A2 June 2025 Paper 2 Q2
2. An archer shoots an arrow towards a target.
In a model
- the arrow is a particle
- the flight path of the arrow is a straight line
- the target is part of a plane
Relative to a fixed origin \(O\)
- the arrow is fired from the point with position vector \(3\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}\)
- the plane containing the target has equation \(2x + 4y - z = 3\)
Use the model to answer parts (a) to (d).
The arrow hits the target at the point with position vector \(6\mathbf{i} - 2\mathbf{j} + \mathbf{k}\)
| Scheme | Marks | AO |
|---|---|---|
| Shortest distance \(= \dfrac{\lvert 2 \times 3 + 4 \times (-5) + (-1) \times 2 - 3\rvert}{\sqrt{2^2 + 4^2 + (-1)^2}} = \ldots\) | M1 | 1.1b |
| awrt 4.15 or \(\dfrac{19}{\sqrt{21}}\) or \(\dfrac{19\sqrt{21}}{21}\) | A1 | 1.1b |
| (2) |
Notes
M1: Correct method to find the shortest distance. Condone \(\pm d\) in the formula. Longer methods are possible. E.g. if done after part (d), then \(\sqrt{19}\cos 17.97\ldots\) is a correct method. Another approach is to form the general line \(\mathbf{r} = \begin{pmatrix}3\\ -5\\ 2\end{pmatrix} + \mu\begin{pmatrix}2\\ 4\\ -1\end{pmatrix}\), substitute into plane equation \(\Rightarrow \mu = \dfrac{19}{21}\) then scale the normal vectors magnitude: \(\dfrac{19}{21} \times \sqrt{2^2 + 4^2 + (-1)^2} = \ldots\)
A1: Correct exact distance or awrt 4.15
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix}6\\ -2\\ 1\end{pmatrix} - \begin{pmatrix}3\\ -5\\ 2\end{pmatrix} = \begin{pmatrix}3\\ 3\\ -1\end{pmatrix}\) or \(\overrightarrow{BA} = \begin{pmatrix}3\\ -5\\ 2\end{pmatrix} - \begin{pmatrix}6\\ -2\\ 1\end{pmatrix} = \begin{pmatrix}-3\\ -3\\ 1\end{pmatrix}\) | M1 | 3.4 |
| \(\mathbf{r} = \begin{pmatrix}6\\ -2\\ 1\end{pmatrix} + \lambda\begin{pmatrix}3\\ 3\\ -1\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\ -5\\ 2\end{pmatrix} + \lambda\begin{pmatrix}3\\ 3\\ -1\end{pmatrix}\) \(\mathbf{r} = \begin{pmatrix}6\\ -2\\ 1\end{pmatrix} + \lambda\begin{pmatrix}-3\\ -3\\ 1\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\ -5\\ 2\end{pmatrix} + \lambda\begin{pmatrix}-3\\ -3\\ 1\end{pmatrix}\) | A1 | 3.3 |
| (2) |
Notes
M1: Finds the direction of the line. Accept either direction and implied by 2 out of three coordinates correct if no method shown.
A1: Correct model for the flight path of the arrow. Must be an equation ie, including the \(\mathbf{r} = \ldots\) (but condone \(l = \ldots\) or other letter).
| Scheme | Marks | AO |
|---|---|---|
| \(\cos\theta = \dfrac{\begin{pmatrix}3\\ 3\\ -1\end{pmatrix}.\begin{pmatrix}2\\ 4\\ -1\end{pmatrix}}{\sqrt{3^2 + 3^2 + (-1)^2} \times \sqrt{2^2 + 4^2 + (-1)^2}} \Rightarrow \theta = \ldots \text{ or } 90 - \theta = \ldots\) | M1 | 3.1b |
| \((\phi =)\ 72^\circ\) | A1 | 1.1b |
| (2) |
Notes
M1: A complete method to find the required angle or angle between normal and plane. Uses the dot product of the direction vector of the line and normal vector of the plane to find the angle (need not subtract from 90 for this mark). Allow if a (single) minor slip is made if the method is clearly attempting to use the correct vectors.
Note may use \(\sin\phi = \dfrac{\begin{pmatrix}3\\ 3\\ -1\end{pmatrix}.\begin{pmatrix}2\\ 4\\ -1\end{pmatrix}}{\sqrt{3^2 + 3^2 + (-1)^2} \times \sqrt{2^2 + 4^2 + (-1)^2}} \Rightarrow \phi = \ldots\) directly.
A1: Correct final angle.
| Scheme | Marks | AO |
|---|---|---|
| Distance \(= \sqrt{3^2 + 3^2 + (-1)^2} = \sqrt{19} =\) awrt 4.36 | B1 | 1.1b |
| (1) |
Notes
B1: \(\sqrt{19}\) or awrt 4.36
| Scheme | Marks | AO |
|---|---|---|
| Not likely to match, as unlikely that the flight path of the arrow will be a straight line. | B1 | 3.2b |
| (1) | ||
| (8 marks) |
Notes
B1: States not likely to match (or equivalent wording) and gives a suitable reason why. This will most commonly be that the arrow will not travel in a straight line, but e.g. accept “the arrow is not a particle so the tip may travel into the target, not stop at the plane”.
Do not accept answers that suggest the arrow will travel less far than the answer to (d). Do not accept “air resistance” arguments unless they specifically refer to the arrow not travelling in a straight line.