| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic vector operations: finding distance using Pythagoras, midpoint formula, and showing two vectors are equal for parallelogram proof. All parts are routine calculations with no problem-solving insight required, making it easier than average but not trivial since it requires multiple standard techniques. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1-(-3))^2 + (-2-(-1))^2 + (5-2)^2 \; (= 26)\) | M1 | Attempt. Allow with one sign error |
| Length \(= \sqrt{26}\) or \(5.10\) or \(5.1\) (2 sf) | A1 [2] | \(\sqrt{\phantom{x}}\) not nec'y |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-1\\-1.5\\3.5\end{pmatrix}\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\vec{BA} = \begin{pmatrix}4\\-1\\3\end{pmatrix}\) | M1 | or quote result for \(\vec{BA}\) from (ii) or (i)(a). SC: Incorrect but equal vectors \(BA\) & \(PQ\) with correct conclusion: SC B1 |
| \(\vec{PQ} = \begin{pmatrix}5\\1\\3\end{pmatrix} - \begin{pmatrix}1\\2\\0\end{pmatrix} = \begin{pmatrix}4\\-1\\3\end{pmatrix}\) | M1 | or similar methods with \(AQ\) & \(BP\), or \(AB\) and \(QP\) etc. Allow find eg \(AB\) and \(PQ\), or \(\vec{BA} = \vec{PQ}\) with arrows, or \( |
| \(BA = PQ\) and \(BA \parallel PQ\), and hence \(ABPQ\) is a parallelogram (AG) | A1 [3] | Both statements needed, dep M1M1. Just \( |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1-(-3))^2 + (-2-(-1))^2 + (5-2)^2 \; (= 26)$ | M1 | Attempt. Allow with one sign error |
| Length $= \sqrt{26}$ or $5.10$ or $5.1$ (2 sf) | A1 **[2]** | $\sqrt{\phantom{x}}$ not nec'y |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-1\\-1.5\\3.5\end{pmatrix}$ | B1 **[1]** | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\vec{BA} = \begin{pmatrix}4\\-1\\3\end{pmatrix}$ | M1 | or quote result for $\vec{BA}$ from (ii) or (i)(a). SC: Incorrect but equal vectors $BA$ & $PQ$ with correct conclusion: SC B1 |
| $\vec{PQ} = \begin{pmatrix}5\\1\\3\end{pmatrix} - \begin{pmatrix}1\\2\\0\end{pmatrix} = \begin{pmatrix}4\\-1\\3\end{pmatrix}$ | M1 | or similar methods with $AQ$ & $BP$, or $AB$ and $QP$ etc. Allow find eg $AB$ and $PQ$, or $\vec{BA} = \vec{PQ}$ with arrows, or $|BA|=|PQ|$ & $|BP|=|AQ|$ shown & stated, or $BA \parallel PQ$ & $BP \parallel AQ$ shown & stated. Allow without method SC Lengths only seen: M1M0 |
| $BA = PQ$ and $BA \parallel PQ$, and hence $ABPQ$ is a parallelogram **(AG)** | A1 **[3]** | Both statements needed, dep M1M1. Just $|BA|=|PQ|$: A0 |
# Mark Scheme Extraction2 The points $A$ and $B$ have position vectors $\left( \begin{array} { c } 1 \\ - 2 \\ 5 \end{array} \right)$ and $\left( \begin{array} { c } - 3 \\ - 1 \\ 2 \end{array} \right)$ respectively.\\
(i) Find the exact length of $A B$.\\
(ii) Find the position vector of the midpoint of $A B$.
The points $P$ and $Q$ have position vectors $\left( \begin{array} { l } 1 \\ 2 \\ 0 \end{array} \right)$ and $\left( \begin{array} { l } 5 \\ 1 \\ 3 \end{array} \right)$ respectively.\\
(iii) Show that $A B P Q$ is a parallelogram.
\hfill \mbox{\textit{OCR H240/02 2018 Q2 [6]}}