| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Angle between two vectors |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: dot product for angle (part i), scalar multiplication to find vectors with given magnitudes (part ii), and finding a unit vector (part iii). All parts involve routine calculations with no problem-solving insight required, making it slightly easier than average, though the 3D context and multiple steps keep it close to typical A-level standard. |
| 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 |
|---|---|---|
| (i) \(3\mathbf{i} - 4k, 2\mathbf{i} + 3\mathbf{i} - 6k\) | M1 | Uses \(x_1x_2 + y_1y_2 + z_1z_2\) |
| Dot product = \(6 + 24 = 30\) | M1 | Method for modulus |
| \(= \sqrt{25} × \sqrt{49} \cos \theta\) | M1 A1 | Links everything correctly. co [4] |
| \(\to\) angle = \(31°\) or 0.54(1) radians | ||
| (ii) \(\mathbf{OA} = (3\mathbf{i} - 4k) × (15 ÷ 5)\) | M1 A1 | M mark for ×(15 ÷ 5) or ×(14 + 7). A1 for OA. A1 for OB [3] |
| \(\to 9\mathbf{i} - 12k\) | ||
| \(\mathbf{OB} = (2\mathbf{i} + 3\mathbf{i} - 6k) × (14 ÷ 7)\) | A1 | |
| \(\to 4\mathbf{i} + 6\mathbf{i} - 12k\) | ||
| (iii) \(\mathbf{AB} = \mathbf{b} - \mathbf{a} = -5\mathbf{i} + 6\mathbf{i}\) | M1 | Correct use for either AB or BA. Complete method for unit vector. co [3] |
| \(\to\) Magnitude of \(\sqrt{61}\) or 7.81 | M1 | |
| \(\to\) Unit vector of \((-5\mathbf{i} + 6\mathbf{i}) ÷ \sqrt{61}\) | A1 |
(i) $3\mathbf{i} - 4k, 2\mathbf{i} + 3\mathbf{i} - 6k$ | M1 | Uses $x_1x_2 + y_1y_2 + z_1z_2$
Dot product = $6 + 24 = 30$ | M1 | Method for modulus
$= \sqrt{25} × \sqrt{49} \cos \theta$ | M1 A1 | Links everything correctly. co [4]
$\to$ angle = $31°$ or 0.54(1) radians | |
(ii) $\mathbf{OA} = (3\mathbf{i} - 4k) × (15 ÷ 5)$ | M1 A1 | M mark for ×(15 ÷ 5) or ×(14 + 7). A1 for OA. A1 for OB [3]
$\to 9\mathbf{i} - 12k$ | |
$\mathbf{OB} = (2\mathbf{i} + 3\mathbf{i} - 6k) × (14 ÷ 7)$ | A1 |
$\to 4\mathbf{i} + 6\mathbf{i} - 12k$ | |
(iii) $\mathbf{AB} = \mathbf{b} - \mathbf{a} = -5\mathbf{i} + 6\mathbf{i}$ | M1 | Correct use for either AB or BA. Complete method for unit vector. co [3]
$\to$ Magnitude of $\sqrt{61}$ or 7.81 | M1 |
$\to$ Unit vector of $(-5\mathbf{i} + 6\mathbf{i}) ÷ \sqrt{61}$ | A1 |\begin{enumerate}[label=(\roman*)]
\item Find the angle between the vectors $\mathbf{3i} - \mathbf{4k}$ and $\mathbf{2i} + \mathbf{3j} - \mathbf{6k}$. [4]
\end{enumerate}
The vector $\overrightarrow{OA}$ has a magnitude of $15$ units and is in the same direction as the vector $\mathbf{3i} - \mathbf{4k}$. The vector $\overrightarrow{OB}$ has a magnitude of $14$ units and is in the same direction as the vector $\mathbf{2i} + \mathbf{3j} - \mathbf{6k}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Express $\overrightarrow{OA}$ and $\overrightarrow{OB}$ in terms of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$. [3]
\item Find the unit vector in the direction of $\overrightarrow{AB}$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2012 Q8 [10]}}