| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Ratio division of line segment |
| Difficulty | Easy -1.2 This is a straightforward application of the section formula for dividing a line segment in a given ratio. Students need only recall that Q divides PR in ratio 1:2, then apply the standard formula PQ = (1/3)PR, leading directly to the required result through basic vector arithmetic. It's a routine bookwork-style question with no problem-solving or novel insight required. |
| Spec | 1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts any one of \((\pm\overrightarrow{PQ}=)\pm(\mathbf{q}-\mathbf{p})\), \((\pm\overrightarrow{PR}=)\pm(\mathbf{r}-\mathbf{p})\), \((\pm\overrightarrow{QR}=)\pm(\mathbf{r}-\mathbf{q})\) | M1 | Attempts relevant vectors by subtracting either way; may be implied by sight of any one of \(\pm(\mathbf{q}-\mathbf{p}), \pm(\mathbf{r}-\mathbf{p}), \pm(\mathbf{r}-\mathbf{q})\) |
| e.g. \(\mathbf{r}-\mathbf{q}=2(\mathbf{q}-\mathbf{p})\), \(\mathbf{r}-\mathbf{p}=3(\mathbf{q}-\mathbf{p})\), leading to \(\mathbf{q}=\mathbf{p}+\frac{1}{3}(\mathbf{r}-\mathbf{p})\) or \(\mathbf{q}=\mathbf{r}+\frac{2}{3}(\mathbf{p}-\mathbf{r})\) | dM1 | Uses given information and writes correctly in vector form that if rearranged would give printed answer |
| \(\Rightarrow \mathbf{r}-\mathbf{q}=2\mathbf{q}-2\mathbf{p} \Rightarrow 2\mathbf{p}+\mathbf{r}=3\mathbf{q} \Rightarrow \mathbf{q}=\frac{1}{3}(\mathbf{r}+2\mathbf{p})\)* | A1* | Fully correct work leading to given answer; allow \(OQ = \ldots\) as long as \(OQ\) defined as \(\mathbf{q}\) |
# Question 2 (Vectors):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts any one of $(\pm\overrightarrow{PQ}=)\pm(\mathbf{q}-\mathbf{p})$, $(\pm\overrightarrow{PR}=)\pm(\mathbf{r}-\mathbf{p})$, $(\pm\overrightarrow{QR}=)\pm(\mathbf{r}-\mathbf{q})$ | M1 | Attempts relevant vectors by subtracting either way; may be implied by sight of any one of $\pm(\mathbf{q}-\mathbf{p}), \pm(\mathbf{r}-\mathbf{p}), \pm(\mathbf{r}-\mathbf{q})$ |
| e.g. $\mathbf{r}-\mathbf{q}=2(\mathbf{q}-\mathbf{p})$, $\mathbf{r}-\mathbf{p}=3(\mathbf{q}-\mathbf{p})$, leading to $\mathbf{q}=\mathbf{p}+\frac{1}{3}(\mathbf{r}-\mathbf{p})$ or $\mathbf{q}=\mathbf{r}+\frac{2}{3}(\mathbf{p}-\mathbf{r})$ | dM1 | Uses given information and writes correctly in vector form that if rearranged would give printed answer |
| $\Rightarrow \mathbf{r}-\mathbf{q}=2\mathbf{q}-2\mathbf{p} \Rightarrow 2\mathbf{p}+\mathbf{r}=3\mathbf{q} \Rightarrow \mathbf{q}=\frac{1}{3}(\mathbf{r}+2\mathbf{p})$* | A1* | Fully correct work leading to given answer; allow $OQ = \ldots$ as long as $OQ$ defined as $\mathbf{q}$ |
---\begin{enumerate}
\item Relative to a fixed origin, points $P , Q$ and $R$ have position vectors $\mathbf { p } , \mathbf { q }$ and $\mathbf { r }$ respectively.
\end{enumerate}
Given that
\begin{itemize}
\item $\quad P , Q$ and $R$ lie on a straight line
\item $Q$ lies one third of the way from $P$ to $R$\\
show that
\end{itemize}
$$\mathbf { q } = \frac { 1 } { 3 } ( \mathbf { r } + 2 \mathbf { p } )$$
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q2 [3]}}