Vector arithmetic and proof - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Vector arithmetic and proof for Edexcel GCSE Maths (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.

28 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

\(\mathbf{p} = \begin{pmatrix}4\\-3\end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix}-2\\5\end{pmatrix}\)

Work out \(2\mathbf{p} - 3\mathbf{q}\) as a column vector.

Model answer

\(2\mathbf{p} = \begin{pmatrix}8\\-6\end{pmatrix}\), \(3\mathbf{q} = \begin{pmatrix}-6\\15\end{pmatrix}\)
\(2\mathbf{p} - 3\mathbf{q} = \begin{pmatrix}8 - (-6)\\-6 - 15\end{pmatrix} = \begin{pmatrix}14\\-21\end{pmatrix}\)

Explanation

Multiplying a vector by a scalar multiplies both components. Then subtract component by component. The most common mistake is with signs: subtracting −6 is the same as adding 6.

Mark scheme

1 mark
M1 for \(2\mathbf{p} = \begin{pmatrix}8\\-6\end{pmatrix}\) or \(3\mathbf{q} = \begin{pmatrix}-6\\15\end{pmatrix}\)
or for one correct component in the final answer; correct answer with no working scores full marks
1 mark
A1 for \(\begin{pmatrix}14\\-21\end{pmatrix}\)
cao

Question 2

2 marks

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

Kai works out \(3\mathbf{a} - 2\mathbf{b}\). Here is his working.

\(3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix}6\\-15\end{pmatrix} - \begin{pmatrix}-2\\8\end{pmatrix} = \begin{pmatrix}4\\-7\end{pmatrix}\)

Kai's answer is wrong.
What mistake has Kai made? Work out the correct answer.

Model answer

Kai's \(3\mathbf{a}\) and \(2\mathbf{b}\) are correct, but \(\begin{pmatrix}6 + (-2)\\-15 + 8\end{pmatrix} = \begin{pmatrix}4\\-7\end{pmatrix}\): he has added them instead of subtracting.
Correct: \(\begin{pmatrix}6 - (-2)\\-15 - 8\end{pmatrix} = \begin{pmatrix}8\\-23\end{pmatrix}\)

Explanation

Subtracting a vector means subtracting each component. Subtracting a negative number is the same as adding: 6 − (−2) = 8. A good check is that subtracting \(2\mathbf{b}\) should move the answer in the opposite direction to \(2\mathbf{b}\).

Mark scheme

1 mark
C1 for a correct explanation, e.g. Kai has added the vectors instead of subtracting \(2\mathbf{b}\)
accept "he did 6 − 2 instead of 6 − (−2), and −15 + 8 instead of −15 − 8"
1 mark
B1 for \(\begin{pmatrix}8\\-23\end{pmatrix}\)
cao

Question 3

3 marks

\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\)
P is a point on the line segment AB.
\(\overrightarrow{OP} = 0.3\mathbf{a} + 0.7\mathbf{b}\)

Find the ratio AP : PB.

Model answer

\(\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = 0.3\mathbf{a} + 0.7\mathbf{b} - \mathbf{a} = 0.7\mathbf{b} - 0.7\mathbf{a} = 0.7(\mathbf{b} - \mathbf{a})\)
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), so \(\overrightarrow{AP} = 0.7\overrightarrow{AB}\).
AP is 0.7 of AB, so PB is 0.3 of AB.
AP : PB = 0.7 : 0.3 = 7 : 3

Explanation

Find \(\overrightarrow{AP}\) and compare it with \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). Since \(\overrightarrow{AP} = 0.7\overrightarrow{AB}\), P is 70% of the way from A to B. A common mistake is to read the ratio straight from the coefficients as 3 : 7, which is the wrong way round: the coefficient of \(\mathbf{b}\) tells you how far P is from A.

Mark scheme

1 mark
M1 for \(\overrightarrow{OP} = \mathbf{a} + k(\mathbf{b} - \mathbf{a})\)
oe, e.g. \(\overrightarrow{AP} = \overrightarrow{OP} - \mathbf{a} = 0.7\mathbf{b} - 0.7\mathbf{a}\); correct answer with no working scores full marks
1 mark
M1 for \(k = 0.7\) or \(\overrightarrow{AP} = 0.7\overrightarrow{AB}\)
1 mark
A1 for 7 : 3
oe, e.g. 0.7 : 0.3

Question 4

4 marks

The diagram shows triangle ABC.
\(\overrightarrow{AB} = \mathbf{p}\) and \(\overrightarrow{AC} = \mathbf{q}\)
D is the point on BC such that BD : DC = 3 : 1.
E is the midpoint of AC.

Find \(\overrightarrow{DE}\) in terms of \(\mathbf{p}\) and \(\mathbf{q}\). Give your answer in its simplest form.

Model answer

\(\overrightarrow{BC} = -\mathbf{p} + \mathbf{q} = \mathbf{q} - \mathbf{p}\)
BD : DC = 3 : 1, so DC is \(\frac{1}{4}\) of BC: \(\overrightarrow{DC} = \frac{1}{4}(\mathbf{q} - \mathbf{p})\)
E is the midpoint of AC, so \(\overrightarrow{CE} = -\frac{1}{2}\mathbf{q}\)
\(\overrightarrow{DE} = \overrightarrow{DC} + \overrightarrow{CE} = \frac{1}{4}\mathbf{q} - \frac{1}{4}\mathbf{p} - \frac{1}{2}\mathbf{q} = -\frac{1}{4}\mathbf{p} - \frac{1}{4}\mathbf{q}\)

Explanation

The ratio 3 : 1 splits BC into 4 equal parts, and DC is 1 of them. Choose a short route from D to E along known vectors (D to C, then C to E) and take care with directions: C to E is against \(\mathbf{q}\). Collect the \(\mathbf{p}\) terms and the \(\mathbf{q}\) terms to simplify.

Mark scheme

1 mark
M1 for \(\overrightarrow{BC} = \mathbf{q} - \mathbf{p}\)
oe; correct answer with no working scores full marks
1 mark
M1 for \(\overrightarrow{DC} = \frac{1}{4}(\mathbf{q} - \mathbf{p})\) or \(\overrightarrow{AD} = \frac{1}{4}\mathbf{p} + \frac{3}{4}\mathbf{q}\)
oe, e.g. \(\overrightarrow{BD} = \frac{3}{4}(\mathbf{q} - \mathbf{p})\)
1 mark
M1 for a correct route for \(\overrightarrow{DE}\), e.g. \(\overrightarrow{DC} + \overrightarrow{CE} = \frac{1}{4}(\mathbf{q} - \mathbf{p}) - \frac{1}{2}\mathbf{q}\)
or \(\overrightarrow{DE} = \frac{1}{2}\mathbf{q} - (\frac{1}{4}\mathbf{p} + \frac{3}{4}\mathbf{q})\)
1 mark
A1 for \(-\frac{1}{4}\mathbf{p} - \frac{1}{4}\mathbf{q}\)
oe, e.g. \(-\frac{1}{4}(\mathbf{p} + \mathbf{q})\)

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