OxfordAQA International GCSE Mathematics: Vectors and vector geometry (9260) – Revision Notes

These notes condense the full explanations in the study guide, so read that first if a step is unfamiliar. They are written for the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), statement G22 in section 3.3.3, Transformations, Matrices and Vectors. G22 appears only in the Extension content column, so all of it is Extension only. The specification says all content can be assessed on the Extension papers.

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Key facts

Idea	What to remember
Column vector (x, y)	x across (right is positive), y up or down (up is positive)
Names	a (underline by hand), →AB (from A to B), or a column
Equal vectors	Same size and direction, wherever they are drawn
Negative	−a reverses direction; →BA = −→AB
Vector between points	→AB = position of B minus position of A
Sum	Add the matching entries
Difference	Subtract the matching entries, or add the negative
Scalar multiple	k(x, y) = (kx, ky); parallel to the original
Commutative	a + b = b + a
Associative	(a + b) + c = a + (b + c)
Not commutative	a − b ≠ b − a; they are negatives of each other

Calculating with columns

Work on the top entries, then the bottom entries. Put brackets round any negative number.

Reminder. a = (6, −2) and b = (1, 4).

a + b  = (7, 2)
2a - 3b = (12, -4) - (3, 12) = (9, -16)

Scalar multiples multiply both entries. Parallel vectors are scalar multiples: (−4, 6) and (6, −9) are parallel because (6, −9) = −1.5(−4, 6).

Drawing vectors

• Sum: draw a, then draw b from the end of a. The sum runs from the start of a to the end of b.
• Difference: a − b = a + (−b). Draw −b from the end of a.
• Scalar multiple: 3a is the same direction and three times the length. −a is the same length and opposite.
• Parallelogram: drawing a and b from the same point and completing the shape gives a + b as one diagonal.

Method: finding a vector in a diagram

1. Choose a path from where you begin to where you finish, using only edges whose vectors you know.
2. Write the vector for each step. Going with an arrow adds it. Going against it subtracts it.
3. If a point splits a line in a ratio, turn the ratio into a fraction of that line first.
4. Collect the a terms and the b terms and give one simplified answer.

Method: ratios on a line

If X is on AB with AX : XB = m : n, then →AX = m/(m + n) of →AB.

Reminder. Triangle OAB has →OA = a and →OB = b. M halves AB, and N is on OB with ON : NB = 2 : 1. Find →MN.

OM = a + (1/2)(b - a) = (1/2)a + (1/2)b
ON = (2/3)b
MN = ON - OM = (2/3)b - (1/2)a - (1/2)b = -(1/2)a + (1/6)b

So →MN = −½a + ⅙b, or (b − 3a)/6.

Method: proofs

To show	Do this	Then say
Two lines are parallel	Show one vector is a number times the other	“So they are parallel”
Three points are collinear	Show →AB = k→BC (or →AB = k→AC)	“Parallel and share a point, so collinear”
A point is a midpoint	Show its vector from O is half the sum of the end points’ vectors from O	“So it is the midpoint”
Two lines are equal in length	Show one vector is ±1 times the other	“So they are equal in length”

Must-know distinctions

• Vector or scalar? A vector has direction. A scalar is only a size.
• Point or vector? A point is a place. A column vector is a movement that can start anywhere.
• Parallel or collinear? Parallel lines never need to meet. Collinear points are on one line, so the vectors also share a point.
• a − b or b − a? They are negatives. Check the order by asking which way the arrow points.
• →AB or →BA? Opposite vectors with opposite signs.

More worked reminders

Reminder: a vector from coordinates. C is (−4, 6) and D is (3, −2). Then →CD = (3 − (−4), −2 − 6) = (7, −8). Its negative is →DC = (−7, 8).

Reminder: solving for unknowns. If m(1, 2) + n(3, −1) = (11, 8), compare entries:

m + 3n  = 11
2m - n  = 8

From the second, n = 2m − 8. Then m + 6m − 24 = 11, so m = 5 and n = 2. Check: (5, 10) + (6, −2) = (11, 8). Always check in the original expression.

Reminder: parallelogram route. ABCD is a parallelogram with →AB = s and →AD = t.

AC = s + t
BD = -s + t = t - s
CA = -s - t

The two diagonals are →AC and →BD. Notice that →BD uses the route B to A to D.

Reminder: collinear with letters. →OP = 2a + b and →OQ = 6a + 3b. Then →OQ = 3→OP, so O, P and Q are collinear and OQ is three times OP.

Language to know

Word	Meaning
Vector	A quantity with size and direction
Scalar	A number with no direction
Column vector	A pair of entries giving movement across and up or down
Scalar multiple	A vector multiplied by a number
Parallel	Same direction or exactly opposite, so one is a scalar multiple of the other
Collinear	On one straight line
Resultant	The single vector equal to a sum of vectors
Midpoint	The point halfway along a line

Exam technique

• Label every vector in a diagram with an arrow or a letter before you start. A clear diagram makes the route method much easier.
• Write each route as a line of working, such as →AM = →AB + ½→BC, before simplifying. A slip in the simplifying is then easier to find.
• When a question gives a ratio, write the fraction of the whole line it represents. A ratio of 2 : 3 means ²⁄₅ of the line from the first point.
• Finish with a sentence. For a proof, say what the working shows in the words of the question.
• If a final answer is a mixture of a and b terms, check you have collected all of each. Substitute a simple pair, such as a = (1, 0) and b = (0, 1), to test an answer.
• Use the marks as a guide to the amount of work. A two-mark vector is usually one route and one simplification.

Quick self-test

1. Find (4, −3) + (−6, 5).
2. Find (9, 2) − (4, −7).
3. Find −3(2, −4).
4. A is (−3, 5) and B is (2, 1). Write →AB as a column.
5. a = (1, 4) and b = (−3, −2). Find 2a + b.
6. Are (−4, 6) and (6, −9) parallel? Give a reason.
7. Simplify →AB + →BC − →DC as one vector.
8. M is the midpoint of PQ, →OP = p and →OQ = q. Find →OM.
9. →PQ = 3a + b and →QR = a − 4b. Find →PR.
10. Find k if k(3, −1) + (1, 4) = (10, 1).
11. Is →AB = (2, −5) parallel to →CD = (−6, 10)? Give a reason.
12. X is on AB with AX : XB = 1 : 4. Write →AX as a fraction of →AB.

Answers

1. (−2, 2)
2. (5, 9)
3. (−6, 12)
4. →AB = (2 − (−3), 1 − 5) = (5, −4)
5. (2, 8) + (−3, −2) = (−1, 6)
6. Yes. (6, −9) = −1.5 × (−4, 6), so one is a scalar multiple of the other.
7. →AB + →BC = →AC. Then −→DC = →CD, so the total is →AC + →CD = →AD.
8. →OM = p + ½(q − p) = ½(p + q)
9. →PR = →PQ + →QR = 4a − 3b
10. 3k + 1 = 10 gives k = 3. Check the lower entry: −3 + 4 = 1.
11. No. (−6, 10) would need a multiple of 3 on the top (−6 = −3 × 2), but −3 × (−5) = 15, not 10.
12. →AX = ⅕→AB. The line is split into 1 + 4 = 5 equal parts.

Where marks are usually lost

• Reading →AB as the vector from B to A.
• Sign errors when subtracting a negative entry, such as 2 − (−7).
• Forgetting to multiply the second entry in a scalar multiple.
• Writing “parallel” from a diagram instead of showing a scalar multiple.
• Proving parallel but not saying that a shared point makes the points collinear.
• Giving an answer such as a + 2b + 3a without collecting like terms.
• Writing a coordinate pair where a vector was asked for.
• Missing the brackets in 2a − 3b when the second vector has negative entries.
• Using one numerical example when the question says “prove” or “show that” for letters.
• Leaving a ratio as a ratio. Convert it to a fraction of the line.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA. Statement G22, section 3.3.3 Transformations, Matrices and Vectors.