Are vectors on the Foundation paper as well as Higher?

Partly. G24 and the column vector half of G25 are crossover content, so describing a translation as a column vector and adding, subtracting or scaling column vectors can appear at either tier. The last clause of G25, using vectors to construct geometric arguments and proofs, is Higher only, which is why questions that ask you to express a vector in terms of a and b and then prove something never reach a Foundation paper.

How many marks are vector questions worth on Edexcel Higher?

A column vector calculation is usually 2-3 marks. The geometric version is typically 4-6 marks and sits late in the paper, split into one part asking for a vector in terms of a and b and one part asking for a proof or a ratio. Not every paper carries the longer version, so the total across a series varies.

Is there a vector formula on the Edexcel formulae sheet?

No. Nothing from G24 or G25 is printed on the Exam Aid sheet issued with 1MA1 papers. The one thing you might reach for a formula for is the magnitude of a column vector, and that is Pythagoras applied to the two components: square them, add, take the square root. Pythagoras itself is on the sheet.

How do you prove three points are collinear using vectors?

Find two vectors joining the three points, for example AB and BC, and show one is a scalar multiple of the other, which makes them parallel. Then state that the two vectors share the point B, so the three points lie on one straight line rather than on two separate parallel lines. Edexcel puts a mark on that closing sentence, so write both halves of it.

Vectors - Edexcel GCSE Higher Maths

Typically 4-6 marks per paper, usually one longer question near the end · Spec 1MA1

A vector has a size and a direction and nothing else, so $\binom{5}{-2}$ means five to the right and two down wherever you start drawing it. Edexcel splits the work across two references. G24 is translations written as 2D column vectors, which both tiers are assessed on. G25 covers adding, subtracting and scaling vectors in column form and in diagrams, and it finishes with a clause that is Higher tier only: using vectors to construct geometric arguments and proofs. That clause is where the topic gets its reputation, because the arithmetic in front of it is short and the proof behind it is not.

Column vector questions can turn up anywhere on Papers 1, 2 and 3, usually for 2-3 marks, and having a calculator changes nothing about them. The geometric question behaves differently. It sits in the last third of the paper, runs to 4-6 marks, and normally arrives in two parts: part (a) wants a vector such as $\overrightarrow{ON}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ , given in its simplest form, and part (b) wants a proof, that two lines are parallel, that three points lie on one straight line, or the ratio in which a point cuts a segment. Part (b) normally carries a communication mark, awarded for the sentence you write rather than for the algebra above it. On the real paper the figure comes with a diagram labelled 'Diagram NOT accurately drawn', so what a length looks like tells you nothing until a ratio in the question says otherwise.

Two habits decide how many of those marks survive. The first is writing the route down before you substitute any letters: $\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON}$ first, then the $\mathbf{a}$ and $\mathbf{b}$ . Examiners see scripts where the right idea is buried in unlabelled working that cannot be awarded method marks, and plenty more where a bracket has gone missing and $\mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a})$ has become $\mathbf{a} + \frac{1}{4}\mathbf{b} - \mathbf{a}$ . The second is converting a ratio into a fraction: AN : NB = 1 : 3 puts N one quarter of the way along AB, not one third. Then factorise, because $\frac{1}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$ hides the multiple that $\frac{1}{4}(\mathbf{a} + \mathbf{b})$ makes obvious, and write the conclusion out in words.

Worked example

OACB is a parallelogram, with vertices in that order round the shape, so its diagonals are OC and AB.

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OB} = \mathbf{b}

. M is the midpoint of OA. N is the point on AB such that AN : NB = 1 : 3. Prove that MN is parallel to OC.

[5 marks]

Travel from A to B through O: $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}$ . The ratio 1 : 3 has four parts, so N is one quarter of the way along and $\overrightarrow{AN} = \frac{1}{4}(\mathbf{b} - \mathbf{a})$ .
M1: correct expression for AB, with the ratio converted to one quarter rather than one third
$\overrightarrow{ON} = \overrightarrow{OA} + \overrightarrow{AN} = \mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a}) = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$ .
M1: a full route to N; the bracket must be expanded before the $\mathbf{a}$ terms are collected
M is the midpoint of OA, so $\overrightarrow{OM} = \frac{1}{2}\mathbf{a}$ and $\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\frac{1}{2}\mathbf{a} + \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$ .
M1: MO is the reverse of OM, so it enters the sum with a minus sign
Collect and factorise: $\overrightarrow{MN} = \frac{1}{4}\mathbf{a} + \frac{1}{4}\mathbf{b} = \frac{1}{4}(\mathbf{a} + \mathbf{b})$ .
A1: simplest form; leaving the two quarters unfactorised keeps the scalar multiple hidden
In the parallelogram $\overrightarrow{AC} = \overrightarrow{OB} = \mathbf{b}$ , so $\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$ . That gives $\overrightarrow{MN} = \frac{1}{4}\overrightarrow{OC}$ . Since MN is a scalar multiple of OC, MN and OC are parallel.
C1: the communication mark, for naming the multiple and stating the conclusion as a sentence

Practice questions

These are original questions in Edexcel 1MA1 Higher style. There are no diagrams, so every figure is spelled out in words: which points are joined, which vectors are given, and exactly where a midpoint or a ratio point sits on a line. Read those sentences twice and sketch the shape before you trace a path. Nothing here needs a calculator, so the column vector work doubles as Paper 1 practice.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [2 marks]

$\mathbf{a} = \binom{3}{-1}$ and $\mathbf{b} = \binom{2}{5}$ . Work out $\mathbf{a} + \mathbf{b}$ as a column vector.

top = bottom =
Question 2 Warm-up [1 mark]

OAB is a triangle. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . Which expression gives $\overrightarrow{AB}$ ?

Answer options for question 2
$\mathbf{b} - \mathbf{a}$ $\mathbf{a} - \mathbf{b}$ $\mathbf{a} + \mathbf{b}$ $-\mathbf{a} - \mathbf{b}$
Question 3 Warm-up [2 marks]

$\mathbf{a} = \binom{1}{-3}$ and $\mathbf{b} = \binom{4}{5}$ . Work out $2\mathbf{a} + \mathbf{b}$ as a column vector.

top = bottom =
Question 4 Exam pace [3 marks]

$\mathbf{p} = \binom{5}{-2}$ and $\mathbf{q} = \binom{-3}{-4}$ . Work out $\mathbf{p} - 3\mathbf{q}$ as a column vector.

top = bottom =
Question 5 Exam pace [1 mark]

Which pair of column vectors is parallel?

Answer options for question 5
$\binom{2}{3}$ and $\binom{6}{9}$ $\binom{2}{3}$ and $\binom{3}{2}$ $\binom{2}{3}$ and $\binom{6}{8}$ $\binom{2}{3}$ and $\binom{4}{5}$
Question 6 Stretch [3 marks]

$\mathbf{a} = \binom{3}{-2}$ and $2\mathbf{a} + 3\mathbf{b} = \binom{0}{5}$ . Write $\mathbf{b}$ as a column vector.

top = bottom =

Common mistakes examiners see

Writing $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$ when $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . That is $\overrightarrow{BA}$ , so every sign after it is reversed and the answer comes out as the negative of the mark scheme's.

Write the two-leg route before the letters: $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}$ , which forces the $-\mathbf{a}$ . The shortcut worth remembering is destination minus start, so AB is $\mathbf{b} - \mathbf{a}$ .
Turning a ratio into the wrong fraction. Reading AN : NB = 1 : 3 as N being one third of the way along AB, or splitting AB into thirds when the question says AP : PB = 2 : 3.

Add the two numbers first and use that as the denominator: 1 : 3 gives four parts, so $\overrightarrow{AN} = \frac{1}{4}\overrightarrow{AB}$ , and 2 : 3 gives five, so $\overrightarrow{AP} = \frac{2}{5}\overrightarrow{AB}$ . Then check which end the ratio starts from.
Losing the bracket when the fraction multiplies a two-term vector, so $\mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a})$ is written as $\mathbf{a} + \frac{1}{4}\mathbf{b} - \mathbf{a}$ and the $\mathbf{a}$ terms wrongly cancel.

Keep the bracket in place until you have written the multiplication out in full, then expand: $\frac{1}{4}\mathbf{b} - \frac{1}{4}\mathbf{a}$ . The fraction hits both terms, exactly as it would in ordinary algebra.
Mishandling a scaled column vector: multiplying only the top component, so $3\binom{4}{-2}$ becomes $\binom{12}{-2}$ , or reading $\binom{3}{4}$ as the fraction three quarters and converting it.

Treat the two components as separate sums that never mix, and multiply both by the scalar. A column vector is a pair of instructions, across then up, and it has no fraction bar.
Doing the algebra correctly and then writing nothing, or writing a vague reason such as 'they have the same gradient' or 'they both have $\mathbf{a}$ and $\mathbf{b}$ in them'. For collinearity, showing two vectors are parallel and stopping there.

Finish with a sentence naming the multiple: ' $\overrightarrow{MN} = \frac{1}{4}\overrightarrow{OC}$ , so MN is parallel to OC'. For three points on a line, add the shared point: parallel plus a common point is what rules out two separate parallel lines.

Frequently asked questions

Are vectors on the Foundation paper as well as Higher?: Partly. G24 and the column vector half of G25 are crossover content, so describing a translation as a column vector and adding, subtracting or scaling column vectors can appear at either tier. The last clause of G25, using vectors to construct geometric arguments and proofs, is Higher only, which is why questions that ask you to express a vector in terms of a and b and then prove something never reach a Foundation paper.
How many marks are vector questions worth on Edexcel Higher?: A column vector calculation is usually 2-3 marks. The geometric version is typically 4-6 marks and sits late in the paper, split into one part asking for a vector in terms of a and b and one part asking for a proof or a ratio. Not every paper carries the longer version, so the total across a series varies.
Is there a vector formula on the Edexcel formulae sheet?: No. Nothing from G24 or G25 is printed on the Exam Aid sheet issued with 1MA1 papers. The one thing you might reach for a formula for is the magnitude of a column vector, and that is Pythagoras applied to the two components: square them, add, take the square root. Pythagoras itself is on the sheet.
How do you prove three points are collinear using vectors?: Find two vectors joining the three points, for example AB and BC, and show one is a scalar multiple of the other, which makes them parallel. Then state that the two vectors share the point B, so the three points lie on one straight line rather than on two separate parallel lines. Edexcel puts a mark on that closing sentence, so write both halves of it.

Vectors - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics