| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| 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 question requiring coordinate setup from a diagram, finding position vectors by subtraction, then using the scalar product formula for angles. While it involves multiple steps and 3D visualization, each step follows standard procedures with no novel problem-solving required. The semicircular prism context adds mild complexity but the actual vector work is routine for A-level. |
| Spec | 1.10b 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 vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{PA} = -6\mathbf{i} - 8\mathbf{j} - 6\mathbf{k}\) | B1 | Co – column vectors ok |
| \(\overrightarrow{PN} = 6\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}\) | B2, 1 | One off for each error (all incorrect sign – just one error) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{PA} \cdot \overrightarrow{PN} = -36 - 16 + 36 = -16\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \(\cos APN = \frac{-16}{\sqrt{136}\sqrt{76}}\) | M1, M1 | Modulus worked correctly for either one. Division of "–16" by "product of moduli" |
| \(\rightarrow APN = 99°\) | A1 | Allow more accuracy |
**(i)**
$\overrightarrow{PA} = -6\mathbf{i} - 8\mathbf{j} - 6\mathbf{k}$ | B1 | Co – column vectors ok
$\overrightarrow{PN} = 6\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}$ | B2, 1 | One off for each error (all incorrect sign – just one error)
**(ii)**
$\overrightarrow{PA} \cdot \overrightarrow{PN} = -36 - 16 + 36 = -16$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$
$\cos APN = \frac{-16}{\sqrt{136}\sqrt{76}}$ | M1, M1 | Modulus worked correctly for either one. Division of "–16" by "product of moduli"
$\rightarrow APN = 99°$ | A1 | Allow more accuracy
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{08729aab-586b-4210-94c9-77b1f6b1d873-2_558_1488_863_331}
The diagram shows a semicircular prism with a horizontal rectangular base $A B C D$. The vertical ends $A E D$ and $B F C$ are semicircles of radius 6 cm . The length of the prism is 20 cm . The mid-point of $A D$ is the origin $O$, the mid-point of $B C$ is $M$ and the mid-point of $D C$ is $N$. The points $E$ and $F$ are the highest points of the semicircular ends of the prism. The point $P$ lies on $E F$ such that $E P = 8 \mathrm {~cm}$.
Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O D , O M$ and $O E$ respectively.\\
(i) Express each of the vectors $\overrightarrow { P A }$ and $\overrightarrow { P N }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(ii) Use a scalar product to calculate angle $A P N$.
\hfill \mbox{\textit{CAIE P1 2008 Q4 [7]}}