| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | March |
| Marks | 8 |
| 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 question requiring standard techniques: finding position vectors using given ratios, expressing vectors in component form, and using scalar product to find an angle. While it involves multiple steps, each step follows routine procedures with no novel insight required. The geometric setup is clearly described, making this slightly easier than average for A-level. |
| 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 vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(CP = \dfrac{3}{5}CA\) soi | M1 | |
| \(CP = \dfrac{3}{5}(4\mathbf{i} - 3\mathbf{k}) = 2.4\mathbf{i} - 1.8\mathbf{k}\) AG | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OP = 2.4\mathbf{i} + 1.2\mathbf{k}\) | B1 | |
| \(BP = 2.4\mathbf{i} - 2.4\mathbf{j} + 1.2\mathbf{k}\) | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BP \cdot CP = 5.76 - 2.16 = 3.6\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \(\ | BP\ | \ |
| \(\cos BPC = \dfrac{3.6}{\sqrt{12.96}\sqrt{9}} \left(= \dfrac{1}{3}\right)\) | M1 | All linked correctly |
| Angle \(BPC = 70.5°\) (or 1.23 rads) cao | A1 [4] |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $CP = \dfrac{3}{5}CA$ soi | M1 | |
| $CP = \dfrac{3}{5}(4\mathbf{i} - 3\mathbf{k}) = 2.4\mathbf{i} - 1.8\mathbf{k}$ **AG** | A1 [2] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OP = 2.4\mathbf{i} + 1.2\mathbf{k}$ | B1 | |
| $BP = 2.4\mathbf{i} - 2.4\mathbf{j} + 1.2\mathbf{k}$ | B1 [2] | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BP \cdot CP = 5.76 - 2.16 = 3.6$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$ |
| $\|BP\|\|CP\| = \sqrt{2.4^2 + 2.4^2 + 1.2^2}\sqrt{2.4^2 + 1.8^2}$ | M1 | Product of moduli |
| $\cos BPC = \dfrac{3.6}{\sqrt{12.96}\sqrt{9}} \left(= \dfrac{1}{3}\right)$ | M1 | All linked correctly |
| Angle $BPC = 70.5°$ (or 1.23 rads) cao | A1 [4] | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{0f58de6c-aba7-4a79-a962-c23be3ee0aa9-3_529_698_260_721}
The diagram shows a pyramid $O A B C$ with a horizontal triangular base $O A B$ and vertical height $O C$. Angles $A O B , B O C$ and $A O C$ are each right angles. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O B$ and $O C$ respectively, with $O A = 4$ units, $O B = 2.4$ units and $O C = 3$ units. The point $P$ on $C A$ is such that $C P = 3$ units.\\
(i) Show that $\overrightarrow { C P } = 2.4 \mathbf { i } - 1.8 \mathbf { k }$.\\
(ii) Express $\overrightarrow { O P }$ and $\overrightarrow { B P }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(iii) Use a scalar product to find angle $B P C$.
\hfill \mbox{\textit{CAIE P1 2016 Q7 [8]}}