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Edexcel GCSE Maths (Non-calculator) Smart practice A mixed round aimed at your weak spots
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Number
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Algebra
Unit 3
Ratio, proportion and rates of change
Unit 4
Geometry and measures
Unit 5
Probability
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Statistics
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Vector arithmetic and proof

Practise 37 exam-style questions on Vector arithmetic and proof for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

  • 37 questions
  • 1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

\(\mathbf{p} = \begin{pmatrix}2\\-4\end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix}-1\\5\end{pmatrix}\)

Kim works out \(3\mathbf{p} - \mathbf{q}\). Here is her working.
\[3\mathbf{p} - \mathbf{q} = \begin{pmatrix}6\\-12\end{pmatrix} - \begin{pmatrix}-1\\5\end{pmatrix} = \begin{pmatrix}5\\-17\end{pmatrix}\]
Kim's answer is wrong.

Describe the mistake Kim has made and work out the correct answer.

Model answer

Kim has made a mistake with the top numbers: she worked out 6 − 1 = 5, but it should be 6 − (−1) = 6 + 1 = 7.
The bottom number is correct: −12 − 5 = −17.
\(3\mathbf{p} - \mathbf{q} = \begin{pmatrix}7\\-17\end{pmatrix}\)

Explanation

When you subtract a column vector, subtract each component separately and take care with negative numbers. Subtracting a negative number is the same as adding, so 6 − (−1) = 7.

Mark scheme
  1. 1 mark

    C1 for explaining the error in the top component: she has worked out 6 − 1 instead of 6 − (−1)

    oe, e.g. 'she subtracted 1 instead of adding 1', 'subtracting −1 means adding 1'

  2. 1 mark

    B1 for \(\begin{pmatrix}7\\-17\end{pmatrix}\)

    cao

Question 2

2 marks

The diagram shows triangle OAB.
M is the midpoint of AB.
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\)

Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Give your answer in its simplest form.

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle OAB. Line segment OM is drawn. M lies on side AB. OA has an arrow pointing from O to A and is labelled with the vector a (printed in bold). OB has an arrow pointing from O to B and is labelled with the vector b (printed in bold).

Model answer

\(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\)
\(\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

Explanation

Find a route from O to M along lines you know: O to A, then half of the way from A to B. The vector from A to B is \(\mathbf{b} - \mathbf{a}\) (back along \(\mathbf{a}\), then along \(\mathbf{b}\)), not \(\mathbf{a} - \mathbf{b}\).

Mark scheme
  1. 1 mark

    M1 for \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), or a correct route such as \(\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})\)

    oe, e.g. \(\mathbf{b} + \frac{1}{2}(\mathbf{a} - \mathbf{b})\); correct answer with no working scores full marks

  2. 1 mark

    A1 for \(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

    oe, e.g. \(\frac{1}{2}(\mathbf{a} + \mathbf{b})\)

Question 3

3 marks

The diagram shows a trapezium OABC.
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OC} = \mathbf{c}\) and \(\overrightarrow{CB} = 2\mathbf{a}\)
M is the midpoint of AB.

Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\). Give your answer in its simplest form.

Geometry diagram, marked "Diagram NOT accurately drawn". Trapezium OABC. Line segment OM is drawn. M lies on side AB. OA has an arrow pointing from O to A and is labelled with the vector a (printed in bold). OC has an arrow pointing from O to C and is labelled with the vector c (printed in bold). CB has an arrow pointing from C to B and is labelled with the vector 2a (printed in bold).

Model answer

\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OC} + \overrightarrow{CB} = -\mathbf{a} + \mathbf{c} + 2\mathbf{a} = \mathbf{a} + \mathbf{c}\)
\(\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c} = \frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\)

Explanation

There is no vector given along AB, so find it with a route round the trapezium: A to O, O to C, C to B. Then go from O to A and half way along AB. Going against an arrow changes the sign of that vector.

Mark scheme
  1. 1 mark

    M1 for \(\overrightarrow{AB} = \mathbf{a} + \mathbf{c}\) (from \(-\mathbf{a} + \mathbf{c} + 2\mathbf{a}\)) or \(\overrightarrow{OB} = \mathbf{c} + 2\mathbf{a}\)

    correct answer with no working scores full marks

  2. 1 mark

    M1 for \(\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{a} + \mathbf{c})\) or \(\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{c} + 2\mathbf{a})\)

    ft their \(\overrightarrow{AB}\)

  3. 1 mark

    A1 for \(\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\)

    oe, e.g. \(\frac{1}{2}(3\mathbf{a} + \mathbf{c})\)

Question 4

4 marks

The diagram shows triangle OAB.
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\)
The line OA is extended to the point P so that OA = AP.
Q is the midpoint of AB.
R is the point on OB such that OR : RB = 2 : 1.

Prove that P, Q and R lie on a straight line.

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle OAB. Line segment AP is drawn. Q lies on side AB. R lies on side OB. OA has an arrow pointing from O to A and is labelled with the vector a (printed in bold). OB has an arrow pointing from O to B and is labelled with the vector b (printed in bold).

Model answer

\(\overrightarrow{OP} = 2\mathbf{a}\), \(\overrightarrow{OQ} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\), \(\overrightarrow{OR} = \frac{2}{3}\mathbf{b}\)
\(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = -\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{b} - 3\mathbf{a})\)
\(\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = -2\mathbf{a} + \frac{2}{3}\mathbf{b} = \frac{2}{3}(\mathbf{b} - 3\mathbf{a})\)
So \(\overrightarrow{PR} = \frac{4}{3}\overrightarrow{PQ}\): PR is parallel to PQ and they share the point P.
Therefore P, Q and R lie on a straight line.

Explanation

Use position vectors from O for all three points, then subtract to find two vectors joining them. Factorising each to a multiple of the same vector, \(\mathbf{b} - 3\mathbf{a}\), shows they are parallel; the common point P completes the proof.

Mark scheme
  1. 1 mark

    P1 for \(\overrightarrow{PQ} = -\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

    oe, e.g. \(-\mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})\); or \(\overrightarrow{QR} = -\frac{1}{2}\mathbf{a} + \frac{1}{6}\mathbf{b}\)

  2. 1 mark

    P1 for \(\overrightarrow{PR} = -2\mathbf{a} + \frac{2}{3}\mathbf{b}\)

    oe, or a second correct vector joining two of P, Q, R

  3. 1 mark

    P1 for showing one is a multiple of the other, e.g. \(\overrightarrow{PQ} = \frac{1}{2}(\mathbf{b} - 3\mathbf{a})\) and \(\overrightarrow{PR} = \frac{2}{3}(\mathbf{b} - 3\mathbf{a})\), or \(\overrightarrow{PR} = \frac{4}{3}\overrightarrow{PQ}\)

  4. 1 mark

    C1 for a complete proof: the vectors are parallel and share a common point (P), so P, Q and R lie on a straight line

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