| Exam Board | OCR |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 13 |
| Topic | Vector Product and Surfaces |
| Type | Vector product calculation |
| Difficulty | Standard +0.8 This FP1 question requires cross products to find perpendicular vectors (standard), then minimizing magnitude subject to constraints (requiring calculus/completing the square), and finally proving impossibility using the scalar triple product or determinant arguments. The multi-step reasoning and proof component elevate it above routine Further Maths exercises, though the techniques themselves are syllabus-standard. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(v = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} × \begin{pmatrix} 1 \\ p \\ p^2 \end{pmatrix}\) | M1 | Understanding that v must be (a multiple of) the vector product |
| Answer: \(\begin{pmatrix} p^2 - p \\ p^2 + 1 \\ -p - 1 \end{pmatrix}\) | A1 | or any non-zero multiple |
| Answer: Or for first 2 marks: \(-1 + a + b = 0\) and \(-1 + ap + bp^2 = 0\) | M1 | Assuming a vector of the form \(\begin{pmatrix} 1 \\ a \\ b \end{pmatrix}\) and understanding that both dot products are zero. Allow one numerical or sign error |
| Answer: \(a = \frac{p^2 + 1}{p(p - 1)}\) and \(b = -\frac{p + 1}{p(p - 1)}\) | A1 | |
| Answer: \(\begin{pmatrix} p^2 - p \\ p^2 + 1 \\ -p - 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ p - 1 \end{pmatrix}\) | M1 | if perpendicular then product = 0 |
| Answer: \(= p^2 - p + 2\) | A1 | |
| Answer: \(b^2 - 4ac = 1 - 8 = -7 < 0\) | M1 | Could be within formula but must note that discriminant is <0 or that \(\sqrt{-7}\) is not real. |
| Answer: so dot product cannot be zero so the vectors cannot be perpendicular | E1 | Must incorporate both ideas AG |
## (ii)
Answer: $v = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} × \begin{pmatrix} 1 \\ p \\ p^2 \end{pmatrix}$ | M1 | Understanding that v must be (a multiple of) the vector product
Answer: $\begin{pmatrix} p^2 - p \\ p^2 + 1 \\ -p - 1 \end{pmatrix}$ | A1 | or any non-zero multiple
Answer: Or for first 2 marks: $-1 + a + b = 0$ and $-1 + ap + bp^2 = 0$ | M1 | Assuming a vector of the form $\begin{pmatrix} 1 \\ a \\ b \end{pmatrix}$ and understanding that both dot products are zero. Allow one numerical or sign error
Answer: $a = \frac{p^2 + 1}{p(p - 1)}$ and $b = -\frac{p + 1}{p(p - 1)}$ | A1 |
Answer: $\begin{pmatrix} p^2 - p \\ p^2 + 1 \\ -p - 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ p - 1 \end{pmatrix}$ | M1 | if perpendicular then product = 0
Answer: $= p^2 - p + 2$ | A1 |
Answer: $b^2 - 4ac = 1 - 8 = -7 < 0$ | M1 | Could be within formula but must note that discriminant is <0 or that $\sqrt{-7}$ is not real.
Answer: so dot product cannot be zero so the vectors cannot be perpendicular | E1 | Must incorporate both ideas AG
---\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $x$, a vector which is perpendicular to the vectors $\begin{pmatrix} x-2 \\ 5 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} x \\ 6 \\ 2 \end{pmatrix}$. [2]
\item Find the shortest possible vector of the form $\begin{pmatrix} 1 \\ a \\ b \end{pmatrix}$ which is perpendicular to the vectors $\begin{pmatrix} x-2 \\ 5 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} x \\ 6 \\ 2 \end{pmatrix}$. [5]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Vector $\mathbf{v}$ is perpendicular to both $\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ p \\ p^2 \end{pmatrix}$ where $p$ is a real number. Show that it is impossible for $\mathbf{v}$ to be perpendicular to the vector $\begin{pmatrix} 1 \\ 1 \\ p-1 \end{pmatrix}$. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 AS 2017 Q8 [13]}}