| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector from magnitude and bearing |
| Difficulty | Moderate -0.8 This is a straightforward multi-part vectors question requiring basic skills: converting magnitude/bearing to components using trigonometry (standard at this level), finding triangle area using ½|a×b|, and using the parallelogram rule. All parts are routine applications of standard techniques with no problem-solving insight required, making it easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Show \(A\) in third quadrant, with length of 8 and relevant angle marked on given axes | B1 | Condone \(A\) being located by correct i and j components instead of length and angle – could be stated as a coordinate or values marked on the axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 8\cos240° = -4\), \(y = 8\sin240° = -4\sqrt{3}\) | M1 | Attempt both components from magnitude of 8 and an angle. Allow M1 for \(8\cos\theta\) and \(8\sin\theta\) attempted. Condone a value for \(\theta\) that may not be consistent with their diagram. Max of M1 only, if \(A\) incorrect on diagram |
| \(A\) is \(-4\mathbf{i} - 4\sqrt{3}\,\mathbf{j}\) | A1 | Obtain one correct component. Condone e.g. \(x = -4\) for \(-4\mathbf{i}\) |
| A1 | Obtain fully correct position vector. Allow 6.93, or better, for \(4\sqrt{3}\). A0 if coordinate or column vector |
| Answer | Marks | Guidance |
|---|---|---|
| area \(= 0.5 \times 8 \times 6 \times \sin120°\) | M1 | Attempt area of triangle, using correct formula. M0 if \(240°\) used. Allow plausible angle i.e. \(30°, 60°, 120°, 150°\). Allow other incorrect angles as long as explicit on their diagram. Allow multi-step methods as long as fully correct method |
| \(= 12\sqrt{3}\) | A1 | Must be exact. WWW e.g. M1A0 for \(12\sqrt{3}\) from \(A\) in second quadrant. M1A0 for \(12\sqrt{3}\) from using \(60°\) without justification that \(\sin120° = \sin60°\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(6\mathbf{i} - (-4\mathbf{i} - 4\sqrt{3}\,\mathbf{j})\) | M1 | Attempt \(6\mathbf{i} - (\text{their } OA)\). Allow BOD for \(6\mathbf{i} - -4\mathbf{i} - 4\sqrt{3}\,\mathbf{j}\), even if final answer is not commensurate with 'invisible brackets' |
| \(C\) is \(10\mathbf{i} + 4\sqrt{3}\,\mathbf{j}\) | A1 | Obtain \(10\mathbf{i} + 4\sqrt{3}\,\mathbf{j}\). Allow 6.93, or better, for \(4\sqrt{3}\). SC B1 for \(2\mathbf{i} - 4\sqrt{3}\,\mathbf{j}\) or \(-10\mathbf{i} - 4\sqrt{3}\,\mathbf{j}\) i.e. a valid parallelogram having misinterpreted \(OABC\) |
## Question 2:
### Part (a)(i)
Show $A$ in third quadrant, with length of 8 and relevant angle marked on given axes | **B1** | Condone $A$ being located by correct **i** and **j** components instead of length and angle – could be stated as a coordinate or values marked on the axes
### Part (a)(ii)
$x = 8\cos240° = -4$, $y = 8\sin240° = -4\sqrt{3}$ | **M1** | Attempt both components from magnitude of 8 and an angle. Allow M1 for $8\cos\theta$ and $8\sin\theta$ attempted. Condone a value for $\theta$ that may not be consistent with their diagram. Max of M1 only, if $A$ incorrect on diagram
$A$ is $-4\mathbf{i} - 4\sqrt{3}\,\mathbf{j}$ | **A1** | Obtain one correct component. Condone e.g. $x = -4$ for $-4\mathbf{i}$
| **A1** | Obtain fully correct position vector. Allow 6.93, or better, for $4\sqrt{3}$. A0 if coordinate or column vector
### Part (b)
area $= 0.5 \times 8 \times 6 \times \sin120°$ | **M1** | Attempt area of triangle, using correct formula. M0 if $240°$ used. Allow plausible angle i.e. $30°, 60°, 120°, 150°$. Allow other incorrect angles as long as explicit on their diagram. Allow multi-step methods as long as fully correct method
$= 12\sqrt{3}$ | **A1** | Must be exact. WWW e.g. M1A0 for $12\sqrt{3}$ from $A$ in second quadrant. M1A0 for $12\sqrt{3}$ from using $60°$ without justification that $\sin120° = \sin60°$
### Part (c)
$6\mathbf{i} - (-4\mathbf{i} - 4\sqrt{3}\,\mathbf{j})$ | **M1** | Attempt $6\mathbf{i} - (\text{their } OA)$. Allow BOD for $6\mathbf{i} - -4\mathbf{i} - 4\sqrt{3}\,\mathbf{j}$, even if final answer is not commensurate with 'invisible brackets'
$C$ is $10\mathbf{i} + 4\sqrt{3}\,\mathbf{j}$ | **A1** | Obtain $10\mathbf{i} + 4\sqrt{3}\,\mathbf{j}$. Allow 6.93, or better, for $4\sqrt{3}$. **SC** B1 for $2\mathbf{i} - 4\sqrt{3}\,\mathbf{j}$ or $-10\mathbf{i} - 4\sqrt{3}\,\mathbf{j}$ i.e. a valid parallelogram having misinterpreted $OABC$
---2 The point $A$ is such that the magnitude of $\overrightarrow { O A }$ is 8 and the direction of $\overrightarrow { O A }$ is $240 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show the point $A$ on the axes provided in the Printed Answer Booklet.
\item Find the position vector of point $A$.
Give your answer in terms of $\mathbf { i }$ and $\mathbf { j }$.
The point $B$ has position vector $6 \mathbf { i }$.
\end{enumerate}\item Find the exact area of triangle $A O B$.
The point $C$ is such that $O A B C$ is a parallelogram.
\item Find the position vector of $C$.
Give your answer in terms of $\mathbf { i }$ and $\mathbf { j }$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2019 Q2 [8]}}