Cambridge O Level Mathematics: Vectors in Two Dimensions: Column Vectors, Magnitude and Vector Geometry (4024) – Revision Notes

Use this page in the last weeks before the exam. For full explanations and worked examples, read the study guide. It follows the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, sections 7.2, 7.3 and 7.4, which are the same for every candidate. Test yourself with the practice set, see the course hub, tick off outcomes on the checklist and find your weak spots with a free diagnostic.

Key definitions

Term	Meaning
Vector	A quantity with size (magnitude) and direction
Column vector (x, y)	x units across (right positive), y units up (up positive)
Scalar	An ordinary number used to multiply a vector
Directed line segment	A line with an arrow; represents a vector
Position vector	Vector from the origin O to a point: OA = a
Magnitude	Length of a vector, written |a| or |AB|
Parallel vectors	One is a scalar multiple of the other
Collinear points	Points on one straight line

Vectors are printed as AB or a. In handwriting, use an arrow over AB and underline a.

Formulas and rules

Result	Rule
Add	(a₁, a₂) + (b₁, b₂) = (a₁ + b₁, a₂ + b₂)
Subtract	(a₁, a₂) − (b₁, b₂) = (a₁ − b₁, a₂ − b₂)
Scalar multiple	k(x, y) = (kx, ky)
Reverse	BA = −AB
Triangle rule	AB + BC = AC
Between two points	AB = OB − OA = b − a
Midpoint of AB	OM = ½(a + b)
Magnitude	|(x, y)| = √(x² + y²)
Scaled magnitude	|k a| = |k| × |a|
Parallel test	AB = k CD for a number k
Ratio	If AP : PB = m : n, then AP = m/(m + n) of AB

Method in steps

Find the column vector between two points

1. Write both coordinates.
2. Subtract: end point minus start point, x then y.
3. Check the direction: a move left or down gives a negative entry.

Express a vector in terms of a and b

1. Sketch and mark the given vectors with arrows.
2. Choose a route from the start to the end using known vectors.
3. Travelling against an arrow means a minus sign.
4. Collect the a terms and the b terms.

Show that three points are collinear

1. Find two vectors joining the points, for example PQ and QR.
2. Show that one is a scalar multiple of the other.
3. State the common point.
4. Write the conclusion in words.

Use similarity from a vector result

1. Show a side is a scalar multiple of another side, with scalar k.
2. State that the sides are parallel and the triangles similar.
3. Lengths scale by k, areas by k².

Small worked reminders

Translation. A(5, 2) is mapped to A’(1, 6). The vector is (1 − 5, 6 − 2) = (−4, 4).

Combining. p = (1, 4) and q = (−3, 2). Then 2p + 3q = (2, 8) + (−9, 6) = (−7, 14).

Magnitude. |(8, 15)| = √(64 + 225) = √289 = 17. And |(1, 1)| = √2, which is not a whole number, so leave it in surd form unless a decimal is asked for.

Position vectors. OA = (2, −1) and OB = (6, 3). Then AB = (4, 4) and the midpoint has position vector ½(8, 2) = (4, 1).

Ratio. AP : PB = 1 : 3, so OP = a + ¼(b − a) = ¾a + ¼b.

Parallel. The vectors 3a − 6b and a − 2b: the first is 3 times the second, so they are parallel.

Collinear. PQ = 2a + b and QR = 6a + 3b = 3(2a + b). PQ and QR are parallel with Q in common, so P, Q and R lie on a straight line.

Extra worked reminders

Two-step route. OABC is a parallelogram with OA = a and OC = c. The midpoint M of AB lies on the side AB, and AB = OC = c, so AM = ½c. Then:

OM = OA + AM = a + ½c
CM = CO + OA + AM = −c + a + ½c = a − ½c

Magnitude of a combination. a = (3, 4) and b = (5, 12). Then |a| = 5 and |b| = 13, but a + b = (8, 16) has magnitude √(64 + 256) = √320 = 8√5. This is not 5 + 13 = 18.

Finding an unknown. (k, 9) is parallel to (2, 3). Since (2, 3) × 3 = (6, 9), k = 6. In general, compare the entries that you can match, then scale the other one.

Answer type. Always ask whether the answer has the right type. A vector question asked for a vector should end in a column or in letters. A question asked for a length should end in a single positive number.

Common question styles

You are told	You are asked	Start with
Coordinates of two points	The vector or its length	End minus start, then Pythagoras
OA = a and OB = b with a point on AB	Position vector of the point	AB = b − a, then the fraction
Two vectors in a and b	Whether they are parallel	Factorise out a common multiple
Three position vectors	Whether the points are collinear	Find two joining vectors
A line parallel to a side	Area or length of a smaller triangle	Scale factor k, then k² for area
Two routes to one point	An equation for an unknown	Compare the a parts and the b parts

Checking your work

• Test a route result with numbers: if you claim AB + BC = AC, add the columns and see.
• For a point on a line, the coefficients of a and b should add to 1.
• A magnitude must be positive. If you get a negative, a sign has slipped before squaring.
• If a proof says “show that”, write every line, including the final sentence.
• Sketch a quick diagram even when the question gives none. Arrows make route errors easy to see.

Must-know distinctions

• Vector against magnitude. AB is a vector with direction. |AB| is a single non-negative number.
• Position vector against displacement. OA starts at the origin. AB can be anywhere.
• Parallel against collinear. Parallel needs a scalar multiple. Collinear also needs a shared point.
• AB against BA. Opposite direction, opposite signs.
• Ratio part against fraction. In a ratio 2 : 3, the first part is 2/5 of the whole.
• Length scale against area scale. A vector multiple k means lengths scale by k and areas by k².

Quick self-test

1. Write the column vector that translates (3, −2) to (−1, 4).
2. Given a = (5, −1) and b = (−2, 3), find 2a − b.
3. Find |(−9, 12)|.
4. The vector (k, 4) has magnitude 5 and k is positive. Find k.
5. OA = (4, 7) and OB = (−2, 3). Find AB.
6. Find the position vector of the midpoint of the line joining (2, 8) and (6, −2).
7. In triangle OAB, OA = a and OB = b. Express AB.
8. Is the vector 6a − 4b parallel to 9a − 6b? Give a reason.
9. AP : PB = 3 : 2. Write AP as a fraction of AB.
10. Two triangles are similar, and lengths in the small one are 3/4 of those in the large one. The large area is 80 cm². Find the small area.
11. Find the vector that takes a point back to where it started after the translations (2, −5) and then (−6, 1).
12. OP = 3a + b and OQ = 9a + 3b. Show that O, P and Q are collinear.

Answers

1. (−1 − 3, 4 − (−2)) = (−4, 6).
2. (10, −2) − (−2, 3) = (12, −5).
3. √(81 + 144) = √225 = 15.
4. k² + 16 = 25, so k = 3.
5. AB = OB − OA = (−2 − 4, 3 − 7) = (−6, −4).
6. ½((2, 8) + (6, −2)) = ½(8, 6) = (4, 3).
7. AB = OB − OA = b − a.
8. Yes. 9a − 6b = 1.5(6a − 4b), a scalar multiple, so parallel.
9. 3/5.
10. Area ratio is (3/4)² = 9/16, so the small area is 9/16 × 80 = 45 cm².
11. The two translations together give (2 − 6, −5 + 1) = (−4, −4). The vector that undoes them is (4, 4).
12. OQ = 3 OP, so OP and OQ are parallel. They share the point O, so O, P and Q are collinear.

Where marks are usually lost

• Reversing a subtraction and giving OA − OB for AB, which flips both signs.
• Dropping a minus sign when subtracting a negative entry, such as 5 − (−2).
• Leaving a magnitude as a bare √ without simplifying when the sum is a perfect square, or rounding a surd too early in a multi-step question.
• Writing “|AB| = (3, 4)” so that a vector and its magnitude are confused.
• Finishing a collinearity proof with the working but no sentence naming the shared point.
• Showing parallel vectors and stating “so collinear” without the common point.
• Reading a ratio 2 : 3 as 2/3 of the line instead of 2/5.
• Scaling areas by the length factor rather than its square.
• Using letters that the question did not define, instead of a and b.
• Mixing up the order of letters, so that the final answer points the wrong way.
• Leaving an unsimplified answer such as (2/5)a + (3/5)b − (1/5)b when asked for a single expression.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, published by Cambridge International. Sections 7.2, 7.3 and 7.4.