| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Vector between two points |
| Difficulty | Easy -1.2 This is a straightforward two-part question testing basic vector operations: finding a displacement vector and its magnitude (using Pythagoras), then finding direction using inverse tangent. Both are routine AS-level procedures requiring only direct application of standard formulas with no problem-solving or conceptual challenges. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\binom{24}{7}\) | B1 | Not necessary for B1B1 |
| \(25\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{7}{24}\) or \(\sin\theta = \frac{7}{25}\) or \(\cos\theta = \frac{24}{25}\) | M1FT | FT their PQ if possible; Cosine rule/sine rule acceptable, formula must be used correctly |
| \(16.3°\) | A1 | awrt 16.3 |
## Question 4(a):
$\binom{24}{7}$ | **B1** | Not necessary for B1B1
$25$ | **B1** |
## Question 4(b):
$\tan\theta = \frac{7}{24}$ or $\sin\theta = \frac{7}{25}$ or $\cos\theta = \frac{24}{25}$ | **M1FT** | FT their **PQ** if possible; Cosine rule/sine rule acceptable, formula must be used correctly
$16.3°$ | **A1** | awrt 16.3
---4 The position vector of $P$ is $\mathbf { p } = \binom { 4 } { 3 }$ and the position vector of $Q$ is $\mathbf { q } = \binom { 28 } { 10 }$.
\begin{enumerate}[label=(\alph*)]
\item Determine the magnitude of $\overrightarrow { \mathrm { PQ } }$.
\item Determine the angle between $\overrightarrow { \mathrm { PQ } }$ and the positive $x$-direction.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2022 Q4 [4]}}