| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Collinearity and ratio division |
| Difficulty | Standard +0.3 This is a straightforward vectors question requiring standard techniques: part (a) uses collinearity (finding scalar multiple between vectors AB and AC), and part (b) involves setting up a parallel vector condition and solving. Both parts are routine applications of A-level vector methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts two of: \(\pm\overrightarrow{AB} = \pm(-4\mathbf{i}+7\mathbf{j}+\mathbf{k})\), \(\pm\overrightarrow{AC} = \pm(-20\mathbf{i}+(p+3)\mathbf{j}+5\mathbf{k})\), \(\pm\overrightarrow{BC} = \pm(-16\mathbf{i}+(p-4)\mathbf{j}+4\mathbf{k})\) | M1 | Attempts two of the three relevant vectors by subtracting either way; method implied by 2 correct components |
| Uses two vectors to find \(p\), e.g. \(p+3 = 5\times7\) | dM1 | Key step: vectors are parallel so multiples of each other (multiple \(\neq 1\)) e.g. \(p+3=5\times7\), \(p-4=\frac{4}{5}(p+3)\), \(p-4=4\times7\) |
| \(p = 32\) | A1 | Condone \(32\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(\overrightarrow{OD} = \lambda\overrightarrow{OB} = 4\lambda\mathbf{j}+6\lambda\mathbf{k}\) and attempts \(\overrightarrow{CD} = 16\mathbf{i}+(4\lambda-\text{"32"})\mathbf{j}+(6\lambda-10)\mathbf{k}\) | M1 | Note correct vector is \(20\mathbf{j}+30\mathbf{k}\) |
| Correct attempt at \(\lambda\) using \(\overrightarrow{CD}\) parallel to \(\overrightarrow{OA}\); e.g. \(4\lambda-32=-12 \Rightarrow \lambda=\ldots\) OR \(6\lambda-10=20 \Rightarrow \lambda=\ldots\) | dM1 | \(\overrightarrow{OA} = 4\mathbf{i}-3\mathbf{j}+5\mathbf{k}\) |
| \(\lvert\overrightarrow{OD}\rvert = 5\times\sqrt{4^2+6^2} = 10\sqrt{13}\) | A1 |
# Question 13:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts two of: $\pm\overrightarrow{AB} = \pm(-4\mathbf{i}+7\mathbf{j}+\mathbf{k})$, $\pm\overrightarrow{AC} = \pm(-20\mathbf{i}+(p+3)\mathbf{j}+5\mathbf{k})$, $\pm\overrightarrow{BC} = \pm(-16\mathbf{i}+(p-4)\mathbf{j}+4\mathbf{k})$ | M1 | Attempts two of the three relevant vectors by subtracting either way; method implied by 2 correct components |
| Uses two vectors to find $p$, e.g. $p+3 = 5\times7$ | dM1 | Key step: vectors are parallel so multiples of each other (multiple $\neq 1$) e.g. $p+3=5\times7$, $p-4=\frac{4}{5}(p+3)$, $p-4=4\times7$ |
| $p = 32$ | A1 | Condone $32\mathbf{j}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $\overrightarrow{OD} = \lambda\overrightarrow{OB} = 4\lambda\mathbf{j}+6\lambda\mathbf{k}$ and attempts $\overrightarrow{CD} = 16\mathbf{i}+(4\lambda-\text{"32"})\mathbf{j}+(6\lambda-10)\mathbf{k}$ | M1 | Note correct vector is $20\mathbf{j}+30\mathbf{k}$ |
| Correct attempt at $\lambda$ using $\overrightarrow{CD}$ parallel to $\overrightarrow{OA}$; e.g. $4\lambda-32=-12 \Rightarrow \lambda=\ldots$ OR $6\lambda-10=20 \Rightarrow \lambda=\ldots$ | dM1 | $\overrightarrow{OA} = 4\mathbf{i}-3\mathbf{j}+5\mathbf{k}$ |
| $\lvert\overrightarrow{OD}\rvert = 5\times\sqrt{4^2+6^2} = 10\sqrt{13}$ | A1 | |\begin{enumerate}
\item Relative to a fixed origin $O$
\end{enumerate}
\begin{itemize}
\item the point $A$ has position vector $4 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k }$
\item the point $B$ has position vector $4 \mathbf { j } + 6 \mathbf { k }$
\item the point $C$ has position vector $- 16 \mathbf { i } + p \mathbf { j } + 10 \mathbf { k }$\\
where $p$ is a constant.\\
Given that $A , B$ and $C$ lie on a straight line,\\
(a) find the value of $p$.
\end{itemize}
The line segment $O B$ is extended to a point $D$ so that $\overrightarrow { C D }$ is parallel to $\overrightarrow { O A }$ (b) Find $| \overrightarrow { O D } |$, writing your answer as a fully simplified surd.
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q13 [6]}}