June 2023 Paper 2 Q18
18 In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors representing due east and due north respectively.
A particle, \(T\), is moving on a plane at a constant speed.
The path followed by \(T\) makes the exact shape of a triangle \(ABC\).
\(T\) moves around \(ABC\) in an anticlockwise direction as shown in the diagram below.
On its journey from \(A\) to \(B\) the velocity vector of \(T\) is \(\left(3\mathbf{i} + \sqrt{3}\mathbf{j}\right)\) m s−1
Show that the acute angle \(ABC = 60^\circ\) [2 marks]
\(T\) returns to its initial position after 9 seconds.
Vertex \(B\) lies at position vector \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) metres with respect to a fixed origin \(O\)
Find the position vector of \(C\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains correct speed of \(\sqrt{12}\) OE AWRT 3.46 Condone missing units | B1 | 1.1b |
| (1) |
Typical solution
Speed \(= \sqrt{(3)^2 + \left(\sqrt{3}\right)^2} = 2\sqrt{3}\) m s−1
| Scheme | Marks | AO |
|---|---|---|
| Uses \(\tan^{-1}\dfrac{\sqrt{3}}{3}\) or \(\tan^{-1}\dfrac{3}{\sqrt{3}}\) to find the angle between one of the velocity vectors relative to the \(\mathbf{i}\) direction or the \(\mathbf{j}\) direction. Sight of sine rule or cosine rule using a magnitude for \(AC\) scores M0 R0 | M1 | 3.1a |
| Completes a reasoned argument to obtain 30° for both angles relative to the \(\mathbf{i}\) direction and adds them together to obtain angle \(ABC\) = 60° or Completes a reasoned argument to obtain 60° for both angles relative to the \(\mathbf{j}\) direction and adds them together and subtracts them from 180° to obtain angle \(ABC\) = 60° Solution must include clear reference to angle \(ABC\) or indicate angle \(ABC\) with a letter on a diagram. When using trigonometric ratios the vectors \(\mathbf{i}\) and \(\mathbf{j}\) must not be included. | R1 | 2.1 |
| (2) |
Typical solution
Angle between \(AB\) and \(\mathbf{i}\) direction
\[= \tan^{-1}\frac{\sqrt{3}}{3} = 30^\circ\]Angle between \(BC\) and \(\mathbf{i}\) direction
\[= \tan^{-1}\frac{\sqrt{3}}{3} = 30^\circ\]Angle ABC = 30° + 30° = 60°
| Scheme | Marks | AO |
|---|---|---|
| Deduces time taken from \(B\) to \(C\) is 3 seconds. | R1 | 2.2a |
| Obtains an expression for displacement from \(B\) to \(C\) of the form \(t\begin{bmatrix} -3 \\ \sqrt{3} \end{bmatrix}\) where \(1 \lt t \leqslant 9\) | M1 | 3.1a |
| Obtains \(\begin{bmatrix} -8 \\ 3\sqrt{3} \end{bmatrix}\) | A1 | 1.1b |
| (3) | ||
| (6 marks) |
Typical solution
Time taken from \(B\) to \(C = \dfrac{9}{3} = 3\)
\[\overrightarrow{OC} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 3\begin{bmatrix} -3 \\ \sqrt{3} \end{bmatrix}\]\[= \begin{bmatrix} -8 \\ 3\sqrt{3} \end{bmatrix}\]