A2 June 2024 Paper 2 Q5
5 Vectors, \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), are given by \(\mathbf{a} = \mathbf{i} + (1 - p)\mathbf{j} + (p + 2)\mathbf{k}\), \(\mathbf{b} = 2\mathbf{i} + \mathbf{j} + \mathbf{k}\) and \(\mathbf{c} = \mathbf{i} + 14\mathbf{j} + (p - 3)\mathbf{k}\) where \(p\) is a constant.
You are given that \(\mathbf{a} \times \mathbf{b}\) is perpendicular to \(\mathbf{c}\).
Determine the possible values of \(p\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 1 \\ 1 - p \\ p + 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\) \(= \begin{pmatrix} (1 - p) \times 1 - (p + 2) \times 1 \\ (p + 2) \times 2 - 1 \times 1 \\ 1 \times 1 - (1 - p) \times 2 \end{pmatrix}\) | M1* | 3.1a |
| \(= \begin{pmatrix} -1 - 2p \\ 2p + 3 \\ 2p - 1 \end{pmatrix}\) | A1 | 1.1 |
| \(\begin{pmatrix} 1 \\ 14 \\ p - 3 \end{pmatrix} \cdot \begin{pmatrix} -1 - 2p \\ 2p + 3 \\ 2p - 1 \end{pmatrix}\) \(= 1(-1 - 2p) + 14(2p + 3) + (p - 3)(2p - 1)\) | M1 | 1.1 |
| \(\mathbf{c}\) perpendicular to \(\mathbf{a} \times \mathbf{b} \Rightarrow \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) = 0\) soi | B1 | 3.1a |
| \(\Rightarrow -1 - 2p + 28p + 42 + 2p^2 - 7p + 3 = 0\) \(\Rightarrow 2p^2 + 19p + 44 = 0\) | depM1* | 1.1 |
| \(\therefore p = -4\) or \(p = -\dfrac{11}{2}\) cao | A1 | 1.1 |
| [6] |
Notes
M1*: Forming the cross-product. Two components correct (possibly unsimplified) or all correct but for wrong sign globally.
or expanding \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 - p & p + 2 \\ 2 & 1 & 1 \end{vmatrix}\)
M1: Correctly forming the scalar product of \(\mathbf{c}\) and their \(\mathbf{a} \times \mathbf{b}\) (even if one or both incorrect)
If M0M0 then SCB1 for correctly forming scalar product of any two vectors containing \(p\).
B1: Rule for two vectors being perpendicular soi.
depM1*: Rearranging to 3-term quadratic equation in \(p\). “= 0” can be implied by correct solution.
A1: \((p + 4)(2p + 11) = 0\)