A2 June 2025 Paper 2 Q1
1 In this question you must show detailed reasoning.
Vectors \(\mathbf{a}\) and \(\mathbf{b}\) are given by \(\mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -1 \\ 3 \\ -2 \end{pmatrix}\).
Determine, in either order
- \(\mathbf{a}.\mathbf{b}\)
- \(\mathbf{a} \times \mathbf{b}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathbf{a} \cdot \mathbf{b} = 2 \cdot (-1) + (-4) \cdot 3 + 3 \cdot (-2)\) \(= -2 - 12 - 6\) \(= -20\) | B1 | 1.1 |
| \(\begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} (-4) \cdot (-2) - 3 \cdot 3 \\ 3 \cdot (-1) - 2 \cdot (-2) \\ 2 \cdot 3 - (-1) \cdot (-4) \end{pmatrix}\) \(= \begin{pmatrix} 8 - 9 \\ -3 - (-4) \\ 6 - 4 \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\) | A1 | 1.1 |
| [3] |
Notes
B1: Some evidence of correct calculation must be shown (i.e. any intermediate step before \(-20\)) but ignore poor/incorrect notation e.g. \(\begin{pmatrix} -2 \\ -12 \\ -6 \end{pmatrix}\) seen in working.
M1: Working must be correct for at least 2 components.
May see components given separately e.g. allow M1 for any two of: \(\begin{vmatrix} -4 & 3 \\ 3 & -2 \end{vmatrix}\), \(\begin{vmatrix} 2 & -1 \\ 3 & -2 \end{vmatrix}\), \(\begin{vmatrix} 2 & -1 \\ -4 & 3 \end{vmatrix}\) or for \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -4 & 3 \\ -1 & 3 & -2 \end{vmatrix}\)
A1: If M0 then SCB1 for \(\begin{pmatrix} \pm 1 \\ \pm 1 \\ \pm 2 \end{pmatrix}\) with working shown (max 2/3).
If no working for either product (B0M0) then SCB1 for both correct (max 1/3).