| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Vector operations and magnitudes |
| Difficulty | Standard +0.3 This is a straightforward vector question involving basic operations (finding vectors between points, magnitudes, and possibly ratios). The question appears incomplete but based on the setup with two position vectors, it likely asks for standard calculations like finding AB, its magnitude, or a point dividing AB in a given ratio. These are routine A-level vector operations requiring direct application of formulas with minimal problem-solving, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{AX} = \begin{pmatrix}6\\2\\3\end{pmatrix}\), and one of \(\mathbf{AB} = \begin{pmatrix}18\\6\\9\end{pmatrix}\), \(\mathbf{XB} = \begin{pmatrix}12\\4\\6\end{pmatrix}\), \(\mathbf{BX} = \begin{pmatrix}-12\\-4\\-6\end{pmatrix}\) | B1B1 | |
| State \(\mathbf{AB} = 3\mathbf{AX}\) (or \(\mathbf{XB} = 2\mathbf{AX}\) or \(\mathbf{AB} = \frac{3}{2}\mathbf{XB}\) etc) hence straight line OR \(\frac{\mathbf{AX \cdot AB}}{ | \mathbf{AX} |
| Answer | Marks |
|---|---|
| \(\mathbf{CX} = \begin{pmatrix}-3\\6\\2\end{pmatrix}\) | B1 |
| \(\mathbf{CX \cdot AX} = -18 + 12 + 6\) | M1 |
| \(= 0\) (hence \(CX\) is perpendicular to \(AX\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \( | \mathbf{CX} | = \sqrt{3^2 + 6^2 + 2^2}\), \( |
| Area \(\triangle ABC = \frac{1}{2} \times 21 \times 7 = 73\frac{1}{2}\) | M1A1 | Accept answers which round to 73.5 |
## Question 10(i):
$\mathbf{AX} = \begin{pmatrix}6\\2\\3\end{pmatrix}$, and one of $\mathbf{AB} = \begin{pmatrix}18\\6\\9\end{pmatrix}$, $\mathbf{XB} = \begin{pmatrix}12\\4\\6\end{pmatrix}$, $\mathbf{BX} = \begin{pmatrix}-12\\-4\\-6\end{pmatrix}$ | **B1B1** |
State $\mathbf{AB} = 3\mathbf{AX}$ (or $\mathbf{XB} = 2\mathbf{AX}$ or $\mathbf{AB} = \frac{3}{2}\mathbf{XB}$ etc) hence straight line **OR** $\frac{\mathbf{AX \cdot AB}}{|\mathbf{AX}||\mathbf{AB}|} = 1\ (\rightarrow \theta = 0)$ or $\frac{\mathbf{AX \cdot BX}}{|\mathbf{AX}||\mathbf{BX}|} = -1\ (\rightarrow \theta = 180)$ hence straight line | **B1** | WWW. A conclusion (i.e. a straight line) is required.
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## Question 10(ii):
$\mathbf{CX} = \begin{pmatrix}-3\\6\\2\end{pmatrix}$ | **B1** |
$\mathbf{CX \cdot AX} = -18 + 12 + 6$ | **M1** |
$= 0$ (hence $CX$ is perpendicular to $AX$) | **A1** |
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## Question 10(iii):
$|\mathbf{CX}| = \sqrt{3^2 + 6^2 + 2^2}$, $|\mathbf{AB}| = \sqrt{18^2 + 6^2 + 9^2}$, both attempted | **M1** |
Area $\triangle ABC = \frac{1}{2} \times 21 \times 7 = 73\frac{1}{2}$ | **M1A1** | Accept answers which round to 73.5
---10 Relative to an origin $O$, the position vectors of the points $A , B$ and $X$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
- 8 \\
- 4 \\
2
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
10 \\
2 \\
11
\end{array} \right) \quad \text { and } \quad \overrightarrow { O X } = \left( \begin{array} { r }
- 2 \\
- 2 \\
5
\end{array} \right)$$
(i) Find $\overrightarrow { A X }$ and show that $A X B$ is a straight line.\\
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The position vector of a point $C$ is given by $\overrightarrow { O C } = \left( \begin{array} { r } 1 \\ - 8 \\ 3 \end{array} \right)$.\\
(ii) Show that $C X$ is perpendicular to $A X$.\\
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(iii) Find the area of triangle $A B C$.\\
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\includegraphics[max width=\textwidth, alt={}, center]{17e813c6-890f-4198-b20a-557b133e8c34-18_949_1087_260_529}
The diagram shows part of the curve $y = ( x - 1 ) ^ { - 2 } + 2$, and the lines $x = 1$ and $x = 3$. The point $A$ on the curve has coordinates $( 2,3 )$. The normal to the curve at $A$ crosses the line $x = 1$ at $B$.\\
(i) Show that the normal $A B$ has equation $y = \frac { 1 } { 2 } x + 2$.\\
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(ii) Find, showing all necessary working, the volume of revolution obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
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\hfill \mbox{\textit{CAIE P1 2019 Q10 [9]}}