| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Conditions for vector product to be zero |
| Difficulty | Standard +0.3 This is a straightforward Further Maths vector product question with two standard parts: (a) computing a cross product and finding triangle area using the formula Ā½|aĆb|, and (b) setting up three equations from aĆb=0 and solving for Ī». Both parts are routine applications of vector product techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 4.04g Vector product: a x b perpendicular vector8.04a Vector product: definition, magnitude/direction, component form8.04c Areas using vector product: triangles and parallelograms |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | š¢ = = 3(n + 5)2 + 3(n + 5) + 7 = 3n2 + 33n + 97= |
| Answer | Marks |
|---|---|
| š | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | Expanding and considering at least one term mod 10 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | The sequence is periodic (with period 5) |
| Answer | Marks |
|---|---|
| possibilities. | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.4 | Allow repeating/cyclic |
Question 2:
2 | (a) | š¢ = = 3(n + 5)2 + 3(n + 5) + 7 = 3n2 + 33n + 97=
š + 5
= 3n2 + 3n + 7 (mod 10) ļŗ š¢ (mod 10)
š | M1
A1
[2] | 3.1a
1.1 | Expanding and considering at least one term mod 10
or evaluating š¢ from n=1 to n=10
š
Correct conclusion from correct algebraic work
2 | (b) | The sequence is periodic (with period 5)
Any shorter periodicity would have to be a factor of 5. Since
the sequence is not constant, there are no smaller
possibilities. | B1
B1
[2] | 1.1
2.4 | Allow repeating/cyclic
Allow statement only that sequence is non-constant or
sight of {3, 5, (3, 7, 7, ā¦)}.2 The points $A$ and $B$ have position vectors $\mathbf { a }$ and $\mathbf { b }$ relative to the origin $O$. It is given that $\mathbf { a } = \left( \begin{array} { c } 2 \\ 4 \\ 3 \lambda \end{array} \right)$ and $\mathbf { b } = \left( \begin{array} { r } \lambda \\ - 4 \\ 6 \end{array} \right)$, where $\lambda$ is a real parameter.
\begin{enumerate}[label=(\alph*)]
\item In the case when $\lambda = 3$, determine the area of triangle $O A B$.
\item Determine the value of $\lambda$ for which $\mathbf { a } \times \mathbf { b } = \mathbf { 0 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2024 Q2 [6]}}