Vector arithmetic
Vectors
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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The vectors are $\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$. Find $\mathbf{a} + \mathbf{b}$, $\mathbf{a} - \mathbf{b}$ and $|\mathbf{a}|$.
3 marks
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Find the magnitude of $2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$.
3 marks
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The points $A$ and $B$ have position vectors $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix}$. Find the displacement vector $\overrightarrow{AB}$ and the position vector of the midpoint of $AB$.
4 marks
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Find the unit vector in the direction of $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$, and find the vector of magnitude $6$ in the same direction.
4 marks
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The points are $A(1, 0, 2)$, $B(3, 1, 5)$ and $C(0, 4, 1)$. Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$, and find $|\overrightarrow{AB}|$.
5 marks
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The point $P$ divides the line $AB$ in the ratio $2:1$, where $A$ is $(1, 2)$ and $B$ is $(7, 8)$. Find the position vector of $P$.
5 marks
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The vectors are $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$. Find $|\mathbf{a} + \mathbf{b}|$, $|\mathbf{a} - \mathbf{b}|$ and $\mathbf{a} \cdot \mathbf{b}$.
6 marks
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The points are $A(1, 2, 0)$, $B(3, 1, 4)$ and $C(5, 0, 2)$. Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$, show that they are not parallel, and find the area of triangle $ABC$.
7 marks
Mark scheme
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Answer. $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$, $\mathbf{a} - \mathbf{b} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$ and $|\mathbf{a}| = \sqrt{10}$.
- Add and subtract the components.
- $|\mathbf{a}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$.
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Answer. $7$
- $\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7$.
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Answer. $\overrightarrow{AB} = \begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$ and the midpoint is $\begin{pmatrix} 2.5 \\ 1 \\ 4 \end{pmatrix}$.
- Subtract the position vector of $A$ from the position vector of $B$.
- The midpoint is the average of the two position vectors.
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Answer. The unit vector is $\begin{pmatrix} \dfrac{2}{3} \\ -\dfrac{1}{3} \\ \dfrac{2}{3} \end{pmatrix}$ and the vector of magnitude $6$ is $\begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}$.
- $|\mathbf{a}| = 3$.
- Divide by $3$ to get the unit vector.
- Multiply the unit vector by $6$, which is the same as multiplying $\mathbf{a}$ by $2$.
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Answer. $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix} -1 \\ 4 \\ -1 \end{pmatrix}$ and $|\overrightarrow{AB}| = \sqrt{14}$.
- Subtract the coordinates of $A$ from those of $B$ and of $C$.
- $|\overrightarrow{AB}| = \sqrt{4 + 1 + 9} = \sqrt{14}$.
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Answer. $\begin{pmatrix} 5 \\ 6 \end{pmatrix}$
- $P$ is two thirds of the way from $A$ to $B$.
- $\overrightarrow{OP} = \dfrac{1 \times \overrightarrow{OA} + 2 \times \overrightarrow{OB}}{3} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}$.
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Answer. $|\mathbf{a} + \mathbf{b}| = 5$, $|\mathbf{a} - \mathbf{b}| = \sqrt{5}$ and $\mathbf{a} \cdot \mathbf{b} = 5$.
- $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, so its magnitude is $5$.
- $\mathbf{a} - \mathbf{b} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$, so its magnitude is $\sqrt{5}$.
- The scalar product is $2 + 3 = 5$.
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Answer. $\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}$, and the area is $3\sqrt{5}$.
- $2\overrightarrow{AB} = \begin{pmatrix} 4 \\ -2 \\ 8 \end{pmatrix}$, which is not $\overrightarrow{AC}$, so the vectors are not parallel.
- The cross product is $\begin{pmatrix} 6 \\ 12 \\ 0 \end{pmatrix}$.
- Its magnitude is $\sqrt{180} = 6\sqrt{5}$.
- The area of the triangle is half of that, $3\sqrt{5}$.