AS June 2023 Paper 1 Q4
4 The vector \(\mathbf{p}\), all of whose components are positive, is given by \(\mathbf{p} = \begin{pmatrix} a^2 \\ a - 5 \\ 26 \end{pmatrix}\) where \(a\) is a constant.
You are given that \(\mathbf{p}\) is perpendicular to the vector \(\begin{pmatrix} 2 \\ 6 \\ -3 \end{pmatrix}\).
Determine the value of \(a\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{p}.\begin{pmatrix} 2 \\ 6 \\ -3 \end{pmatrix} = 0\) | B1 | 2.2a |
| \(2a^2 + 6(a - 5) - 3 \times 26 = 0\) | M1 | 3.1a |
| \(\Rightarrow 2a^2 + 6a - 108 = 0\) \(\Rightarrow a = 6\) or \(a = -9\) | A1 | 1.1 |
| \(a = -9\) leads to negative \(y\) component \((-14)\) so \(a = 6\) is the only solution | A1 | 2.3 |
| [4] |
Notes
B1: Knowledge that perpendicularity implies that scalar product = 0, used anywhere in solution
M1: Using the scalar product to set up a quadratic equation in \(a\)
A1: Correctly solving the equation (could be BC). Might only see \(a = 6\) here.
\(a^2 + 3a - 54 = (a - 6)(a + 9) = 0\)
\(a^2 + 3a - 54 = \left(a + \dfrac{3}{2}\right)^2 - \dfrac{225}{4}\)
A1: \(a = -9\) or brackets \((a - 6)(a + 9)\) must be seen in solution and then \(a = -9\) explicitly rejected with rationale
Accept “all components must be positive” for rationale.
Do NOT accept “\(a\) must be positive” as sole rationale