A2 June 2022 Q3
3. With respect to a fixed origin \(O\), the points \(A\) and \(B\) have coordinates \((2, 2, -1)\) and \((4, 2p, 1)\) respectively, where \(p\) is a constant.
For each of the following, determine the possible values of \(p\) for which,
| Scheme | Marks | AO |
|---|---|---|
| \(\left(|OB| =\right)\sqrt{4^2 + (2p)^2 + 1^2}\ \left(= \sqrt{17 + 4p^2}\right)\) | B1 | 1.1b |
| \(\cos 45 = \dfrac{4}{\sqrt{17 + 4p^2}} \Rightarrow p = \ldots\) | M1 | 3.1a |
| \(p = \pm\dfrac{\sqrt{15}}{2}\) | A1 | 1.1b |
| (3) |
Notes
B1: Correct expression for the magnitude for \(\overrightarrow{OB}\) (may be seen in formula)
M1: A complete method to find a value for \(p\). E.g. Sets \(\cos 45 = 4/\)their magnitude of \(\overrightarrow{OB}\) and solves to find a value for \(p\). Note this may arise from attempts using dot products, but the same equation is reached and a full method to find \(p\) is required.
A1: \(p = \pm\dfrac{\sqrt{15}}{2}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{OA} \times \overrightarrow{OB} = \begin{pmatrix}2 + 2p\\ -6\\ 4p - 8\end{pmatrix}\) | B1 | 1.1b |
| E.g. Sets \(\begin{pmatrix}2 + 2p\\ -6\\ 4p - 8\end{pmatrix} = \begin{pmatrix}4\lambda\\ -p\lambda\\ 2\lambda\end{pmatrix}\) and solves to find a value for \(p\) | M1 | 3.1a |
| \(p = 3\) only | A1 | 2.2a |
| (3) |
Notes
B1: Correct vector product, allow if seen anywhere in the question. Alternatively, if method 2 below is used, the cross product is not necessary, and this mark may be awarded for a correct equation in \(p\) from either dot product.
M1: A complete method to find a value for \(p\). E.g.
- Set the vector product equal to a multiple of the parallel vector and solves to find a value for \(p\),
- Attempts dot products of both \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) with \(\begin{pmatrix}4\\ -p\\ 2\end{pmatrix}\), solves and finds an answer from both,
- Finds cross product of \(\overrightarrow{OA} \times \overrightarrow{OB}\) with \(\begin{pmatrix}4\\ -p\\ 2\end{pmatrix}\) and sets at least one coefficient to zero to find \(p\).
A1cso: A complete argument leading to \(p = 3\) only, which must be consistent with their work. Where a method leads to more than one value for \(p\) the candidate will need to check which values hold and give the answer \(p = 3\) only. No method needs be seen for this but other values must be rejected.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{2}\left|\overrightarrow{OA} \times \overrightarrow{OB}\right| = 3\sqrt{2} \Rightarrow (2 + 2p)^2 + (-6)^2 + (4p - 8)^2 = \left(6\sqrt{2}\right)^2\) | M1 | 3.1a |
| Solves a 3TQ to find a value for \(p\) \(20p^2 - 56p + 32 = 0 \Rightarrow p = \ldots\) | dM1 | 1.1b |
| \(p = 2,\ \dfrac{4}{5}\) | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Notes
M1: A complete method to set up a polynomial in \(p\). E.g. sets half magnitude of their vector product \(= 3\sqrt{2}\) and reaches a quadratic expression in \(p\). An alternative approach is:
\[\cos\angle AOB = \frac{2 \times 4 + 2 \times 2p - 1 \times 1}{\sqrt{4 + 4 + 1}\sqrt{16 + 4p^2 + 1}} = \frac{7 + 4p}{3\sqrt{17 + 4p^2}}\]\[\Rightarrow 3\sqrt{2} = \frac{1}{2}OA.OB\sin\angle AOB = \frac{1}{2}3\sqrt{17 + 4p^2}\sqrt{1 - \frac{(7 + 4p)^2}{9\left(17 + 4p^2\right)}}\]\[\Rightarrow 72 = 9\left(17 + 4p^2\right) - \left(49 + 56p + 16p^2\right)\]dM1: Dependent on the previous method mark. Solve a 3TQ to find a value for \(p\).
A1: \(p = 2,\ \dfrac{4}{5}\) only. Note: allow this mark if their \(\overrightarrow{OA} \times \overrightarrow{OB}\) was correct apart from the \(\mathbf{j}\) component sign.