| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Collinearity and ratio division |
| Difficulty | Standard +0.3 This is a standard 3D vectors question requiring routine application of midpoint and ratio division formulas, followed by showing line intersection by equating parametric forms. The cuboid structure provides clear geometric relationships, and the techniques are well-practiced A-level material with no novel insight required. |
| Spec | 1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{OM}=\frac{1}{2}(\mathbf{b}+\mathbf{c})\) or \(\mathbf{b}+\frac{1}{2}(-\mathbf{b}+\mathbf{c})\) oe | B1 | Can be implied |
| \(\overrightarrow{AM}\) or \(\overrightarrow{MA}\) attempted in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\); \(\left(=\pm\left(\frac{1}{2}(\mathbf{b}+\mathbf{c})-\mathbf{a}\right)\text{ oe}\right)\) | M1 | May be included in working, eg \(\overrightarrow{AX}=\frac{2}{3}\left(\frac{1}{2}(\mathbf{b}+\mathbf{c})-\mathbf{a}\right)\). Not necessarily correct |
| \(\overrightarrow{OX}=\mathbf{a}+\frac{2}{3}\overrightarrow{AM}\) or \(\overrightarrow{OM}+\frac{1}{3}\overrightarrow{MA}\) oe, attempted in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) | M1 | Not necessarily correct |
| \(\overrightarrow{OX}=\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})\) | A1 | or equivalent simplified form |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})\) | B1 | |
| \(\overrightarrow{AX} = \frac{2}{3}\overrightarrow{AM} = \frac{2}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})\) | M1 | for \(\overrightarrow{AM} = (\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})\) implied |
| \(\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}\overrightarrow{AM} = \mathbf{a} + \frac{2}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})\) | M1 | |
| \(= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{BM} = \frac{1}{2}(-\mathbf{b} + \mathbf{c})\) | B1 | Implied |
| \(\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{BM} = -\mathbf{a} + \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c}\) | M1 | |
| \(\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}(-\mathbf{a} + \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c})\) | M1 | |
| \(= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})\) | B1 | |
| \(\overrightarrow{XM} = \frac{1}{3}\overrightarrow{AM} = \frac{1}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})\) | M1 | for \(\overrightarrow{AM} = (\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})\) implied |
| \(\overrightarrow{OX} = \overrightarrow{OM} - \frac{1}{3}\overrightarrow{AM} = \frac{1}{2}(\mathbf{b} + \mathbf{c}) - \frac{1}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})\) | M1 | equivalent to \(\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MA}\) |
| \(= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OF} = \mathbf{a} + \mathbf{b} + \mathbf{c}\) | B1* | |
| Hence \(X\) lies on \(OF\), so \(AM\) and \(OF\) intersect | B1dep [2] | Both statements needed. NB dep on B1 |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{OM}=\frac{1}{2}(\mathbf{b}+\mathbf{c})$ or $\mathbf{b}+\frac{1}{2}(-\mathbf{b}+\mathbf{c})$ oe | **B1** | Can be implied |
| $\overrightarrow{AM}$ or $\overrightarrow{MA}$ attempted in terms of $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$; $\left(=\pm\left(\frac{1}{2}(\mathbf{b}+\mathbf{c})-\mathbf{a}\right)\text{ oe}\right)$ | **M1** | May be included in working, eg $\overrightarrow{AX}=\frac{2}{3}\left(\frac{1}{2}(\mathbf{b}+\mathbf{c})-\mathbf{a}\right)$. Not necessarily correct |
| $\overrightarrow{OX}=\mathbf{a}+\frac{2}{3}\overrightarrow{AM}$ or $\overrightarrow{OM}+\frac{1}{3}\overrightarrow{MA}$ oe, attempted in terms of $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ | **M1** | Not necessarily correct |
| $\overrightarrow{OX}=\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$ | **A1** | or equivalent simplified form |
## Question 9(a) continued:
**Method 1:**
| $\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})$ | B1 | |
|---|---|---|
| $\overrightarrow{AX} = \frac{2}{3}\overrightarrow{AM} = \frac{2}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})$ | M1 | for $\overrightarrow{AM} = (\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})$ implied |
| $\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}\overrightarrow{AM} = \mathbf{a} + \frac{2}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})$ | M1 | |
| $= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ | A1 | |
**Method 2:**
| $\overrightarrow{BM} = \frac{1}{2}(-\mathbf{b} + \mathbf{c})$ | B1 | Implied |
|---|---|---|
| $\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{BM} = -\mathbf{a} + \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c}$ | M1 | |
| $\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}(-\mathbf{a} + \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c})$ | M1 | |
| $= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ | A1 | |
**Method 3:**
| $\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})$ | B1 | |
|---|---|---|
| $\overrightarrow{XM} = \frac{1}{3}\overrightarrow{AM} = \frac{1}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})$ | M1 | for $\overrightarrow{AM} = (\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})$ implied |
| $\overrightarrow{OX} = \overrightarrow{OM} - \frac{1}{3}\overrightarrow{AM} = \frac{1}{2}(\mathbf{b} + \mathbf{c}) - \frac{1}{3}(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a})$ | M1 | equivalent to $\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MA}$ |
| $= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ | A1 | |
---
## Question 9(b):
| $\overrightarrow{OF} = \mathbf{a} + \mathbf{b} + \mathbf{c}$ | B1* | |
|---|---|---|
| Hence $X$ lies on $OF$, so $AM$ and $OF$ intersect | B1dep [2] | Both statements needed. NB dep on B1 |
---9 Points $A , B$ and $C$ have position vectors $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$ relative to an origin $O$ in 3-dimensional space. Rectangles $O A D C$ and $B E F G$ are the base and top surface of a cuboid.\\
\includegraphics[max width=\textwidth, alt={}, center]{7298e7b9-ad52-480c-bc2b-8289aeab9ebb-07_522_812_952_280}
\begin{itemize}
\item The point $M$ is the midpoint of $B C$.
\item The point $X$ lies on $A M$ such that $A X = 2 X M$.
\begin{enumerate}[label=(\alph*)]
\item Find $\overrightarrow { O X }$ in terms of $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$, simplifying your answer.
\item Hence show that the lines $O F$ and $A M$ intersect.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2021 Q9 [6]}}