| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Vector between two points |
| Difficulty | Moderate -0.8 This is a straightforward AS-level vectors question testing basic operations: finding a vector between two points (subtraction), magnitude calculation, and collinearity. Parts (a)-(b) are routine recall, while (c)(i) requires recognizing collinearity means vectors are scalar multiples. Part (c)(ii) is slightly more conceptual but follows directly from the collinearity result. Overall easier than average due to standard techniques and clear structure. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (-8\mathbf{i}+9\mathbf{j})-(10\mathbf{i}-3\mathbf{j})\) | M1 | Subtraction either way; cannot award for adding vectors |
| \(= -18\mathbf{i}+12\mathbf{j}\) | A1 | cao; condone \(\begin{pmatrix}-18\\12\end{pmatrix}\); do not allow \(\begin{pmatrix}-18\mathbf{i}\\12\mathbf{j}\end{pmatrix}\) or \((-18,12)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | \overrightarrow{AB} | = \sqrt{"18"^2 + "12"^2}\ \{=\sqrt{468}\}\) |
| \(= 6\sqrt{13}\) | A1 | cao; may come from \(\begin{pmatrix}\pm18\\\pm12\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = \lambda\overrightarrow{BC} \Rightarrow -18\mathbf{i}+12\mathbf{j} = 6\lambda\mathbf{i}+\lambda(p-9)\mathbf{j}\) with components equated | M1 | Key step: \(BCA\) forms straight line; condone sign slips; award for \(\pm\frac{p-9}{6} = \pm\frac{2}{3}\) leading to \(p=\ldots\) |
| (i) \(p = 5\) | A1 | Implied by \(p=5\) unless from clearly incorrect work |
| (ii) ratio \(= 2:3\) | B1 (A1 on EPEN) | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\vec{AB} = \alpha\vec{AC} \Rightarrow -18\mathbf{i}+12\mathbf{j} = -12\alpha\mathbf{i}+\alpha(p+3)\mathbf{j}\) with components equated leading to a value for \(\alpha\) and \(p\) | M1 | Award for attempt to use vectors |
| gradient \(BC\) = gradient \(BA = -\frac{2}{3}\), e.g. \(\frac{p-9}{6} = \frac{9--3}{-8-10}\) leading to \(p=\ldots\) | M1 | Alternative method |
| Triangles \(BCM\) and \(BAN\) similar with lengths ratio 1:3, e.g. \(p = 9-\frac{1}{3}\times12\) or \(p=-3+\frac{2}{3}\times12\) | M1 | Alternative method |
| Line \(AB\) has equation \(y=-\frac{2}{3}x+\frac{11}{3}\), sub in \(x=-2\) leading to \(p=\ldots\) | M1 | Alternative method |
| \(\frac{p+3}{12}=\frac{2}{3}\) or \(\frac{p+3}{2}=9-p\) leading to \(p=\ldots\) | M1 | Alternative method |
| \(p=5\) | A1 | Correct answer implies both marks unless from clearly incorrect work |
| States ratio \(= 2:3\) or equivalent such as \(1:1.5\) or \(22:33\) | B1 | Note 3:2 is incorrect but condone \(\{\text{Area}\}AOB:\{\text{Area}\}AOC = 3:2\). Area \(AOC=22\), area \(AOB=33\), area \(BOC=11\) |
# Question 13:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (-8\mathbf{i}+9\mathbf{j})-(10\mathbf{i}-3\mathbf{j})$ | M1 | Subtraction either way; cannot award for adding vectors |
| $= -18\mathbf{i}+12\mathbf{j}$ | A1 | cao; condone $\begin{pmatrix}-18\\12\end{pmatrix}$; do not allow $\begin{pmatrix}-18\mathbf{i}\\12\mathbf{j}\end{pmatrix}$ or $(-18,12)$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|\overrightarrow{AB}| = \sqrt{"18"^2 + "12"^2}\ \{=\sqrt{468}\}$ | M1 | Pythagoras on vector from (a); note $-18$ commonly squared as $18$ |
| $= 6\sqrt{13}$ | A1 | cao; may come from $\begin{pmatrix}\pm18\\\pm12\end{pmatrix}$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \lambda\overrightarrow{BC} \Rightarrow -18\mathbf{i}+12\mathbf{j} = 6\lambda\mathbf{i}+\lambda(p-9)\mathbf{j}$ with components equated | M1 | Key step: $BCA$ forms straight line; condone sign slips; award for $\pm\frac{p-9}{6} = \pm\frac{2}{3}$ leading to $p=\ldots$ |
| (i) $p = 5$ | A1 | Implied by $p=5$ unless from clearly incorrect work |
| (ii) ratio $= 2:3$ | B1 (A1 on EPEN) | 2.2a |
## Question 13 (Collinear Points):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\vec{AB} = \alpha\vec{AC} \Rightarrow -18\mathbf{i}+12\mathbf{j} = -12\alpha\mathbf{i}+\alpha(p+3)\mathbf{j}$ with components equated leading to a value for $\alpha$ and $p$ | M1 | Award for attempt to use vectors |
| gradient $BC$ = gradient $BA = -\frac{2}{3}$, e.g. $\frac{p-9}{6} = \frac{9--3}{-8-10}$ leading to $p=\ldots$ | M1 | Alternative method |
| Triangles $BCM$ and $BAN$ similar with lengths ratio 1:3, e.g. $p = 9-\frac{1}{3}\times12$ or $p=-3+\frac{2}{3}\times12$ | M1 | Alternative method |
| Line $AB$ has equation $y=-\frac{2}{3}x+\frac{11}{3}$, sub in $x=-2$ leading to $p=\ldots$ | M1 | Alternative method |
| $\frac{p+3}{12}=\frac{2}{3}$ or $\frac{p+3}{2}=9-p$ leading to $p=\ldots$ | M1 | Alternative method |
| $p=5$ | A1 | Correct answer implies both marks unless from clearly incorrect work |
| States ratio $= 2:3$ or equivalent such as $1:1.5$ or $22:33$ | B1 | Note 3:2 is incorrect but condone $\{\text{Area}\}AOB:\{\text{Area}\}AOC = 3:2$. Area $AOC=22$, area $AOB=33$, area $BOC=11$ |
---\begin{enumerate}
\item Relative to a fixed origin $O$
\end{enumerate}
\begin{itemize}
\item point $A$ has position vector $10 \mathbf { i } - 3 \mathbf { j }$
\item point $B$ has position vector $- 8 \mathbf { i } + 9 \mathbf { j }$
\item point $C$ has position vector $- 2 \mathbf { i } + p \mathbf { j }$ where $p$ is a constant\\
(a) Find $\overrightarrow { A B }$\\
(b) Find $| \overrightarrow { A B } |$ giving your answer as a fully simplified surd.
\end{itemize}
Given that points $A , B$ and $C$ lie on a straight line,\\
(c) (i) find the value of $p$,\\
(ii) state the ratio of the area of triangle $A O C$ to the area of triangle $A O B$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q13 [7]}}