| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Area of triangle or parallelogram using vectors |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard vector techniques: dot product for angle (part a is 'show that' so calculation is guided), Pythagorean identity for sine, and area formula for parallelogram. All steps are routine applications of formulas with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4\mathbf{i}-3\mathbf{j})-(2\mathbf{i}+4\mathbf{j})(=2\mathbf{i}-7\mathbf{j})\) | M1 | Correct method to find either \(\overrightarrow{AB}\) or \(\overrightarrow{BA}\) |
| \(\sqrt{53}\) or \(53\) from \(\pm(2\mathbf{i}-7\mathbf{j})\) | A1 | cao |
| \( | \overrightarrow{OA} | =\sqrt{20}\), \( |
| \(\cos AOB = \dfrac{(\sqrt{20})^2+5^2-(\sqrt{53})^2}{2(\sqrt{20})(5)}\) | M1 | Correct use of cosine rule for their \(OA\), \(OB\) and \(AB\); \(\cos AOB\) may not be the subject, but substitutions must be correct for their values |
| \(\cos AOB = \left(\dfrac{20+25-53}{10\sqrt{20}}\right) = -\dfrac{4}{5(2\sqrt{5})} = -\dfrac{2\sqrt{5}}{25}\) | A1 | AG – sufficient working must be shown; Condone this result from calculator without intermediate working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin^2 AOB = 1-\left(-\dfrac{2\sqrt{5}}{25}\right)^2\) | M1 | Using the identity \(\cos^2 X + \sin^2 X = 1\) with \(\cos X = -\dfrac{2\sqrt{5}}{25}\) |
| \(\sin^2 AOB = \dfrac{121}{125} \Rightarrow \sin AOB = \dfrac{11\sqrt{5}}{25}\) | A1 | Or exact equivalent – justification not required for taking the positive square root |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area of \(OACB = 2\left(\dfrac{1}{2}(\sqrt{20})(5)\left(\dfrac{11\sqrt{5}}{25}\right)\right)\) | M1 | Use of \(A = ab\sin C\) (or equivalent) with \(OA\) and \(OB\) and \(\sin AOB\); May not use exact values here |
| \(22\) | A1 | cao; Condone awrt \(22.0\) |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4\mathbf{i}-3\mathbf{j})-(2\mathbf{i}+4\mathbf{j})(=2\mathbf{i}-7\mathbf{j})$ | M1 | Correct method to find either $\overrightarrow{AB}$ or $\overrightarrow{BA}$ |
| $\sqrt{53}$ or $53$ from $\pm(2\mathbf{i}-7\mathbf{j})$ | A1 | cao |
| $|\overrightarrow{OA}|=\sqrt{20}$, $|\overrightarrow{OB}|=5$ | B1 | Correct lengths for $OA$ and $OB$ (or their squares) |
| $\cos AOB = \dfrac{(\sqrt{20})^2+5^2-(\sqrt{53})^2}{2(\sqrt{20})(5)}$ | M1 | Correct use of cosine rule for their $OA$, $OB$ and $AB$; $\cos AOB$ may not be the subject, but substitutions must be correct for their values |
| $\cos AOB = \left(\dfrac{20+25-53}{10\sqrt{20}}\right) = -\dfrac{4}{5(2\sqrt{5})} = -\dfrac{2\sqrt{5}}{25}$ | A1 | **AG** – sufficient working must be shown; Condone this result from calculator without intermediate working |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin^2 AOB = 1-\left(-\dfrac{2\sqrt{5}}{25}\right)^2$ | M1 | Using the identity $\cos^2 X + \sin^2 X = 1$ with $\cos X = -\dfrac{2\sqrt{5}}{25}$ |
| $\sin^2 AOB = \dfrac{121}{125} \Rightarrow \sin AOB = \dfrac{11\sqrt{5}}{25}$ | A1 | Or exact equivalent – justification not required for taking the positive square root |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of $OACB = 2\left(\dfrac{1}{2}(\sqrt{20})(5)\left(\dfrac{11\sqrt{5}}{25}\right)\right)$ | M1 | Use of $A = ab\sin C$ (or equivalent) with $OA$ and $OB$ and $\sin AOB$; May not use exact values here |
| $22$ | A1 | cao; Condone awrt $22.0$ |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-06_648_586_255_244}
The diagram shows the parallelogram $O A C B$ where $\overrightarrow { O A } = 2 \mathbf { i } + 4 \mathbf { j }$ and $\overrightarrow { O B } = 4 \mathbf { i } - 3 \mathbf { j }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos A O B = - \frac { 2 \sqrt { 5 } } { 25 }$.
\item Hence find the exact value of $\sin A O B$.
\item Determine the area of $O A C B$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [9]}}