| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Vector geometry in 3D shapes |
| Difficulty | Standard +0.8 This is a substantial 3D vector geometry problem requiring spatial visualization, coordinate setup, similar triangles reasoning to find the height, vector arithmetic across multiple points, and scalar product application for angles. While the individual techniques are standard A-level content, the multi-part nature, 3D spatial reasoning with a pyramid, and integration of geometric insight with vector methods makes this moderately challenging—above average but not requiring novel mathematical insight. |
| 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 multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{10-a}{10} = \dfrac{6}{10}\) oe | M1 | or \(PDE\) is isos hence \(PD = 6\) (M1) |
| \(a = 4\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{BG} = -10\mathbf{j} - 10\mathbf{i} + 4\mathbf{k} + 6\mathbf{j} = -10\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\) | B2,1 | Any acceptable notation; loses 1 for each error |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{BG}.\overrightarrow{BA} = 40\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \(\cos GBA = \dfrac{40}{\sqrt{132}\sqrt{100}}\) | M1 DM1 | Modulus worked correctly for either; must be using \(\pm\overrightarrow{BG} \cdot \pm\overrightarrow{AB}\) |
| \(GBA = 69.6°\) | A1 | Must be the acute angle |
## Question 9:
**Part (i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{10-a}{10} = \dfrac{6}{10}$ oe | M1 | **or** $PDE$ is isos hence $PD = 6$ (M1) |
| $a = 4$ | A1 | **AG** |
**Part (ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{BG} = -10\mathbf{j} - 10\mathbf{i} + 4\mathbf{k} + 6\mathbf{j} = -10\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}$ | B2,1 | Any acceptable notation; loses 1 for each error |
**Part (iii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{BG}.\overrightarrow{BA} = 40$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$ |
| $\cos GBA = \dfrac{40}{\sqrt{132}\sqrt{100}}$ | M1 DM1 | Modulus worked correctly for either; must be using $\pm\overrightarrow{BG} \cdot \pm\overrightarrow{AB}$ |
| $GBA = 69.6°$ | A1 | Must be the acute angle |
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{ae57d8f1-5a0d-426c-952d-e8b99c6aeaba-4_582_1072_255_541}
The diagram shows a pyramid $O A B C P$ in which the horizontal base $O A B C$ is a square of side 10 cm and the vertex $P$ is 10 cm vertically above $O$. The points $D , E , F , G$ lie on $O P , A P , B P , C P$ respectively and $D E F G$ is a horizontal square of side 6 cm . The height of $D E F G$ above the base is $a \mathrm {~cm}$. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O C$ and $O D$ respectively.\\
(i) Show that $a = 4$.\\
(ii) Express the vector $\overrightarrow { B G }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(iii) Use a scalar product to find angle $G B A$.
\hfill \mbox{\textit{CAIE P1 2010 Q9 [8]}}