| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Linear combinations of vectors |
| Difficulty | Moderate -0.8 This is a straightforward vector question requiring basic operations: finding a position vector using vector addition, calculating a unit vector, and solving a simple system of linear equations from vector components. All techniques are routine for A-level Pure Mathematics 1, with no problem-solving insight needed beyond direct application of standard methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}-1\\1\\-2\end{pmatrix}\) and \(\overrightarrow{AC} = \begin{pmatrix}-2\\2\\-4\end{pmatrix}\) | M1 | For \(\pm(\mathbf{b} - \mathbf{a})\) (not \(\mathbf{b} + \mathbf{a}\)) |
| \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix}2\\3\\-6\end{pmatrix}\) | A1 | Co |
| Unit vector \(= \frac{1}{7}\begin{pmatrix}2\\3\\-6\end{pmatrix}\) | M1, A1\(\sqrt{}\) [4] | Division by the modulus; \(\sqrt{}\) for his \(\overrightarrow{OC}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m\begin{pmatrix}4\\1\\-2\end{pmatrix} + n\begin{pmatrix}3\\2\\-4\end{pmatrix} = \begin{pmatrix}1\\4\\k\end{pmatrix}\) | M1 | Forming 2 simultaneous equations |
| \(\rightarrow 4m + 3n = 1\) and \(m + 2n = 4\) | A1 | co |
| \(\rightarrow m = -2\) and \(n = 3\) | M1A1 | |
| \(\rightarrow k = -8\) | [4] | Equation for \(k\) in terms of \(m\) and \(n\); co |
## Question 9(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}-1\\1\\-2\end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix}-2\\2\\-4\end{pmatrix}$ | M1 | For $\pm(\mathbf{b} - \mathbf{a})$ (not $\mathbf{b} + \mathbf{a}$) |
| $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix}2\\3\\-6\end{pmatrix}$ | A1 | Co |
| Unit vector $= \frac{1}{7}\begin{pmatrix}2\\3\\-6\end{pmatrix}$ | M1, A1$\sqrt{}$ [4] | Division by the modulus; $\sqrt{}$ for his $\overrightarrow{OC}$ |
## Question 9(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m\begin{pmatrix}4\\1\\-2\end{pmatrix} + n\begin{pmatrix}3\\2\\-4\end{pmatrix} = \begin{pmatrix}1\\4\\k\end{pmatrix}$ | M1 | Forming 2 simultaneous equations |
| $\rightarrow 4m + 3n = 1$ and $m + 2n = 4$ | A1 | co |
| $\rightarrow m = -2$ and $n = 3$ | M1A1 | |
| $\rightarrow k = -8$ | [4] | Equation for $k$ in terms of $m$ and $n$; co |
---9 Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
4 \\
1 \\
- 2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { r }
3 \\
2 \\
- 4
\end{array} \right) .$$
(i) Given that $C$ is the point such that $\overrightarrow { A C } = 2 \overrightarrow { A B }$, find the unit vector in the direction of $\overrightarrow { O C }$.
The position vector of the point $D$ is given by $\overrightarrow { O D } = \left( \begin{array} { l } 1 \\ 4 \\ k \end{array} \right)$, where $k$ is a constant, and it is given that $\overrightarrow { O D } = m \overrightarrow { O A } + n \overrightarrow { O B }$, where $m$ and $n$ are constants.\\
(ii) Find the values of $m , n$ and $k$.
\hfill \mbox{\textit{CAIE P1 2007 Q9 [8]}}