Finding the vector between two points

Points A and B have position vectors a = (2 over 7) and b = (5 over 3). Find the vector AB.

1. Apply the formula: AB = b - a
2. Subtract component by component: (5 - 2 over 3 - 7) = (3 over -4)

So the vector AB is (3 over -4). It tells you that to get from A to B, you move 3 units right and 4 units down. Notice how BA would be (-3 over 4), the exact negative.

Magnitude of a vector

The magnitude (or modulus) of a vector is its length. For a column vector (x over y), the magnitude is sqrt(x² + y²). You'll recognise this as Pythagoras applied to the horizontal and vertical components.

The notation uses vertical bars: |a| means "the magnitude of vector a".

Worked example 2: Calculating magnitude

Find the magnitude of the vector v = (6 over -8).

1. Square each component: 6² = 36, (-8)² = 64
2. Add them: 36 + 64 = 100
3. Take the square root: sqrt(100) = 10

So |v| = 10. The magnitude is always positive, regardless of whether the components are negative.

Unit vectors

A unit vector has magnitude 1. It points in a specific direction but travels exactly one unit of distance. Think of it as the "pure direction" version of any vector.

To find the unit vector in the direction of v, divide v by its magnitude:

Unit vector = v / |v|

Worked example 3: Finding a unit vector

Find the unit vector in the direction of v = (3 over 4).

1. Find the magnitude: |v| = sqrt(9 + 16) = sqrt(25) = 5
2. Divide each component by 5: (3/5 over 4/5) = (0.6 over 0.8)

You can verify: sqrt(0.6² + 0.8²) = sqrt(0.36 + 0.64) = sqrt(1) = 1. It works.

Unit vectors appear in questions that ask you to express a vector "in the direction of" another vector with a given magnitude. The method is always the same: find the unit vector first, then multiply by the desired magnitude.

Adding and subtracting vectors

Vector addition and subtraction are done component by component. If a = (a₁ over a₂) and b = (b₁ over b₂), then:

• a + b = (a₁ + b₁ over a₂ + b₂)
• a - b = (a₁ - b₁ over a₂ - b₂)

Geometrically, addition follows the "nose-to-tail" rule. Place the start of b at the end of a, and the resultant vector runs from the start of a to the end of b. Subtraction is the same as adding the negative: a - b = a + (-b).

Worked example 4: Combining vectors

Given p = (4 over -1) and q = (-2 over 5), find:

(a) p + q

(b) p - 2q

1. (a) Add component by component: (4 + (-2) over -1 + 5) = (2 over 4)
2. (b) First find 2q = (-4 over 10). Then subtract: (4 - (-4) over -1 - 10) = (8 over -11)

Part (b) combines scalar multiplication and subtraction in one step. These multi-operation questions are common on both Paper 1 and Paper 2 of the Cambridge IGCSE Additional Mathematics exam, and the key is taking it one operation at a time.

Scalar multiplication

Multiplying a vector by a scalar (a plain number) scales each component by that number. If k is a scalar and a = (a₁ over a₂), then ka = (ka₁ over ka₂).

What does this look like geometrically?

• If k > 1, the vector stretches longer in the same direction
• If 0 < k < 1, the vector shrinks but keeps its direction
• If k < 0, the vector reverses direction and scales by |k|
• If k = 0, you get the zero vector (0 over 0)

An essential consequence: two vectors are parallel if and only if one is a scalar multiple of the other. This fact is the engine behind most geometric proof questions.

Worked example 5: Proving parallelism

Show that the vectors u = (6 over -9) and v = (-2 over 3) are parallel.

1. Check whether u = kv for some scalar k
2. From the first component: 6 = k(-2), so k = -3
3. Check the second component: -9 = (-3)(3) = -9. Confirmed.

Since u = -3v, the vectors are parallel. The negative scalar tells you they point in opposite directions.

Geometric applications: midpoints and ratios

Vectors give you a clean way to find positions along a line segment without plotting anything on graph paper.

If M is the midpoint of AB, then the position vector of M is:

m = (a + b) / 2

More generally, if a point P divides AB in the ratio m : n, then:

p = a + [m / (m + n)] (AB) = a + [m / (m + n)] (b - a)

Worked example 6: Point dividing a segment

A has position vector (1 over 2) and B has position vector (7 over 8). Point P divides AB in the ratio 2 : 1. Find the position vector of P.

1. Find AB: b - a = (7 - 1 over 8 - 2) = (6 over 6)
2. The fraction along the segment is 2/(2+1) = 2/3
3. Find the position vector: p = (1 over 2) + (2/3)(6 over 6) = (1 over 2) + (4 over 4) = (5 over 6)

You can verify by checking that AP : PB = 2 : 1. AP = (5 - 1 over 6 - 2) = (4 over 4), and PB = (7 - 5 over 8 - 6) = (2 over 2). Since (4 over 4) = 2(2 over 2), the ratio is indeed 2 : 1.

Geometric proofs with vectors

Proof questions are where vectors really earn their keep. The typical pattern goes like this: you're given a geometric figure with some vectors labelled, and you need to prove that certain lines are parallel, or that three points are collinear (lie on the same straight line).

Your toolkit for these proofs is small but effective:

• Parallel lines: Show that one vector is a scalar multiple of the other
• Collinear points: Show that vectors AB and AC (or AB and BC) are parallel. If they share a common point and are parallel, the three points must lie on one line.
• Route finding: Express an unknown vector as a chain of known vectors, using the nose-to-tail rule

Worked example 7: Collinearity proof

OA = a and OB = b. Point C has position vector 3a - 2b. Show that A, B, and C are collinear when a = (2 over 1) and b = (-1 over 2).

1. Find AB = b - a = (-1 - 2 over 2 - 1) = (-3 over 1)
2. Find AC = c - a = (3a - 2b) - a = 2a - 2b = 2(a - b) = 2(2 - (-1) over 1 - 2) = 2(3 over -1) = (6 over -2)
3. Check parallelism: AC = (6 over -2) = -2(-3 over 1) = -2 AB

Since AC is a scalar multiple of AB and they share the point A, the three points A, B, and C are collinear. The scalar -2 tells you that C lies on the line through A and B, beyond A in the direction away from B, at twice the distance AB from A.

Common mistakes and how to sidestep them

Self-check questions

Try these yourself before checking the answers. Working through them with pen and paper is the best way to build confidence.

1. Points A and B have position vectors (3 over -2) and (-1 over 6). Find the vector AB and its magnitude.
2. Find the unit vector in the direction of (5 over -12).
3. Given a = (2 over -3) and b = (4 over 1), find 3a - 2b.
4. P has position vector (1 over 5) and Q has position vector (9 over 1). Find the position vector of the point that divides PQ in the ratio 3 : 1.
5. Show that u = (8 over -6) and v = (12 over -9) are parallel, and state the scalar linking them.

Exam strategy for vector questions

Vector questions on the Cambridge IGCSE Additional Mathematics papers usually fall into two styles. The first style gives you coordinates and asks for calculations: find a vector, its magnitude, or a unit vector. These are direct and reward careful arithmetic. The second style presents a geometric figure with vectors labelled in terms of a and b, and asks you to prove something about the figure.

For calculation questions, set your working out clearly in column vector form. Write each step on a new line so you can spot sign errors before they propagate. If a question asks for magnitude, leave your answer in surd form unless told otherwise.

For proof questions, your opening move should always be to express the target vector as a chain of known vectors. Write the route explicitly: "AC = AO + OC" or "AC = AB + BC". Once you have the expression, simplify and look for scalar multiples. State your conclusion clearly: "Since XY = 2PQ, the lines XY and PQ are parallel." Don't leave the examiner to infer it.

One final thought. Vectors reward neatness more than most topics. A small sign error in the second component cascades through every subsequent line. Take an extra ten seconds to double-check your subtraction at the start, and you'll save yourself minutes of confusion later.