Vectors - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Vectors for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
2–10 marks each
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18 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

3 marks

The non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel. The scalars \(\lambda\) and \(\mu\) are such that
\[(2\lambda + \mu - 7)\mathbf{a} + (\lambda - 2\mu + 4)\mathbf{b} = \mathbf{0}\]
Find \(\lambda\) and \(\mu\).

Model answer

\(\mathbf{a}\) and \(\mathbf{b}\) are non-zero and not parallel, so both coefficients are zero:
\(2\lambda + \mu = 7\) and \(\lambda - 2\mu = -4\)
From the first, \(\mu = 7 - 2\lambda\), so \(\lambda - 14 + 4\lambda = -4 \Rightarrow \lambda = 2\)
\(\mu = 7 - 4 = 3\)

Explanation

If \(p\mathbf{a} + q\mathbf{b} = \mathbf{0}\) with \(q ≠ 0\), then \(\mathbf{b}\) would be a multiple of \(\mathbf{a}\), so for non-parallel vectors both coefficients must be zero. This gives two simultaneous equations with solution \(\lambda = 2\), \(\mu = 3\). This is the step used in every vector ratio problem; the common error is setting the two coefficients equal to each other instead of to zero.

Mark scheme

1 mark
B1 States or uses that both coefficients must be zero, because \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel: \(2\lambda + \mu - 7 = 0\) and \(\lambda - 2\mu + 4 = 0\); correct values \(\lambda = 2\), \(\mu = 3\) with no working score full marks
An answer of \(\lambda = 2\), \(\mu = 3\) with no working scores 3 marks.
1 mark
M1 Solves their two linear equations simultaneously to find a value of \(\lambda\) or \(\mu\)
A calculator may be used to solve the simultaneous equations.
1 mark
A1 \(\lambda = 2\) and \(\mu = 3\)
Both needed.

Question 2

4 marks

Relative to a fixed origin, the points \(A\), \(B\) and \(C\) have position vectors \((\mathbf{j} + 2\mathbf{k})\), \((3\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})\) and \((4\mathbf{i} - \mathbf{j} + 7\mathbf{k})\) respectively.

Determine, with a reason, whether angle \(BAC\) is acute, a right angle or obtuse.

Model answer

\(\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j}\), \(AB^2 = 9 + 16 = 25\)
\(\overrightarrow{AC} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\), \(AC^2 = 16 + 4 + 25 = 45\)
\(\overrightarrow{BC} = \mathbf{i} - 6\mathbf{j} + 5\mathbf{k}\), \(BC^2 = 1 + 36 + 25 = 62\)
Cosine rule: \(\cos BAC = \dfrac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC} = \dfrac{25 + 45 - 62}{2 \times 5 \times 3\sqrt{5}} = \dfrac{8}{30\sqrt{5}} > 0\)
Since \(BC^2 = 62 < AB^2 + AC^2 = 70\), \(\cos BAC > 0\) and angle \(BAC\) is acute (about \(83.2^\circ\)).

Explanation

By the cosine rule, the angle at \(A\) is acute exactly when the square of the opposite side \(BC\) is less than \(AB^2 + AC^2\), right-angled when they are equal and obtuse when it is greater. Here \(62 < 70\), so the angle is acute, a little under \(90^\circ\). Common errors are comparing the wrong side with the sum of the other two and using the lengths instead of their squares.

Mark scheme

1 mark
M1 Finds at least two of \(\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j}\), \(\overrightarrow{AC} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\), \(\overrightarrow{BC} = \mathbf{i} - 6\mathbf{j} + 5\mathbf{k}\) and their magnitudes (or squared magnitudes); a conclusion that angle \(BAC\) is acute with no working scores 0 marks
An answer of 'acute' with no working scores 0 marks: a reason is required.
1 mark
A1 \(AB^2 = 25\), \(AC^2 = 45\) and \(BC^2 = 62\)
All three needed. Accept \(AB = 5\), \(AC = 3\sqrt{5}\), \(BC = \sqrt{62}\).
1 mark
M1 Compares \(BC^2\) with \(AB^2 + AC^2\), or uses the cosine rule: \(\cos BAC = \dfrac{25 + 45 - 62}{2 \times 5 \times 3\sqrt{5}}\)
Follow through their lengths. BC must be the side opposite angle \(BAC\).
1 mark
A1 \(62 < 25 + 45 = 70\) (or \(\cos BAC = \dfrac{8}{30\sqrt{5}} > 0\), angle awrt \(83.2^\circ\), i.e. [83.15, 83.25)), so angle \(BAC\) is acute
Needs correct values, the comparison and the conclusion.

Question 3

6 marks

In the triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point \(P\) is the midpoint of \(OA\), the point \(Q\) lies on \(AB\) such that \(AQ : QB = 2 : 1\), and the point \(R\) lies on \(OB\) extended beyond \(B\) such that \(OR = 2OB\).

Show that \(P\), \(Q\) and \(R\) lie on a straight line, and find the ratio \(PQ : QR\).

Model answer

\(\overrightarrow{OP} = \dfrac{1}{2}\mathbf{a}\), \(\overrightarrow{OR} = 2\mathbf{b}\)
\(\overrightarrow{OQ} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\)
\(\overrightarrow{PQ} = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b} - \dfrac{1}{2}\mathbf{a} = -\dfrac{1}{6}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\)
\(\overrightarrow{QR} = 2\mathbf{b} - \dfrac{1}{3}\mathbf{a} - \dfrac{2}{3}\mathbf{b} = -\dfrac{1}{3}\mathbf{a} + \dfrac{4}{3}\mathbf{b}\)
\(\overrightarrow{QR} = 2\overrightarrow{PQ}\), so \(PQ\) is parallel to \(QR\); they share the point \(Q\), so \(P\), \(Q\) and \(R\) are collinear.
\(PQ : QR = 1 : 2\)

Explanation

Find the position vectors of \(P\), \(Q\) and \(R\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), then show that \(\overrightarrow{QR}\) is a scalar multiple of \(\overrightarrow{PQ}\). Parallel vectors through a common point lie on one line, and the scalar 2 gives the ratio \(PQ : QR = 1 : 2\). Common errors are taking \(\overrightarrow{AQ} = \dfrac{1}{3}\overrightarrow{AB}\) and not stating that the two parallel vectors share a point.

Mark scheme

1 mark
M1 Finds \(\overrightarrow{OQ} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: this is a 'show that' question, and the ratio must come from the vectors.
1 mark
A1 \(\overrightarrow{OQ} = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\)
1 mark
M1 Uses \(\overrightarrow{OP} = \dfrac{1}{2}\mathbf{a}\) and \(\overrightarrow{OR} = 2\mathbf{b}\) to find two of \(\overrightarrow{PQ}\), \(\overrightarrow{QR}\), \(\overrightarrow{PR}\)
1 mark
A1 \(\overrightarrow{PQ} = -\dfrac{1}{6}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\) and \(\overrightarrow{QR} = -\dfrac{1}{3}\mathbf{a} + \dfrac{4}{3}\mathbf{b}\) (or \(\overrightarrow{PR} = -\dfrac{1}{2}\mathbf{a} + 2\mathbf{b}\))
Both of the two vectors used must be correct.
1 mark
A1* \(\overrightarrow{QR} = 2\overrightarrow{PQ}\) (or \(\overrightarrow{PR} = 3\overrightarrow{PQ}\)), so \(PQ\) and \(QR\) are parallel, and as they share the point \(Q\), \(P\), \(Q\) and \(R\) are collinear
cso. Needs the scalar multiple, 'parallel' and the common point (or equivalent wording) with a conclusion. 'Parallel' alone is not enough for collinear.
1 mark
B1 \(PQ : QR = 1 : 2\)
Accept equivalent ratios. \(1 : 3\) (the ratio \(PQ : PR\)) is B0.

Question 4

8 marks

Figure 1 shows a triangle \(OAB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\).

The point \(C\) lies on \(AB\) such that \(AC : CB = 1 : 3\). The point \(D\) is the midpoint of \(OC\). The line through \(A\) and \(D\) meets \(OB\) at the point \(E\).

Find \(\overrightarrow{OC}\) and \(\overrightarrow{OD}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

You must use a vector method to find the ratio \(OE : EB\) and the ratio \(AD : DE\).

Model answer

\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
\(\overrightarrow{OC} = \mathbf{a} + \dfrac{1}{4}(\mathbf{b} - \mathbf{a}) = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\)
\(\overrightarrow{OD} = \dfrac{1}{2}\overrightarrow{OC} = \dfrac{3}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\)
\(\overrightarrow{AD} = \overrightarrow{OD} - \mathbf{a} = -\dfrac{5}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\)
\(E\) is on \(AD\) extended: \(\overrightarrow{OE} = \mathbf{a} + \lambda\left(-\dfrac{5}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\right) = \left(1 - \dfrac{5}{8}\lambda\right)\mathbf{a} + \dfrac{1}{8}\lambda\mathbf{b}\)
\(E\) is on \(OB\): \(\overrightarrow{OE} = \mu\mathbf{b}\)
\(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, so \(1 - \dfrac{5}{8}\lambda = 0\) and \(\dfrac{1}{8}\lambda = \mu\)
\(\lambda = \dfrac{8}{5}\), \(\mu = \dfrac{1}{5}\)
\(\overrightarrow{OE} = \dfrac{1}{5}\mathbf{b}\), so \(OE : EB = 1 : 4\)
\(\overrightarrow{AE} = \dfrac{8}{5}\overrightarrow{AD}\), so \(AD : DE = 1 : \dfrac{3}{5} = 5 : 3\)

Explanation

Write the position of \(E\) in two ways, as a point on the line \(AD\) and as a point on \(OB\), and equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\), which is allowed because \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel. This gives \(\lambda = \dfrac{8}{5}\) and \(\mu = \dfrac{1}{5}\), so \(OE : EB = 1 : 4\) and \(AD : DE = 5 : 3\). Common errors are taking \(\overrightarrow{AC} = \dfrac{1}{3}\overrightarrow{AB}\) from the ratio \(1 : 3\) and using the same letter for both scalars.

Mark scheme

1 mark
M1 Uses \(\overrightarrow{OC} = \overrightarrow{OA} + \dfrac{1}{4}\overrightarrow{AB}\) with \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\); or correct answers of \(\overrightarrow{OC} = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\) and \(\overrightarrow{OD} = \dfrac{3}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\) with no working
Correct answers of \(\overrightarrow{OC} = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\) and \(\overrightarrow{OD} = \dfrac{3}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\) with no working score 3 marks; the ratios with no vector working score 0 marks, as the question demands a vector method. Using \(\dfrac{1}{3}\overrightarrow{AB}\) (from the ratio 1 : 3) is M0.
1 mark
A1 \(\overrightarrow{OC} = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\)
1 mark
A1 \(\overrightarrow{OD} = \dfrac{3}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\)
A1ft: half of their \(\overrightarrow{OC}\); needs the M mark.
1 mark
M1 Finds \(\overrightarrow{AD} = -\dfrac{5}{8}\mathbf{a} + \dfrac{1}{8}\mathbf{b}\) (ft) and writes \(\overrightarrow{OE}\) in two ways: \(\overrightarrow{OE} = \mathbf{a} + \lambda\overrightarrow{AD}\) and \(\overrightarrow{OE} = \mu\mathbf{b}\)
Different letters must be used for the two scalars.
1 mark
M1 (dM1) Equates the coefficients of \(\mathbf{a}\) and of \(\mathbf{b}\) (as \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel): \(1 - \dfrac{5}{8}\lambda = 0\) and \(\dfrac{1}{8}\lambda = \mu\), and solves
Depends on the previous M mark.
1 mark
A1 \(\lambda = \dfrac{8}{5}\) and \(\mu = \dfrac{1}{5}\), i.e. \(\overrightarrow{OE} = \dfrac{1}{5}\mathbf{b}\)
1 mark
A1 \(OE : EB = 1 : 4\)
cao. Needs both M marks; it follows from \(\mu = \dfrac{1}{5}\). Accept equivalent ratios, e.g. \(\dfrac{1}{5} : \dfrac{4}{5}\).
1 mark
A1 \(AD : DE = 5 : 3\)
Needs both M marks. From \(\overrightarrow{AE} = \dfrac{8}{5}\overrightarrow{AD}\), so \(DE = \dfrac{3}{5}AD\). Accept equivalent ratios.

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