| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Geometric properties using vectors |
| Difficulty | Moderate -0.8 This is a straightforward AS-level vectors question requiring basic vector magnitude calculations to show isosceles property, and simple vector addition for the parallelogram. Both parts use routine techniques with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple steps and understanding of geometric properties. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ | \overrightarrow{PQ}\ | = \sqrt{1^2+7^2} = \sqrt{50}\) |
| \(\overrightarrow{PR} = (\mathbf{i}+7\mathbf{j})+(4\mathbf{i}-12\mathbf{j}) = 5\mathbf{i}-5\mathbf{j}\) | M1 (AO 1.1a) | Attempt to add vectors |
| \(\ | \overrightarrow{PR}\ | = \sqrt{5^2+5^2} = \sqrt{50}\), so the triangle is isosceles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PQRS parallelogram so \(\overrightarrow{PS} = \overrightarrow{QR} = 4\mathbf{i}-12\mathbf{j}\) | M1 (AO 3.1a) | Using the properties of the parallelogram. SPECIAL CASES: Allow SC1 for correct answer for either PQSR or PSQR |
| Position vector: \(\overrightarrow{OS} = \overrightarrow{OP}+\overrightarrow{PS} = (-3\mathbf{i}-\mathbf{j})+(4\mathbf{i}-12\mathbf{j}) = \mathbf{i}-13\mathbf{j}\) | A1 [2] (AO 1.1) | cao. If PQSR used: \(\overrightarrow{OS}=3\mathbf{i}+\mathbf{j}\); if PSQR used: \(\overrightarrow{OS}=-7\mathbf{i}+11\mathbf{j}\) |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\|\overrightarrow{PQ}\| = \sqrt{1^2+7^2} = \sqrt{50}$ | B1 (AO 1.1a) | Allow for $PQ^2$; allow finding $\overrightarrow{RP} = -5\mathbf{i}+5\mathbf{j}$ |
| $\overrightarrow{PR} = (\mathbf{i}+7\mathbf{j})+(4\mathbf{i}-12\mathbf{j}) = 5\mathbf{i}-5\mathbf{j}$ | M1 (AO 1.1a) | Attempt to add vectors |
| $\|\overrightarrow{PR}\| = \sqrt{5^2+5^2} = \sqrt{50}$, so the triangle is isosceles | A1 [3] (AO 2.2a) | Must deduce the triangle is isosceles from correct working |
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## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| PQRS parallelogram so $\overrightarrow{PS} = \overrightarrow{QR} = 4\mathbf{i}-12\mathbf{j}$ | M1 (AO 3.1a) | Using the properties of the parallelogram. SPECIAL CASES: Allow SC1 for correct answer for either PQSR or PSQR |
| Position vector: $\overrightarrow{OS} = \overrightarrow{OP}+\overrightarrow{PS} = (-3\mathbf{i}-\mathbf{j})+(4\mathbf{i}-12\mathbf{j}) = \mathbf{i}-13\mathbf{j}$ | A1 [2] (AO 1.1) | cao. If PQSR used: $\overrightarrow{OS}=3\mathbf{i}+\mathbf{j}$; if PSQR used: $\overrightarrow{OS}=-7\mathbf{i}+11\mathbf{j}$ |
---3 Fig. 3 shows a triangle PQR . The vector $\overrightarrow { \mathrm { PQ } }$ is $\mathbf { i } + 7 \mathbf { j }$ and the vector $\overrightarrow { \mathrm { QR } }$ is $4 \mathbf { i } - 12 \mathbf { j }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a1b6c827-7d74-4527-9b60-58872e3d5ef7-3_412_234_1736_244}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the triangle PQR is isosceles.
The point P has position vector $- 3 \mathbf { i } - \mathbf { j }$. The point S is added so that PQRS is a parallelogram.
\item Find the position vector of S .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q3 [5]}}