Moderate -0.3 This is a straightforward application of the perpendicularity condition (dot product equals zero) leading to a quadratic equation. The constraint that all components are positive provides a simple check to select the correct root. While it requires solving a quadratic, the setup is direct and the algebra is routine for Further Maths students.
4 The vector \(\mathbf { p }\), all of whose components are positive, is given by \(\mathbf { p } = \left( \begin{array} { c } a ^ { 2 } \\ a - 5 \\ 26 \end{array} \right)\) where \(a\) is a constant.
You are given that \(\mathbf { p }\) is perpendicular to the vector \(\left( \begin{array} { c } 2 \\ 6 \\ - 3 \end{array} \right)\).
Determine the value of \(a\).
Knowledge that perpendicularity implies scalar product = 0, used anywhere in solution
\(2a^2 + 6(a-5) - 3 \times 26 = 0\)
M1
Using scalar product to set up a quadratic equation in \(a\)
\(\Rightarrow 2a^2 + 6a - 108 = 0 \Rightarrow a = 6\) or \(a = -9\)
A1
Correctly solving the equation (could be BC). \(a^2 + 3a - 54 = (a-6)(a+9) = 0\)
\(a = -9\) leads to negative \(y\) component (\(-14\)) so \(a = 6\) is the only solution
A1
\(a = -9\) or brackets \((a-6)(a+9)\) must be seen and \(a = -9\) explicitly rejected with rationale. Accept "all components must be positive". Do NOT accept "\(a\) must be positive" as sole rationale
# Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{p} \cdot \begin{pmatrix} 2 \\ 6 \\ -3 \end{pmatrix} = 0$ | **B1** | Knowledge that perpendicularity implies scalar product = 0, used anywhere in solution |
| $2a^2 + 6(a-5) - 3 \times 26 = 0$ | **M1** | Using scalar product to set up a quadratic equation in $a$ |
| $\Rightarrow 2a^2 + 6a - 108 = 0 \Rightarrow a = 6$ or $a = -9$ | **A1** | Correctly solving the equation (could be BC). $a^2 + 3a - 54 = (a-6)(a+9) = 0$ |
| $a = -9$ leads to negative $y$ component ($-14$) so $a = 6$ is the only solution | **A1** | $a = -9$ or brackets $(a-6)(a+9)$ must be seen and $a = -9$ explicitly rejected with rationale. Accept "all components must be positive". Do NOT accept "$a$ must be positive" as sole rationale |
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4 The vector $\mathbf { p }$, all of whose components are positive, is given by $\mathbf { p } = \left( \begin{array} { c } a ^ { 2 } \\ a - 5 \\ 26 \end{array} \right)$ where $a$ is a constant.\\
You are given that $\mathbf { p }$ is perpendicular to the vector $\left( \begin{array} { c } 2 \\ 6 \\ - 3 \end{array} \right)$.\\
Determine the value of $a$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q4 [4]}}