AS June 2025 Paper 1 Q2
2 Two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), are given by \(\mathbf{a} = \begin{pmatrix} 2 \\ -3 \\ 13 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -4 \\ 6 \\ p \end{pmatrix}\) where \(p\) is a constant.
- \(\mathbf{a}.\mathbf{b}\)
- \(\mathbf{a} \times \mathbf{b}\)
- \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular
- \(\mathbf{a}\) and \(\mathbf{b}\) are parallel
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{a}.\mathbf{b} = -8 - 18 + 13p = 13p - 26\) | B1 | 1.1 |
| \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 2 \\ -3 \\ 13 \end{pmatrix} \times \begin{pmatrix} -4 \\ 6 \\ p \end{pmatrix} = \begin{pmatrix} -3 \times p - 13 \times 6 \\ 13 \times -4 - 2 \times p \\ 2 \times 6 - -3 \times -4 \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} -3p - 78 \\ -2p - 52 \\ 0 \end{pmatrix}\) | A1 | 1.1 |
| [3] |
Notes
B1: Must be simplified correctly
M1: Using formula for cross product. Condone one error (can be implied by 2 of the 3 components given correctly in final answer).
A1: Needs to be simplified correctly
If M0 then SCB1 for \(\begin{pmatrix} 3p + 78 \\ 2p + 52 \\ 0 \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| Perpendicular: \((13p - 26 = 0 \Rightarrow)\ p = 2\) | B1FT | 2.2a |
| Parallel: (need \(-3p - 78 = 0\) and \(-2p - 52 = 0\) \(\Rightarrow\)) \(p = -26\) | B1FT | 2.2a |
| [2] |
Notes
B1FT: Follow through on answer to 2(a)
No working necessary
B1FT: Could use scaling, or \(\mathbf{a}.\mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\)
Allow FT from part (a) as long as the first two components give the same value of \(p\) (ignore \(z\) component).
SCB1 (or B1FT) for both answers correct but with no indication of which is which.
Do not allow SCB1 if they are incorrectly identified