| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| 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 vectors question requiring basic coordinate setup and standard scalar product application. Part (i) involves simple ratio work on a diagonal (OE is 2 units along OB where OB has length 10), and part (ii) is a routine scalar product calculation to find an angle. The rectangular base and clear vertical height make visualization easy, and both parts follow standard textbook procedures with no novel insight required. |
| 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 vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{OE} = \frac{2}{10}(8\mathbf{i} + 6\mathbf{j}) = 1.6\mathbf{i} + 1.2\mathbf{j}\) AG | M1A1 | Evidence of \(OB = 10\) or other valid method (e.g. trigonometry) is required |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{OD} = 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k}\) | B1 | Allow reversal of one or both of OD, BD |
| \(\mathbf{BD} = -8\mathbf{i} - 6\mathbf{j} + 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k}\), \(\mathbf{OE} = -6.4\mathbf{i} - 4.8\mathbf{j} + 7\mathbf{k}\) | M1A1 | For M mark allow sign errors. Also if 2 out of 3 components correct |
| Correct method for \(\pm\mathbf{OD}.\pm\mathbf{BD}\) (using their answers) | M1 | Expect \(1.6 \times -6.4 + 1.2 \times -4.8 + 49 = 33\) or \(\frac{825}{25}\) |
| Correct method for \( | \mathbf{OD} | \) or \( |
| \(\cos BDO =\) their \(\frac{\mathbf{OD.BD}}{ | \mathbf{OD} | \times |
| \(64.8°\) Allow \(1.13\) (rad) | A1 | Can't score A1 if 1 vector only is reversed unless explained well |
| 7 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{OE} = \frac{2}{10}(8\mathbf{i} + 6\mathbf{j}) = 1.6\mathbf{i} + 1.2\mathbf{j}$ AG | M1A1 | Evidence of $OB = 10$ or other valid method (e.g. trigonometry) is required |
| | **2** | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{OD} = 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k}$ | B1 | Allow reversal of one or both of **OD**, **BD** |
| $\mathbf{BD} = -8\mathbf{i} - 6\mathbf{j} + 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k}$, $\mathbf{OE} = -6.4\mathbf{i} - 4.8\mathbf{j} + 7\mathbf{k}$ | M1A1 | For M mark allow sign errors. Also if 2 out of 3 components correct |
| Correct method for $\pm\mathbf{OD}.\pm\mathbf{BD}$ (using their answers) | M1 | Expect $1.6 \times -6.4 + 1.2 \times -4.8 + 49 = 33$ or $\frac{825}{25}$ |
| Correct method for $|\mathbf{OD}|$ or $|\mathbf{BD}|$ (using their answers) | M1 | Expect $\sqrt{1.6^2 + 1.2^2 + 7^2}$ or $\sqrt{6.4^2 + 4.8^2 + 7^2} = \sqrt{53}$ or $\sqrt{113}$ |
| $\cos BDO =$ their $\frac{\mathbf{OD.BD}}{|\mathbf{OD}| \times |\mathbf{BD}|}$ | DM1 | Expect $\frac{33}{77.4}$. Dep. on all previous M marks and either B1 or A1 |
| $64.8°$ Allow $1.13$ (rad) | A1 | Can't score A1 if 1 vector only is reversed unless explained well |
| | **7** | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{ea402a1d-3632-4637-9198-2365715b5246-14_670_857_260_644}
The diagram shows a pyramid $O A B C D$ with a horizontal rectangular base $O A B C$. The sides $O A$ and $A B$ have lengths of 8 units and 6 units respectively. The point $E$ on $O B$ is such that $O E = 2$ units. The point $D$ of the pyramid is 7 units vertically above $E$. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A$, $O C$ and $E D$ respectively.\\
(i) Show that $\overrightarrow { O E } = 1.6 \mathbf { i } + 1.2 \mathbf { j }$.\\
(ii) Use a scalar product to find angle $B D O$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q9 [9]}}