| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Geometric properties using vectors |
| Difficulty | Moderate -0.8 This is a straightforward multi-part vectors question requiring basic operations: vector subtraction, dot product to verify perpendicularity, and area calculation using perpendicular vectors. All steps are routine AS-level techniques with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{PQ}=(3-9)\mathbf{i}+(-5+8)\mathbf{j}\) | M1 | Attempts subtraction either way round. Cannot be awarded for adding vectors. May be implied by one correct component or sight of \(\mp6\mathbf{i}\pm3\mathbf{j}\) |
| \(=-6\mathbf{i}+3\mathbf{j}\) | A1 | Correct answer. Allow \(-6\mathbf{i}+3\mathbf{j}\) or \(\begin{pmatrix}-6\\3\end{pmatrix}\) but not \(\begin{pmatrix}-6\mathbf{i}\\3\mathbf{j}\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of \(PQ=\frac{-5--8}{3-9}\left(=-\frac{1}{2}\right)\) and Gradient of \(QR=\frac{18}{9}(=2)\), or \( | \overrightarrow{PQ} | =\sqrt{(-6)^2+3^2}\ (=3\sqrt{5})\) and \( |
| e.g. shows \(-\frac{1}{2}\times2=-1\) and deduces angle \(PQR=90°\), or shows \( | \overrightarrow{PQ} | ^2+ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \( | \overrightarrow{PQ} | =\sqrt{(-6)^2+3^2}\) and either \( |
| \( | \overrightarrow{PQ} | =\sqrt{45}\ (=3\sqrt{5})\) and either \( |
| e.g. Area \(=\frac{1}{2}\times(9\sqrt{5}+27\sqrt{5})\times\sqrt{45}\) or \(\frac{1}{2}\times4\times9\sqrt{5}\times3\sqrt{5}\) | dM1 | Correct method to find area of trapezium. Dependent on first M mark and method to find lengths must be correct |
| \(=270\) | A1 |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{PQ}=(3-9)\mathbf{i}+(-5+8)\mathbf{j}$ | M1 | Attempts subtraction either way round. Cannot be awarded for adding vectors. May be implied by one correct component or sight of $\mp6\mathbf{i}\pm3\mathbf{j}$ |
| $=-6\mathbf{i}+3\mathbf{j}$ | A1 | Correct answer. Allow $-6\mathbf{i}+3\mathbf{j}$ or $\begin{pmatrix}-6\\3\end{pmatrix}$ but not $\begin{pmatrix}-6\mathbf{i}\\3\mathbf{j}\end{pmatrix}$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $PQ=\frac{-5--8}{3-9}\left(=-\frac{1}{2}\right)$ and Gradient of $QR=\frac{18}{9}(=2)$, **or** $|\overrightarrow{PQ}|=\sqrt{(-6)^2+3^2}\ (=3\sqrt{5})$ and $|\overrightarrow{QR}|=\sqrt{9^2+18^2}\ (=9\sqrt{5})$ and $|\overrightarrow{PR}|=\sqrt{3^2+21^2}\ (=15\sqrt{2})$ | M1 | Attempts gradient of $PQ$ and gradient of $QR$, or finds lengths $PQ$, $QR$ and $PR$ or $PQ^2$, $QR^2$ and $PR^2$. Do not allow sign slips |
| e.g. shows $-\frac{1}{2}\times2=-1$ and deduces angle $PQR=90°$, **or** shows $|\overrightarrow{PQ}|^2+|\overrightarrow{QR}|^2=|\overrightarrow{PR}|^2$ and deduces angle $PQR=90°$ | A1* | Correct working AND conclusion that angle $PQR=90°$. Must show product $=-1$ or verify Pythagoras. Using scalar dot product must show $\begin{pmatrix}-6\\3\end{pmatrix}\begin{pmatrix}9\\18\end{pmatrix}=0$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $|\overrightarrow{PQ}|=\sqrt{(-6)^2+3^2}$ **and** either $|\overrightarrow{QR}|=\sqrt{9^2+18^2}$ **or** $|\overrightarrow{PS}|=\sqrt{27^2+54^2}$ | M1 | Correct use of Pythagoras to find length of $PQ$ and at least one of $QR$ or $PS$. Must be used or seen in (c) |
| $|\overrightarrow{PQ}|=\sqrt{45}\ (=3\sqrt{5})$ **and** either $|\overrightarrow{QR}|=\sqrt{405}\ (=9\sqrt{5})$ **or** $|\overrightarrow{PS}|=27\sqrt{5}$ | A1ft | Correct length of $PQ$ and at least one of $QR$ or $PS$. $QR$ must be $\sqrt{405}$. Lengths do not need to be simplified but must be exact |
| e.g. Area $=\frac{1}{2}\times(9\sqrt{5}+27\sqrt{5})\times\sqrt{45}$ or $\frac{1}{2}\times4\times9\sqrt{5}\times3\sqrt{5}$ | dM1 | Correct method to find area of trapezium. Dependent on first M mark and method to find lengths must be correct |
| $=270$ | A1 | |\begin{enumerate}
\item Relative to a fixed origin $O$,
\end{enumerate}
\begin{itemize}
\item point $P$ has position vector $9 \mathbf { i } - 8 \mathbf { j }$
\item point $Q$ has position vector $3 \mathbf { i } - 5 \mathbf { j }$\\
(a) Find $\overrightarrow { P Q }$
\end{itemize}
Given that $R$ is the point such that $\overrightarrow { Q R } = 9 \mathbf { i } + 18 \mathbf { j }$\\
(b) show that angle $P Q R = 90 ^ { \circ }$
Given also that $S$ is the point such that $\overrightarrow { P S } = 3 \overrightarrow { Q R }$\\
(c) find the exact area of $P Q R S$
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q3 [8]}}