Vectors
Geometry and measures
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Write the column vector from $(1, 2)$ to $(4, 6)$.
2 marks
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$\mathbf{a} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$. Find $\mathbf{a} + \mathbf{b}$.
2 marks
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$\mathbf{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$. Find $3\mathbf{a}$ and $\mathbf{a} - \mathbf{b}$.
3 marks
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Find the magnitude of $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$.
3 marks
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$OABC$ is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. Find $\overrightarrow{OB}$ and the position vector of the midpoint of $OB$.
4 marks
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$\mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$. Find the magnitude of $\mathbf{a}$, and a vector parallel to $\mathbf{a}$ with magnitude $15$.
4 marks
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In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$. Find $\overrightarrow{AB}$, $\overrightarrow{AM}$ and $\overrightarrow{OM}$. If $\mathbf{a} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}$, find $\overrightarrow{OM}$ as a column vector.
5 marks
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$A$ is $(2, 1)$, $B$ is $(6, 3)$ and $C$ is $(5, 6)$. Find $\overrightarrow{AB}$, $\overrightarrow{BC}$ and $\overrightarrow{AC}$. Show that $AB$ is not parallel to $BC$, and find the midpoint of $AC$.
Points A(2, 1), B(6, 3) and C(5, 6) 6 marks
Mark scheme
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Answer. $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$
- Subtract the coordinates.
- $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$
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Answer. $\begin{pmatrix} 5 \\ -1 \end{pmatrix}$
- Add the components.
- $\begin{pmatrix} 5 \\ -1 \end{pmatrix}$
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Answer. $\begin{pmatrix} 6 \\ -3 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$
- $3\mathbf{a} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$
- $\mathbf{a} - \mathbf{b} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}$
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Answer. $10$
- $\sqrt{6^{2} + 8^{2}}$
- $10$
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Answer. $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$, and the midpoint is $\dfrac{\mathbf{a} + \mathbf{c}}{2}$
- The diagonal is the sum of the two side vectors.
- The midpoint is half of that vector.
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Answer. Magnitude $5$. The vector is $\begin{pmatrix} 9 \\ 12 \end{pmatrix}$.
- $\sqrt{3^{2} + 4^{2}} = 5$
- $3\mathbf{a} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}$, which has magnitude $15$.
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Answer. $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, $\overrightarrow{AM} = \dfrac{\mathbf{b} - \mathbf{a}}{2}$, $\overrightarrow{OM} = \dfrac{\mathbf{a} + \mathbf{b}}{2} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$
- $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
- $\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{\mathbf{a} + \mathbf{b}}{2}$
- $\dfrac{1}{2}\begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$
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Answer. $\begin{pmatrix} 4 \\ 2 \end{pmatrix}$, $\begin{pmatrix} -1 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$. The vectors are not multiples. Midpoint $(3.5, 3.5)$.
- $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$.
- $\overrightarrow{AC} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$
- $4$ and $2$ are not in the same ratio as $-1$ and $3$, so the vectors are not parallel.
- Midpoint $= \left(\dfrac{2 + 5}{2}, \dfrac{1 + 6}{2}\right) = (3.5, 3.5)$.