| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Ratio division of line segment |
| Difficulty | Standard +0.3 This is a standard A-level vectors question involving midpoints, ratio division, and equating two expressions for the same point. Part (a) requires basic vector addition, part (b) involves parametric form (which is given in the answer), and part (c) requires equating coefficients—all routine techniques for vectors at this level. Slightly easier than average due to scaffolding and the answer being provided in part (b). |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{CM} = \overrightarrow{CA} + \overrightarrow{AM} = \overrightarrow{CA} + \frac{1}{2}\overrightarrow{AB} \Rightarrow \overrightarrow{CM} = -\mathbf{a} + \frac{1}{2}(\mathbf{b}-\mathbf{a})\) | M1 | 3.1a — Valid attempt to find \(\overrightarrow{CM}\) using a combination of known vectors a and b |
| \(\overrightarrow{CM} = \overrightarrow{CB} + \overrightarrow{BM} = \overrightarrow{CB} + \frac{1}{2}\overrightarrow{BA} \Rightarrow \overrightarrow{CM} = (-2\mathbf{a}+\mathbf{b}) + \frac{1}{2}(\mathbf{a}-\mathbf{b})\) | M1 | Alternative route |
| \(\overrightarrow{CM} = -\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) | A1 | 1.1b — Simplified correct answer; needs to be simplified and seen in (a) only |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{ON} = \overrightarrow{OC} + \overrightarrow{CN} \Rightarrow \overrightarrow{ON} = \overrightarrow{OC} + \lambda\overrightarrow{CM}\) | M1 | 1.1b — Uses \(\overrightarrow{ON} = \overrightarrow{OC} + \lambda\overrightarrow{CM}\) |
| \(\overrightarrow{ON} = 2\mathbf{a} + \lambda\left(-\frac{3}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}\right) \Rightarrow \overrightarrow{ON} = \left(2-\frac{3}{2}\lambda\right)\mathbf{a}+\frac{1}{2}\lambda\mathbf{b}\) | A1* | 2.1 — Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left(2-\frac{3}{2}\lambda\right) = 0 \Rightarrow \lambda = \ldots\) | M1 | 2.2a — Deduces coefficient equals zero and attempts to find \(\lambda\) |
| \(\lambda = \frac{4}{3} \Rightarrow \overrightarrow{ON} = \frac{2}{3}\mathbf{b}\), \(\overrightarrow{NB} = \frac{1}{3}\mathbf{b}\), \(ON:NB = 2:1\) | A1* | 2.1 — Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{ON} = \mu\mathbf{b} \Rightarrow \left(2-\frac{3}{2}\lambda\right)\mathbf{a}+\frac{1}{2}\lambda\mathbf{b} = \mu\mathbf{b}\); coefficient of a: \(\left(2-\frac{3}{2}\lambda\right)=0 \Rightarrow \lambda=\ldots\); coefficient of b: \(\frac{1}{2}\lambda = \mu\); \(\lambda=\frac{4}{3} \Rightarrow \mu=\frac{2}{3}\) | M1 | 2.2a |
| \(\lambda=\frac{4}{3}\) or \(\mu=\frac{2}{3} \Rightarrow \overrightarrow{ON}=\frac{2}{3}\mathbf{b}\), \(\overrightarrow{NB}=\frac{1}{3}\mathbf{b} \Rightarrow ON:NB=2:1\) | A1* | 2.1 — Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{OB} = \overrightarrow{ON}+\overrightarrow{NB} \Rightarrow \mathbf{b} = \left(2-\frac{3}{2}\lambda\right)\mathbf{a}+\frac{1}{2}\lambda\mathbf{b}+K\mathbf{b}\); a: \(\left(2-\frac{3}{2}\lambda\right)=0 \Rightarrow \lambda=\ldots\); b: \(\frac{1}{2}\lambda+K=1\), \(\lambda=\frac{4}{3} \Rightarrow K=\frac{1}{3}\) | M1 | 2.2a |
| \(\lambda=\frac{4}{3}\) or \(K=\frac{1}{3} \Rightarrow \overrightarrow{ON}=\frac{2}{3}\mathbf{b}\), \(\overrightarrow{NB}=\frac{1}{3}\mathbf{b} \Rightarrow ON:NB=2:1\) | A1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{ON}=\mu\mathbf{b}\) and \(\overrightarrow{CN}=k\overrightarrow{CM} \Rightarrow \overrightarrow{CO}+\overrightarrow{ON}=k\overrightarrow{CM}\); \(-2\mathbf{a}+\mu\mathbf{b}=k\left(-\frac{3}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}\right)\); a: \(-2=-\frac{3}{2}k \Rightarrow k=\frac{4}{3}\); b: \(\mu=\frac{1}{2}k \Rightarrow \mu=\frac{1}{2}\left(\frac{4}{3}\right)=\ldots\) | M1 | 2.2a — Complete attempt to find value of \(\mu\) |
| \(\mu=\frac{2}{3} \Rightarrow \overrightarrow{ON}=\frac{2}{3}\mathbf{b}\), \(\overrightarrow{NB}=\frac{1}{3}\mathbf{b} \Rightarrow ON:NB=2:1\) | A1 | 2.1 — Correct proof |
# Question 10:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{CM} = \overrightarrow{CA} + \overrightarrow{AM} = \overrightarrow{CA} + \frac{1}{2}\overrightarrow{AB} \Rightarrow \overrightarrow{CM} = -\mathbf{a} + \frac{1}{2}(\mathbf{b}-\mathbf{a})$ | M1 | 3.1a — Valid attempt to find $\overrightarrow{CM}$ using a combination of known vectors **a** and **b** |
| $\overrightarrow{CM} = \overrightarrow{CB} + \overrightarrow{BM} = \overrightarrow{CB} + \frac{1}{2}\overrightarrow{BA} \Rightarrow \overrightarrow{CM} = (-2\mathbf{a}+\mathbf{b}) + \frac{1}{2}(\mathbf{a}-\mathbf{b})$ | M1 | Alternative route |
| $\overrightarrow{CM} = -\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$ | A1 | 1.1b — Simplified correct answer; needs to be simplified and seen in (a) only |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{ON} = \overrightarrow{OC} + \overrightarrow{CN} \Rightarrow \overrightarrow{ON} = \overrightarrow{OC} + \lambda\overrightarrow{CM}$ | M1 | 1.1b — Uses $\overrightarrow{ON} = \overrightarrow{OC} + \lambda\overrightarrow{CM}$ |
| $\overrightarrow{ON} = 2\mathbf{a} + \lambda\left(-\frac{3}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}\right) \Rightarrow \overrightarrow{ON} = \left(2-\frac{3}{2}\lambda\right)\mathbf{a}+\frac{1}{2}\lambda\mathbf{b}$ | A1* | 2.1 — Correct proof |
## Part (c) — Way 1
| Working | Mark | Guidance |
|---------|------|----------|
| $\left(2-\frac{3}{2}\lambda\right) = 0 \Rightarrow \lambda = \ldots$ | M1 | 2.2a — Deduces coefficient equals zero and attempts to find $\lambda$ |
| $\lambda = \frac{4}{3} \Rightarrow \overrightarrow{ON} = \frac{2}{3}\mathbf{b}$, $\overrightarrow{NB} = \frac{1}{3}\mathbf{b}$, $ON:NB = 2:1$ | A1* | 2.1 — Correct proof |
## Part (c) — Way 2
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{ON} = \mu\mathbf{b} \Rightarrow \left(2-\frac{3}{2}\lambda\right)\mathbf{a}+\frac{1}{2}\lambda\mathbf{b} = \mu\mathbf{b}$; coefficient of **a**: $\left(2-\frac{3}{2}\lambda\right)=0 \Rightarrow \lambda=\ldots$; coefficient of **b**: $\frac{1}{2}\lambda = \mu$; $\lambda=\frac{4}{3} \Rightarrow \mu=\frac{2}{3}$ | M1 | 2.2a |
| $\lambda=\frac{4}{3}$ or $\mu=\frac{2}{3} \Rightarrow \overrightarrow{ON}=\frac{2}{3}\mathbf{b}$, $\overrightarrow{NB}=\frac{1}{3}\mathbf{b} \Rightarrow ON:NB=2:1$ | A1* | 2.1 — Correct proof |
## Part (c) — Way 3
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{OB} = \overrightarrow{ON}+\overrightarrow{NB} \Rightarrow \mathbf{b} = \left(2-\frac{3}{2}\lambda\right)\mathbf{a}+\frac{1}{2}\lambda\mathbf{b}+K\mathbf{b}$; **a**: $\left(2-\frac{3}{2}\lambda\right)=0 \Rightarrow \lambda=\ldots$; **b**: $\frac{1}{2}\lambda+K=1$, $\lambda=\frac{4}{3} \Rightarrow K=\frac{1}{3}$ | M1 | 2.2a |
| $\lambda=\frac{4}{3}$ or $K=\frac{1}{3} \Rightarrow \overrightarrow{ON}=\frac{2}{3}\mathbf{b}$, $\overrightarrow{NB}=\frac{1}{3}\mathbf{b} \Rightarrow ON:NB=2:1$ | A1 | 2.1 |
## Part (c) — Way 4
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{ON}=\mu\mathbf{b}$ and $\overrightarrow{CN}=k\overrightarrow{CM} \Rightarrow \overrightarrow{CO}+\overrightarrow{ON}=k\overrightarrow{CM}$; $-2\mathbf{a}+\mu\mathbf{b}=k\left(-\frac{3}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}\right)$; **a**: $-2=-\frac{3}{2}k \Rightarrow k=\frac{4}{3}$; **b**: $\mu=\frac{1}{2}k \Rightarrow \mu=\frac{1}{2}\left(\frac{4}{3}\right)=\ldots$ | M1 | 2.2a — Complete attempt to find value of $\mu$ |
| $\mu=\frac{2}{3} \Rightarrow \overrightarrow{ON}=\frac{2}{3}\mathbf{b}$, $\overrightarrow{NB}=\frac{1}{3}\mathbf{b} \Rightarrow ON:NB=2:1$ | A1 | 2.1 — Correct proof |
---10.
Figure 7
Figure 7 shows a sketch of triangle $O A B$.\\
The point $C$ is such that $\overrightarrow { O C } = 2 \overrightarrow { O A }$.\\
The point $M$ is the midpoint of $A B$.\\
The straight line through $C$ and $M$ cuts $O B$ at the point $N$.\\
Given $\overrightarrow { O A } = \mathbf { a }$ and $\overrightarrow { O B } = \mathbf { b }$
\begin{enumerate}[label=(\alph*)]
\item Find $\overrightarrow { C M }$ in terms of $\mathbf { a }$ and $\mathbf { b }$
\item Show that $\overrightarrow { O N } = \left( 2 - \frac { 3 } { 2 } \lambda \right) \mathbf { a } + \frac { 1 } { 2 } \lambda \mathbf { b }$, where $\lambda$ is a scalar constant.
\item Hence prove that $O N : N B = 2 : 1$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2019 Q10 [6]}}