| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Angle between two vectors |
| Difficulty | Moderate -0.3 This is a straightforward vectors question requiring standard techniques: finding vectors BA and BC, using the scalar product formula to find the angle, and applying the parallelogram rule. All steps are routine for A-level, though the 3D context and multi-step nature elevate it slightly above pure recall, placing it just below average difficulty. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(\vec{BA} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}\), \(\vec{BC} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}\) | B1 M1 | Correct two vectors for angle \(ABC\). Correct method for one of the sides. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\to \theta = 112.4°\) or \(1.96\) radians | M1 M1 A1 [6] | Correct use for any pair of vectors. Correct method for moduli. All linked correctly. co (67.6° usually gets 4/6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \begin{pmatrix} 8 \\ 1 \\ 8 \end{pmatrix}\) | M1 A1∨ [2] | Correct method. (allow for \(\mathbf{d} = \mathbf{a} + \mathbf{b} - \mathbf{c}\) or for \(\mathbf{d} = \mathbf{a} + \mathbf{c} - \mathbf{b}\) or for \(\mathbf{d} = \mathbf{b} + \mathbf{c} - \mathbf{a}\)). A1∨ for his \(\vec{BC}\). |
**(i)** $\vec{BA} \cdot \vec{BC}$ or $\vec{AB} \cdot \vec{CB}$
$\vec{BA} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$, $\vec{BC} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}$ | B1 M1 | Correct two vectors for angle $ABC$. Correct method for one of the sides.
$\vec{BA} \cdot \vec{BC} = -8 = 3 \times 7 \times \cos\theta$
$\to \theta = 112.4°$ or $1.96$ radians | M1 M1 A1 [6] | Correct use for **any** pair of vectors. Correct method for moduli. All linked correctly. co (67.6° usually gets 4/6)
**(ii)** $\vec{OD} = \vec{OA} + \vec{AD} = \vec{OA} + \vec{BC}$
$= \begin{pmatrix} 8 \\ 1 \\ 8 \end{pmatrix}$ | M1 A1∨ [2] | Correct method. (allow for $\mathbf{d} = \mathbf{a} + \mathbf{b} - \mathbf{c}$ or for $\mathbf{d} = \mathbf{a} + \mathbf{c} - \mathbf{b}$ or for $\mathbf{d} = \mathbf{b} + \mathbf{c} - \mathbf{a}$). A1∨ for his $\vec{BC}$.Relative to the origin $O$, the position vectors of the points $A$, $B$ and $C$ are given by
$$\overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 10 \\ 0 \\ 6 \end{pmatrix}.$$
\begin{enumerate}[label=(\roman*)]
\item Find angle $ABC$. [6]
\end{enumerate}
The point $D$ is such that $ABCD$ is a parallelogram.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the position vector of $D$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q8 [8]}}