| Exam Board | OCR |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 11 |
| Topic | Vectors: Cross Product & Distances |
| Type | Find parameter value for geometric condition |
| Difficulty | Challenging +1.3 This is a Further Maths question requiring proof by induction with vector cross products and solving an inequality with logarithms. While it involves multiple techniques (cross product, induction, logarithms), each step is fairly standard: the induction follows a predictable structure once the recurrence relation is computed, and part (ii) is routine manipulation of inequalities and logs. The cross product calculation is mechanical, making this above-average difficulty but not requiring deep insight. |
| Spec | 4.01a Mathematical induction: construct proofs4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(n = 1, \text{LHS} = \text{RHS} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) | B1 | Not just LHS = RHS. |
| Assume true for \(n = k\) ie \(\mathbf{a}_k = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{k-1} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) | M1 | Must have statement in terms of some other variable than \(n\). |
| \(\mathbf{a}_{k+1} = (\mathbf{a} \times \mathbf{b}) \times \mathbf{b} = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{k-1} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \mathbf{b} \times \mathbf{b}\) | M1 | Uses inductive hypothesis properly |
| \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} -5 \\ -5 \\ -5 \end{pmatrix}\) | M1 | Any two components correct or all 3 wrong sign. |
| \(\frac{1}{4} \left[ -5 \times \begin{pmatrix} 1 \\ -4 \\ 4 \end{pmatrix} \begin{pmatrix} -3 \\ 2 \end{pmatrix} \right] = \frac{1}{16} \begin{pmatrix} -14 \\ -14 \\ -14 \end{pmatrix} = -\frac{14}{16}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = -\frac{7}{8}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) | A1 | Working must be convincing. Do not award A1 if from 3wrong signs twice. |
| \(\mathbf{a}_{k+1} = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^k \times -\frac{7}{8}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) | M1 | |
| \(\mathbf{a}_{k+1} = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{(k+1)-1} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) (AG) | A1 | Simplification with sufficient working to establish truth for \(k + 1\) |
| So true for all positive integer \(n\) | E1 | Clear conclusion for induction process. |
| [8] | ||
| (ii) \( | \mathbf{a}_n | = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{n-1} \sqrt{3}\) or \(0.875^{-1}\sqrt{3}\) |
| \(n - 1 > \frac{\ln\frac{0.001}{\sqrt{3}}}{\ln\frac{7}{8}}\) (n=56.845...) | M1 | Correctly taking logs (any base) and using 3rd law of logs with correct change of inequality sign |
| So smallest value of \(n\) is 57 | A1 | |
| [3] |
**(i)** $n = 1, \text{LHS} = \text{RHS} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ | B1 | Not just LHS = RHS.
Assume true for $n = k$ ie $\mathbf{a}_k = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{k-1} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ | M1 | Must have statement in terms of some other variable than $n$.
$\mathbf{a}_{k+1} = (\mathbf{a} \times \mathbf{b}) \times \mathbf{b} = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{k-1} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \mathbf{b} \times \mathbf{b}$ | M1 | Uses inductive hypothesis properly
$\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} -5 \\ -5 \\ -5 \end{pmatrix}$ | M1 | Any two components correct or all 3 wrong sign.
$\frac{1}{4} \left[ -5 \times \begin{pmatrix} 1 \\ -4 \\ 4 \end{pmatrix} \begin{pmatrix} -3 \\ 2 \end{pmatrix} \right] = \frac{1}{16} \begin{pmatrix} -14 \\ -14 \\ -14 \end{pmatrix} = -\frac{14}{16}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = -\frac{7}{8}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ | A1 | Working must be convincing. Do not award A1 if from 3wrong signs twice.
$\mathbf{a}_{k+1} = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^k \times -\frac{7}{8}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ | M1 |
$\mathbf{a}_{k+1} = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{(k+1)-1} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ (AG) | A1 | Simplification with sufficient working to establish truth for $k + 1$
So true for all positive integer $n$ | E1 | Clear conclusion for induction process.
| [8] |
**(ii)** $|\mathbf{a}_n| = \begin{pmatrix} -\frac{7}{8} \end{pmatrix}^{n-1} \sqrt{3}$ or $0.875^{-1}\sqrt{3}$ | B1 |
$n - 1 > \frac{\ln\frac{0.001}{\sqrt{3}}}{\ln\frac{7}{8}}$ (n=56.845...) | M1 | Correctly taking logs (any base) and using 3rd law of logs with correct change of inequality sign
So smallest value of $n$ is 57 | A1 |
| [3] |8 In this question you must show detailed reasoning.
A sequence of vectors $\mathbf { a } _ { 1 } , \mathbf { a } _ { 2 } , \mathbf { a } _ { 3 } , \ldots$ is defined by
\begin{itemize}
\item $\mathbf { a } _ { 1 } = \left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right)$
\item $\quad \mathbf { a } _ { n + 1 } = \left( \mathbf { a } _ { n } \times \mathbf { b } \right) \times \mathbf { b }$, for integers $n \geqslant 1$, where $\mathbf { b }$ is the vector $\frac { 1 } { 4 } \left( \begin{array} { c } - 3 \\ 1 \\ 2 \end{array} \right)$.\\
(i) Prove by induction that $\mathbf { a } _ { n } = \left( - \frac { 7 } { 8 } \right) ^ { n - 1 } \left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right)$. for all integers $n \geqslant 1$.\\
(ii) Use an algebraic method to find the smallest value of $n$ such that $\left| \mathbf { a } _ { n } \right| < 0.001$.
\end{itemize}
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR FP1 AS 2018 Q8 [11]}}