| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Geometric properties using vectors |
| Difficulty | Standard +0.3 This is a straightforward vectors question requiring routine midpoint calculations, verification of parallelogram properties using vectors, and expressing vectors in terms of given vectors. Part (b) involves more steps but uses standard vector addition techniques without requiring novel insight. Slightly easier than average due to its methodical, step-by-step nature. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0.5,\ 0)\) | B1 | AO 1.1 — Any midpoint correct |
| \(Q(5.5,\ 0.5)\), \(R(4,\ 3.5)\), \(S(-1,\ 3)\) | B1 | AO 1.1 — All midpoints correct. If no labels BOD for 1 or 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient \(PQ=0.1\), or length of \(PQ=\frac{\sqrt{101}}{2}\), or vector \(\overrightarrow{PQ}=\begin{pmatrix}5\\0.5\end{pmatrix}\) | B1 | AO 1.1 — Gradient, length or vector of any one side of PQRS. Or SC1 if KLMN used |
| Gradient \(SR\) is \(\frac{0.5}{5}=0.1=\) gradient \(PQ\), or length of \(SR=\sqrt{0.5^2+5^2}=\frac{\sqrt{101}}{2}\), or vector \(\overrightarrow{SR}=\begin{pmatrix}4-{-1}\\3.5-3\end{pmatrix}=\begin{pmatrix}5\\0.5\end{pmatrix}\); Gradient \(PS=-2\), or length of \(PS=\frac{3\sqrt{5}}{2}\), or vector \(\overrightarrow{PS}=\begin{pmatrix}-1.5\\3\end{pmatrix}\); Gradient \(QR\) is \(\frac{3}{-1.5}=-2\), or length of \(QR=\sqrt{1.5^2+3^2}=\frac{3\sqrt{5}}{2}\), or vector \(\overrightarrow{QR}=\begin{pmatrix}4-5.5\\3.5-0.5\end{pmatrix}=\begin{pmatrix}-1.5\\3\end{pmatrix}\) | M1 | AO 2.2a — Gradient of opposite side of PQRS shown to be equal, or length/vector of opposite side shown to be equal; some working must be seen. Condone confusion of labels for M1/B1. Watch out for valid alternatives e.g. 2 sides equal length and equal gradient |
| Repeat process for other pair of opposite sides and conclude it's a parallelogram | E1 | AO 2.4 — Convincing completion dep on M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{WX} = -3\mathbf{c} - 3\mathbf{a} + 3\mathbf{b} = 3(-\mathbf{a} + \mathbf{b} - \mathbf{c})\) | B1 [1] | Convincing completion; \(\overrightarrow{WX} = \overrightarrow{WV} + \overrightarrow{VT} + \overrightarrow{TX}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AH} = -\mathbf{a} + \mathbf{b}\) | B1 | AO 1.1 |
| \(\overrightarrow{WX} = -3\mathbf{c} - 3\mathbf{a} + 3\mathbf{b}\) | — | — |
| \(\overrightarrow{WE} = -\mathbf{c} - \mathbf{a} + \mathbf{b}\) | — | — |
| \(\overrightarrow{DE} = \mathbf{c} - \mathbf{c} - \mathbf{a} + \mathbf{b} = -\mathbf{a} + \mathbf{b}\); So AH is parallel to DE | E1 [2] | \(\overrightarrow{DE}\) from any correct route must be shown; convincing completion with conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{BC} = \mathbf{a} + \mathbf{c}\) | — | — |
| \(\overrightarrow{GF} = \mathbf{b} - (-\mathbf{c} - \mathbf{a} + \mathbf{b}) = \mathbf{c} + \mathbf{a}\) | B1 | AO 2.2a; \(\overrightarrow{GF}\) from any correct route must be shown |
| \(\overrightarrow{BC} = \overrightarrow{GF}\) so they are parallel | E1 [2] | Convincing completion with conclusion |
## Question 6(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0.5,\ 0)$ | B1 | AO 1.1 — Any midpoint correct |
| $Q(5.5,\ 0.5)$, $R(4,\ 3.5)$, $S(-1,\ 3)$ | B1 | AO 1.1 — All midpoints correct. If no labels BOD for 1 or 2 marks |
**Total: [2]**
---
## Question 6(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient $PQ=0.1$, or length of $PQ=\frac{\sqrt{101}}{2}$, or vector $\overrightarrow{PQ}=\begin{pmatrix}5\\0.5\end{pmatrix}$ | B1 | AO 1.1 — Gradient, length or vector of any one side of PQRS. Or SC1 if KLMN used |
| Gradient $SR$ is $\frac{0.5}{5}=0.1=$ gradient $PQ$, or length of $SR=\sqrt{0.5^2+5^2}=\frac{\sqrt{101}}{2}$, or vector $\overrightarrow{SR}=\begin{pmatrix}4-{-1}\\3.5-3\end{pmatrix}=\begin{pmatrix}5\\0.5\end{pmatrix}$; Gradient $PS=-2$, or length of $PS=\frac{3\sqrt{5}}{2}$, or vector $\overrightarrow{PS}=\begin{pmatrix}-1.5\\3\end{pmatrix}$; Gradient $QR$ is $\frac{3}{-1.5}=-2$, or length of $QR=\sqrt{1.5^2+3^2}=\frac{3\sqrt{5}}{2}$, or vector $\overrightarrow{QR}=\begin{pmatrix}4-5.5\\3.5-0.5\end{pmatrix}=\begin{pmatrix}-1.5\\3\end{pmatrix}$ | M1 | AO 2.2a — Gradient of opposite side of PQRS **shown** to be equal, or length/vector of opposite side **shown** to be equal; some working must be seen. Condone confusion of labels for M1/B1. Watch out for valid alternatives e.g. 2 sides equal length and equal gradient |
| Repeat process for other pair of opposite sides and conclude it's a parallelogram | E1 | AO 2.4 — Convincing completion dep on M1 |
**Total: [3]**
## Question 6(b)(i):
$\overrightarrow{WX} = -3\mathbf{c} - 3\mathbf{a} + 3\mathbf{b} = 3(-\mathbf{a} + \mathbf{b} - \mathbf{c})$ | B1 [1] | Convincing completion; $\overrightarrow{WX} = \overrightarrow{WV} + \overrightarrow{VT} + \overrightarrow{TX}$
## Question 6(b)(ii):
$\overrightarrow{AH} = -\mathbf{a} + \mathbf{b}$ | B1 | AO 1.1
$\overrightarrow{WX} = -3\mathbf{c} - 3\mathbf{a} + 3\mathbf{b}$ | — | —
$\overrightarrow{WE} = -\mathbf{c} - \mathbf{a} + \mathbf{b}$ | — | —
$\overrightarrow{DE} = \mathbf{c} - \mathbf{c} - \mathbf{a} + \mathbf{b} = -\mathbf{a} + \mathbf{b}$; So AH is parallel to DE | E1 [2] | $\overrightarrow{DE}$ from any correct route must be shown; convincing completion with conclusion
## Question 6(b)(iii):
$\overrightarrow{BC} = \mathbf{a} + \mathbf{c}$ | — | —
$\overrightarrow{GF} = \mathbf{b} - (-\mathbf{c} - \mathbf{a} + \mathbf{b}) = \mathbf{c} + \mathbf{a}$ | B1 | AO 2.2a; $\overrightarrow{GF}$ from any correct route must be shown
$\overrightarrow{BC} = \overrightarrow{GF}$ so they are parallel | E1 [2] | Convincing completion with conclusion
---6
\begin{enumerate}[label=(\alph*)]
\item Quadrilateral KLMN has vertices $\mathrm { K } ( - 4,1 ) , \mathrm { L } ( 5 , - 1 ) , \mathrm { M } ( 6,2 )$ and $\mathrm { N } ( 2,5 )$, as shown in Fig. 6.1.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\includegraphics[alt={},max width=\textwidth]{20639e13-01cc-4d96-b694-fb3cf1828f4d-06_567_1004_404_319}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the following midpoints.
\begin{itemize}
\item P , the midpoint of KL
\item Q, the midpoint of LM
\item R, the midpoint of MN
\item S, the midpoint of NK
\item Verify that PQRS is a parallelogram.
\item TVWX is a quadrilateral as shown in Fig. 6.2.
\end{itemize}
Points A and B divide side TV into 3 equal parts. Points C and D divide side VW into 3 equal parts. Points E and F divide side WX into 3 equal parts. Points G and H divide side TX into 3 equal parts.\\
$\overrightarrow { \mathrm { TA } } = \mathbf { a } , \quad \overrightarrow { \mathrm { TH } } = \mathbf { b } , \quad \overrightarrow { \mathrm { VC } } = \mathbf { c }$.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\includegraphics[alt={},max width=\textwidth]{20639e13-01cc-4d96-b694-fb3cf1828f4d-06_577_671_1877_319}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that $\overrightarrow { \mathrm { WX } } = k ( - \mathbf { a } + \mathbf { b } - \mathbf { c } )$, where $k$ is a constant to be determined.
\item Verify that AH is parallel to DE .
\item Verify that BC is parallel to GF .
\end{enumerate}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q6 [10]}}