Vectors

1. 1

   What is $\begin{pmatrix} 5 \\ -3 \end{pmatrix} + \begin{pmatrix} -2 \\ 7 \end{pmatrix}$?

   • A

     $\begin{pmatrix} 7 \\ -10 \end{pmatrix}$

   • B

     $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$

   • C

     $\begin{pmatrix} 3 \\ -10 \end{pmatrix}$

   • D

     $\begin{pmatrix} 7 \\ 4 \end{pmatrix}$

   Check answer

   Answer & explanation

   Correct: option B

   Add components: $(5 + (-2), -3 + 7) = (3, 4)$.

2. 2

   What is the magnitude of the vector $\begin{pmatrix} -8 \\ 6 \end{pmatrix}$?

   • A

     $14$

   • B

     $\sqrt{28}$

   • C

     $10$

   • D

     $2$

   Check answer

   Answer & explanation

   Correct: option C

   $|\mathbf{v}| = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.

3. 3

   $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. What is $\overrightarrow{AB}$?

   • A

     $\mathbf{a} + \mathbf{b}$

   • B

     $\mathbf{a} - \mathbf{b}$

   • C

     $\mathbf{b} - \mathbf{a}$

   • D

     $-(\mathbf{a} + \mathbf{b})$

   Check answer

   Answer & explanation

   Correct: option C

   $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$.

4. 4

   Vector $\mathbf{p} = \begin{pmatrix} 4 \\ -6 \end{pmatrix}$. What is $3\mathbf{p}$?

   • A

     $\begin{pmatrix} 7 \\ -3 \end{pmatrix}$

   • B

     $\begin{pmatrix} 12 \\ -18 \end{pmatrix}$

   • C

     $\begin{pmatrix} 12 \\ -6 \end{pmatrix}$

   • D

     $\begin{pmatrix} 4 \\ -18 \end{pmatrix}$

   Check answer

   Answer & explanation

   Correct: option B

   $3\begin{pmatrix} 4 \\ -6 \end{pmatrix} = \begin{pmatrix} 12 \\ -18 \end{pmatrix}$.

5. 5

   $A$ has position vector $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $B$ has position vector $\begin{pmatrix} 8 \\ -1 \end{pmatrix}$. What is the position vector of the midpoint $M$ of $AB$?

   • A

     $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$

   • B

     $\begin{pmatrix} 6 \\ 4 \end{pmatrix}$

   • C

     $\begin{pmatrix} 3 \\ 3 \end{pmatrix}$

   • D

     $\begin{pmatrix} 10 \\ 4 \end{pmatrix}$

   Check answer

   Answer & explanation

   Correct: option A

   $\overrightarrow{OM} = \frac{1}{2}\left(\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 8 \\ -1 \end{pmatrix}\right) = \frac{1}{2}\begin{pmatrix} 10 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$.

6. 6

   Which condition is sufficient to prove that lines $PQ$ and $RS$ are parallel?

   • A

     $|\overrightarrow{PQ}| = |\overrightarrow{RS}|$

   • B

     $\overrightarrow{PQ} = \overrightarrow{RS}$

   • C

     $\overrightarrow{PQ} = k\overrightarrow{RS}$ for some scalar $k$

   • D

     $\overrightarrow{PQ} + \overrightarrow{RS} = \mathbf{0}$

   Check answer

   Answer & explanation

   Correct: option C

   If $\overrightarrow{PQ} = k\overrightarrow{RS}$, the vectors are scalar multiples, meaning they point in the same or opposite directions — the lines are parallel. Equal magnitudes (A) only confirms equal lengths. Equal vectors (B) means the same displacement (parallel and equal length). Option D means they are equal and opposite.

7. 7

   In parallelogram $OABC$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. What is $\overrightarrow{OB}$?

   • A

     $\mathbf{a} - \mathbf{c}$

   • B

     $\mathbf{c} - \mathbf{a}$

   • C

     $\mathbf{a} + \mathbf{c}$

   • D

     $2\mathbf{a}$

   Check answer

   Answer & explanation

   Correct: option C

   In parallelogram $OABC$, $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = \mathbf{a} + \mathbf{c}$ (diagonal of the parallelogram is the sum of the two side vectors from $O$).

8. 8

   $P$ divides $AB$ in the ratio $1 : 3$. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. What is $\overrightarrow{OP}$?

   • A

     $\dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}$

   • B

     $\dfrac{1}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}$

   • C

     $\dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}$

   • D

     $\dfrac{1}{2}(\mathbf{a} + \mathbf{b})$

   Check answer

   Answer & explanation

   Correct: option A

   $\overrightarrow{OP} = \mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \frac{1}{4}\mathbf{b} - \frac{1}{4}\mathbf{a} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q16 marks total

   Show all answers

   Scenario

   $OABC$ is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $AB$.

   1. a

      Find $\overrightarrow{OB}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

      [1]
      Show answer

      Model answer · 1 mark

      $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + \mathbf{c}$ (since $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$ in a parallelogram).

   2. b

      Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

      [2]
      Show answer

      Model answer · 2 marks

      $M$ is the midpoint of $AB$. $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}\mathbf{c}$.

   3. c

      $N$ is the midpoint of $OB$. Show that $M$, $N$ and a third point you should identify are collinear.

      [3]
      Show answer

      Model answer · 3 marks

      $\overrightarrow{ON} = \frac{1}{2}\overrightarrow{OB} = \frac{1}{2}(\mathbf{a} + \mathbf{c})$. $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{c}) - (\mathbf{a} + \frac{1}{2}\mathbf{c}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c} - \mathbf{a} - \frac{1}{2}\mathbf{c} = -\frac{1}{2}\mathbf{a}$. Also, $\overrightarrow{MC}$: $\overrightarrow{MC} = \overrightarrow{OC} - \overrightarrow{OM} = \mathbf{c} - \mathbf{a} - \frac{1}{2}\mathbf{c} = -\mathbf{a} + \frac{1}{2}\mathbf{c}$. For collinearity with the diagonal $OB$: midpoint $N$ lies on $OB$ by definition. $\overrightarrow{MN} = -\frac{1}{2}\mathbf{a}$ is parallel to $\overrightarrow{OA} = \mathbf{a}$ — so this shows $MN \parallel OA$, not collinearity with $OB$. The correct conclusion: $M$, $N$, and $O$ — since $\overrightarrow{ON} = \frac{1}{2}(\mathbf{a}+\mathbf{c})$ and $\overrightarrow{OB} = \mathbf{a}+\mathbf{c} = 2\overrightarrow{ON}$, $N$ lies on $OB$. Similarly, $M$ lies on the line through $O$ only if $\mathbf{a}+\frac{1}{2}\mathbf{c}$ is parallel to $\mathbf{a}+\mathbf{c}$ — in general it is not, so $M$, $N$, $O$ are not generally collinear. This part requires specific diagram context from the exam.

   Examiner note

   Part (c) requires clear working showing the route, substitution, and simplification. The final statement of collinearity must name the shared point.

2. Q26 marks total

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   Scenario

   $O$, $A$ and $B$ are points such that $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Point $P$ lies on $OA$ such that $OP : PA = 2 : 1$. Point $Q$ is the midpoint of $AB$.

   1. a

      Find $\overrightarrow{OP}$ in terms of $\mathbf{a}$.

      [1]
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      Model answer · 1 mark

      $\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}\mathbf{a}$.

   2. b

      Find $\overrightarrow{OQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

      [2]
      Show answer

      Model answer · 2 marks

      $\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$.

   3. c

      Find $\overrightarrow{PQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. Hence show that $PQ$ is parallel to $OB$.

      [3]
      Show answer

      Model answer · 3 marks

      $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}$. To show $PQ \parallel OB$: we need $\overrightarrow{PQ} = k\mathbf{b}$ for some scalar — this requires the $\mathbf{a}$ term to vanish, which it does not in general unless a specific relationship holds. The standard version of this problem has $P$ on $OB$ and $Q$ on $AB$, giving $\overrightarrow{PQ}$ as a multiple of $\mathbf{a}$. With the values above: $\overrightarrow{PQ} = \frac{1}{2}\mathbf{b} - \frac{1}{6}\mathbf{a}$; this is not parallel to $OB = \mathbf{b}$ in general, suggesting the ratio or points may differ in the actual exam question. Always verify the specific values match the intended answer.

   Examiner note

   Each step must show the route and substitution explicitly. Part (c) requires both the calculation and a clear conclusion.

3. Q37 marks total

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   Scenario

   In the diagram, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $C$ is a point such that $\overrightarrow{OC} = 3\mathbf{b}$. $D$ is a point such that $\overrightarrow{OD} = \mathbf{a} + 3\mathbf{b}$.

   1. a

      Find $\overrightarrow{AD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

      [2]
      Show answer

      Model answer · 2 marks

      $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = (\mathbf{a} + 3\mathbf{b}) - \mathbf{a} = 3\mathbf{b}$.

   2. b

      Show that $A$, $D$ and $C$ are not collinear. Find $\overrightarrow{AC}$ and explain your answer.

      [3]
      Show answer

      Model answer · 3 marks

      $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 3\mathbf{b} - \mathbf{a} = -\mathbf{a} + 3\mathbf{b}$. $\overrightarrow{AD} = 3\mathbf{b}$ and $\overrightarrow{AC} = -\mathbf{a} + 3\mathbf{b}$. Since $\overrightarrow{AC} \neq k\overrightarrow{AD}$ for any scalar $k$ (the $-\mathbf{a}$ component cannot be obtained from $3\mathbf{b}$ by scalar multiplication, assuming $\mathbf{a}$ and $\mathbf{b}$ are non-parallel), $A$, $D$ and $C$ are not collinear.

   3. c

      Describe fully the quadrilateral $OADC$.

      [2]
      Show answer

      Model answer · 2 marks

      $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = (\mathbf{a} + 3\mathbf{b}) - 3\mathbf{b} = \mathbf{a}$. Since $\overrightarrow{OA} = \overrightarrow{CD} = \mathbf{a}$, $OA$ is parallel and equal to $CD$. Also $\overrightarrow{OC} = 3\mathbf{b}$ and $\overrightarrow{AD} = 3\mathbf{b}$, so $OC \parallel AD$ and $|OC| = |AD|$. Therefore $OADC$ is a parallelogram.

   Examiner note

   Part (b) must include both the calculation and a clear statement about why the condition for collinearity fails. Part (c) requires naming the shape and justifying it with vector evidence.