| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Triangle and parallelogram problems |
| Difficulty | Moderate -0.3 This is a straightforward vectors question testing standard techniques: part (i) uses the parallelogram property that opposite sides are equal (routine vector addition), and part (ii) requires finding the angle between two vectors using the scalar product formula. Both are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD}\) or \(\overrightarrow{OB} + \overrightarrow{BC} + \overrightarrow{CD}\) | AEF | M1 |
| \(\overrightarrow{AD} = \overrightarrow{BC}\) or \(\overrightarrow{CD} = \overrightarrow{BA}\) | A1, A1 3 | |
| \((\mathbf{a} + \mathbf{c} - \mathbf{b}) = 2\mathbf{j} + \mathbf{k}\) | ||
| (ii) \(\overrightarrow{AB} \cdot \overrightarrow{CB} = | \overrightarrow{AB} | |
| Scalar product of any 2 vectors | M1, A1 4 | |
| \(94°(94.386...)\) or \(1.65(1.647...)\) | 7 |
(i) $\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD}$ or $\overrightarrow{OB} + \overrightarrow{BC} + \overrightarrow{CD}$ | **AEF** | M1 | Connect $\overrightarrow{OD}$ & 2/3/4 vectors in their diag
$\overrightarrow{AD} = \overrightarrow{BC}$ or $\overrightarrow{CD} = \overrightarrow{BA}$ | | A1, A1 3 | Or similar from their diag [i.e if diag mislabelled, M1A1A0 possible]
$(\mathbf{a} + \mathbf{c} - \mathbf{b}) = 2\mathbf{j} + \mathbf{k}$ | |
(ii) $\overrightarrow{AB} \cdot \overrightarrow{CB} = |\overrightarrow{AB}||\overrightarrow{CB}| \cos \theta$ | M1, M1, M1 | Or $\overrightarrow{AB}.\overrightarrow{BC}$ i.e. scalar prod for correct pair
Scalar product of any 2 vectors | | M1, A1 4 | Or $\overrightarrow{AB}.\overrightarrow{BC}$ i.e. scalar product for correct pair; $2 + 3 - 6 = -1$ is expected; $\sqrt{19}$ or $3$ expected; Accept $86°(85.614...) $ or $1.49(1.647...)$
$94°(94.386...)$ or $1.65(1.647...)$ | | **7**$ABCD$ is a parallelogram. The position vectors of $A$, $B$ and $C$ are given respectively by
$$\mathbf{a} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}, \quad \mathbf{b} = 3\mathbf{i} - 2\mathbf{j}, \quad \mathbf{c} = \mathbf{i} - \mathbf{j} - 2\mathbf{k}.$$
\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $D$. [3]
\item Determine, to the nearest degree, the angle $ABC$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2005 Q5 [7]}}