| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Vector geometry in 3D shapes |
| Difficulty | Moderate -0.3 This is a straightforward 3D vectors question requiring basic position vector manipulation, unit vector calculation, and scalar product application to find an angle. While it involves multiple parts and 3D visualization, each step uses standard techniques with no novel problem-solving required, 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 multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{DF} = -6\mathbf{i} + 2\mathbf{k}\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{EF} = -6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\) | B1 | |
| \( | \overrightarrow{EF} | = \sqrt{(-6)^2 + (-3)^2 + 2^2}\) |
| Unit vector \(= \frac{1}{7}(-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{DF}\cdot\overrightarrow{EF} = (-6\mathbf{i}+2\mathbf{k})\cdot(-6\mathbf{i}-3\mathbf{j}+2\mathbf{k}) = 36 + 4 = 40\) | M1 | |
| \( | \overrightarrow{DF} | = \sqrt{40},\ |
| \(\cos EFD = \frac{40}{7\sqrt{40}}\) | M1 | |
| \(EFD = 25.4°\) | A1 | Special case: use of cosine rule M1 (must evaluate lengths using correct method) A1 only |
| Total: 4 |
## Question 8:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{DF} = -6\mathbf{i} + 2\mathbf{k}$ | B1 | |
| | **Total: 1** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{EF} = -6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$ | B1 | |
| $|\overrightarrow{EF}| = \sqrt{(-6)^2 + (-3)^2 + 2^2}$ | M1 | Must use *their* $\overrightarrow{EF}$ |
| Unit vector $= \frac{1}{7}(-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})$ | A1 | |
| | **Total: 3** | |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{DF}\cdot\overrightarrow{EF} = (-6\mathbf{i}+2\mathbf{k})\cdot(-6\mathbf{i}-3\mathbf{j}+2\mathbf{k}) = 36 + 4 = 40$ | M1 | |
| $|\overrightarrow{DF}| = \sqrt{40},\ |\overrightarrow{EF}| = 7$ | M1 | |
| $\cos EFD = \frac{40}{7\sqrt{40}}$ | M1 | |
| $EFD = 25.4°$ | A1 | Special case: use of cosine rule M1 (must evaluate lengths using correct method) A1 only |
| | **Total: 4** | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{d178603a-f59a-4986-b5ab-b47eceedb2fc-12_595_748_260_699}
The diagram shows a solid figure $O A B C D E F$ having a horizontal rectangular base $O A B C$ with $O A = 6$ units and $A B = 3$ units. The vertical edges $O F , A D$ and $B E$ have lengths 6 units, 4 units and 4 units respectively. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O C$ and $O F$ respectively.\\
(i) Find $\overrightarrow { D F }$.\\
(ii) Find the unit vector in the direction of $\overrightarrow { E F }$.\\
(iii) Use a scalar product to find angle $E F D$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q8 [8]}}