| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Vector between two points |
| Difficulty | Moderate -0.8 This is a straightforward multi-part vectors question requiring only standard techniques: finding vector length using Pythagoras, solving a simple quadratic equation for equal lengths, using the midpoint formula, and basic vector arithmetic. Part (d) requires recognizing a trapezium from the geometric properties, but this follows naturally from the calculations. All steps are routine applications of basic vector methods with no novel problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}\) | B1 | Correct length aef. Condone 4.47 or better. Allow isw eg \(\sqrt{20} = 4\sqrt{5}\). Allow BOD on signs eg \(AB = -2\mathbf{i} - 4\mathbf{j}\) seen |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((p-3)^2 + (p-5)^2 = 20\) | M1 | Attempt correct equation for length \(BC\). Using their attempt at length of \(AB\). Condone error on RHS eg having \(\sqrt{20}\) not 20 |
| \(p^2 - 8p + 7 = 0\), \(p = 7\) | A1 | BC. Solve correct quadratic to obtain at least \(p = 7\). If second value of \(p\) stated then it must be correct |
| \(C\) is \(7\mathbf{i} + 7\mathbf{j}\) | A1 | Correct position vector for \(C\); it could be given as column vector, but not coordinate. No need to discard \(p = 1\). If M0, question is 'determine' so some evidence needed for full marks: either justifying lengths are equal, or use of components of 2 and 4. \(7\mathbf{i}+7\mathbf{j}\) with some explanation B3; \(7\mathbf{i}+7\mathbf{j}\) with no explanation B2; \((7,7)\) with some explanation B2; \((7,7)\) with no explanation B1 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OM\) is \(4\mathbf{i} + 4\mathbf{j}\) OR \(BM\) is \(\mathbf{i} - \mathbf{j}\) | B1 | Correct midpoint soi. Could instead find vector \(BM\). Allow \(M\) seen as coordinate, as it is part of their method and not a requested answer. Condone \(M = 4\mathbf{i} + 4\mathbf{j}\), but penalise clear error eg \(AM = 4\mathbf{i} + 4\mathbf{j}\) is B0. Could be soi on a diagram |
| \(D\) is \(6\mathbf{i} + 2\mathbf{j}\) | B1 | Correct position vector (not coordinate) for \(D\). Do not penalise \(D\) given as coordinate if already penalised in part (b) |
| [2] | Answer only is B0B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Kite | B1* | Mark independently of reason |
| eg two pairs of adjacent sides of same length | B1dep* | Evidence is required to support statements made. \(AD = CD = \sqrt{26}\) (or compare components of vectors); condone not stating \(AB = BC\) as given in question. Sides must be defined as adjacent, so B0 for just 'two pairs of equal sides', but allow BOD if clarified on an explicit diagram seen in part (d) |
| eg diagonals are perpendicular | \(AC\) has gradient of 1, \(BD\) has gradient of \(-1\). If using a geometrical argument, then identify that \(ABC\) is isosceles, \(M\) is mid-point of \(AC\) hence perpendicular bisector | |
| eg \(BD\) being a line of symmetry | \(AM = MC\), with perpendicular argument as above. B0 for reasoning using angles (ie a pair of facing equal angles) unless justified | |
| [2] |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$ | **B1** | Correct length aef. Condone 4.47 or better. Allow isw eg $\sqrt{20} = 4\sqrt{5}$. Allow BOD on signs eg $AB = -2\mathbf{i} - 4\mathbf{j}$ seen |
| **[1]** | | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(p-3)^2 + (p-5)^2 = 20$ | **M1** | Attempt correct equation for length $BC$. Using their attempt at length of $AB$. Condone error on RHS eg having $\sqrt{20}$ not 20 |
| $p^2 - 8p + 7 = 0$, $p = 7$ | **A1** | BC. Solve correct quadratic to obtain at least $p = 7$. If second value of $p$ stated then it must be correct |
| $C$ is $7\mathbf{i} + 7\mathbf{j}$ | **A1** | Correct position vector for $C$; it could be given as column vector, but not coordinate. No need to discard $p = 1$. If **M0**, question is 'determine' so some evidence needed for full marks: either justifying lengths are equal, or use of components of 2 and 4. $7\mathbf{i}+7\mathbf{j}$ with some explanation **B3**; $7\mathbf{i}+7\mathbf{j}$ with no explanation **B2**; $(7,7)$ with some explanation **B2**; $(7,7)$ with no explanation **B1** |
| **[3]** | | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OM$ is $4\mathbf{i} + 4\mathbf{j}$ OR $BM$ is $\mathbf{i} - \mathbf{j}$ | **B1** | Correct midpoint soi. Could instead find vector $BM$. Allow $M$ seen as coordinate, as it is part of their method and not a requested answer. Condone $M = 4\mathbf{i} + 4\mathbf{j}$, but penalise clear error eg $AM = 4\mathbf{i} + 4\mathbf{j}$ is **B0**. Could be soi on a diagram |
| $D$ is $6\mathbf{i} + 2\mathbf{j}$ | **B1** | Correct position vector (not coordinate) for $D$. Do not penalise $D$ given as coordinate if already penalised in part (b) |
| **[2]** | Answer only is **B0B1** | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Kite | **B1*** | Mark independently of reason |
| eg two pairs of adjacent sides of same length | **B1dep*** | Evidence is required to support statements made. $AD = CD = \sqrt{26}$ (or compare components of vectors); condone not stating $AB = BC$ as given in question. Sides must be defined as adjacent, so **B0** for just 'two pairs of equal sides', but allow BOD if clarified on an explicit diagram seen in part (d) |
| eg diagonals are perpendicular | | $AC$ has gradient of 1, $BD$ has gradient of $-1$. If using a geometrical argument, then identify that $ABC$ is isosceles, $M$ is mid-point of $AC$ hence perpendicular bisector |
| eg $BD$ being a line of symmetry | | $AM = MC$, with perpendicular argument as above. **B0** for reasoning using angles (ie a pair of facing equal angles) unless justified |
| **[2]** | | |
---4 It is given that $A B C D$ is a quadrilateral. The position vector of $A$ is $\mathbf { i } + \mathbf { j }$, and the position vector of $B$ is $3 \mathbf { i } + 5 \mathbf { j }$.
\begin{enumerate}[label=(\alph*)]
\item Find the length $A B$.
\item The position vector of $C$ is $p \mathbf { i } + p \mathbf { j }$ where $p$ is a constant greater than 1 .
Given that the length $A B$ is equal to the length $B C$, determine the position vector of $C$.
\item The point $M$ is the midpoint of $A C$.
Given that $\overrightarrow { M D } = 2 \overrightarrow { B M }$, determine the position vector of $D$.
\item State the name of the quadrilateral $A B C D$, giving a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q4 [8]}}