| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Geometric properties using vectors |
| Difficulty | Moderate -0.5 This is a straightforward vector geometry question requiring basic vector manipulation and understanding of collinearity. Part (i) is simple plotting, and part (ii) involves expressing AE or BE in terms of AB, which is a standard technique taught early in C4 vectors with no novel insight required. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Sketch with \(C\) correctly located on sketch | B1 | For \(C\) correctly located on sketch |
| Sketch with \(D\) correctly located on sketch | B1 | For \(D\) correctly located on sketch |
| Sketch with \(E\) correctly located wrt \(O\) and \(D\) | B1 | For \(E\) correctly located wrt \(O\) and \(D\) |
| 3 | ||
| (ii) \(\overrightarrow{AE} = \frac{1}{3}(2a+b) - a = \frac{1}{3}(b-a)\) | M1 | For relevant subtraction involving \(\overrightarrow{OE}\) |
| A1 | For correct expression for \((\pm)\overrightarrow{AE}\) or \(\overrightarrow{EB}\) | |
| A1 | For correct recognition of parallel property | |
| Hence \(AE\) is parallel to \(AB\), i.e. \(E\) lies on the line joining \(A\) to \(B\) | A1 | For complete proof of required result |
| 4 |
**(i)** Sketch with $C$ correctly located on sketch | B1 | For $C$ correctly located on sketch
Sketch with $D$ correctly located on sketch | B1 | For $D$ correctly located on sketch
Sketch with $E$ correctly located wrt $O$ and $D$ | B1 | For $E$ correctly located wrt $O$ and $D$
| **3** |
**(ii)** $\overrightarrow{AE} = \frac{1}{3}(2a+b) - a = \frac{1}{3}(b-a)$ | M1 | For relevant subtraction involving $\overrightarrow{OE}$
| A1 | For correct expression for $(\pm)\overrightarrow{AE}$ or $\overrightarrow{EB}$
| A1 | For correct recognition of parallel property
Hence $AE$ is parallel to $AB$, i.e. $E$ lies on the line joining $A$ to $B$ | A1 | For complete proof of required result
| **4** |4\\
\includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-2_428_572_861_760}
As shown in the diagram the points $A$ and $B$ have position vectors $\mathbf { a }$ and $\mathbf { b }$ with respect to the origin $O$.\\
(i) Make a sketch of the diagram, and mark the points $C , D$ and $E$ such that $\overrightarrow { O C } = 2 \mathbf { a } , \overrightarrow { O D } = 2 \mathbf { a } + \mathbf { b }$ and $\overrightarrow { O E } = \frac { 1 } { 3 } \overrightarrow { O D }$.\\
(ii) By expressing suitable vectors in terms of $\mathbf { a }$ and $\mathbf { b }$, prove that $E$ lies on the line joining $A$ and $B$.
\hfill \mbox{\textit{OCR C4 Q4 [7]}}