| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Vector geometry in 3D shapes |
| Difficulty | Standard +0.3 This is a straightforward 3D vector problem requiring visualization of a prism-like shape, finding position vectors using given constraints (parallel faces, midpoint), then applying the standard scalar product formula for angles. While it involves multiple steps and 3D geometry, the techniques are routine for A-level: no novel insight needed, just careful application of vector addition and the cosine formula. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AM} = 1.5\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\) | B3,2,1 | Loses 1 mark for each error |
| \(\overrightarrow{GM} = 6.5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AM} \cdot \overrightarrow{GM} = 9.75 - 16 - 25 = -31.25\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) on AM and GM |
| \( | \overrightarrow{AM} | \ \cdot\ |
| Equating \(\rightarrow\) Angle \(GMA = 121°\) | A1 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AM} = 1.5\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}$ | B3,2,1 | Loses 1 mark for each error |
| $\overrightarrow{GM} = 6.5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}$ | | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AM} \cdot \overrightarrow{GM} = 9.75 - 16 - 25 = -31.25$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$ on AM and GM |
| $|\overrightarrow{AM}|\ \cdot\ |\overrightarrow{GM}| = \sqrt{(1.5^2+4^2+5^2)} \times \sqrt{(6.5^2+4^2+5^2)}\cos GMA$ | M1 M1 | M1 for product of 2 modulii, M1 all correctly connected |
| Equating $\rightarrow$ Angle $GMA = 121°$ | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{ebf16cae-1e80-44d2-9c51-630f5dc3c11f-12_775_823_260_662}
The diagram shows a three-dimensional shape in which the base $O A B C$ and the upper surface $D E F G$ are identical horizontal squares. The parallelograms $O A E D$ and $C B F G$ both lie in vertical planes. The point $M$ is the mid-point of $A F$.
Unit vectors $\mathbf { i }$ and $\mathbf { j }$ are parallel to $O A$ and $O C$ respectively and the unit vector $\mathbf { k }$ is vertically upwards. The position vectors of $A$ and $D$ are given by $\overrightarrow { O A } = 8 \mathbf { i }$ and $\overrightarrow { O D } = 3 \mathbf { i } + 10 \mathbf { k }$.\\
(i) Express each of the vectors $\overrightarrow { A M }$ and $\overrightarrow { G M }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(ii) Use a scalar product to find angle $G M A$ correct to the nearest degree.\\
\hfill \mbox{\textit{CAIE P1 2019 Q7 [7]}}