| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| 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 coordinate setup, dot product for perpendicularity, and parallel vector conditions. While it has multiple parts and requires careful bookkeeping of cuboid geometry, all techniques are standard P1 material with no novel insights needed. Slightly easier than average due to the structured setup and routine application of vector methods. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{XP} = -4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}\) | B1 | Or PX; Attempt scalar product with OP/PO and set \(= 0\) |
| \([-4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}]\cdot(p\mathbf{j} + 2\mathbf{k}) = 0\) | M1 | (\(= 0\) could be implied) |
| \(p^2 - 5p + 4 = 0\) | A1 | |
| \(p = 1\) or \(4\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{XP} = -4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \rightarrow | \mathbf{XP} | = \sqrt{16+16+4}\) |
| Unit vector \(= 1/6(-4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})\) oe | A1 [2] |
| Answer | Marks |
|---|---|
| \(\mathbf{AG} = -4\mathbf{i} + 15\mathbf{j} + 2\mathbf{k}\) | B1 |
| \(\mathbf{XQ} = \lambda\mathbf{AG}\) soi | M1 |
| \(\lambda = 2/3 \rightarrow \mathbf{XQ} = -\dfrac{8}{3}\mathbf{i} + 10\mathbf{j} + \dfrac{4}{3}\mathbf{k}\) | A1 [3] |
## Question 9:
### Part (i):
$\mathbf{XP} = -4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}$ | **B1** | Or **PX**; Attempt scalar product with **OP/PO** and set $= 0$
$[-4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}]\cdot(p\mathbf{j} + 2\mathbf{k}) = 0$ | **M1** | ($= 0$ could be implied)
$p^2 - 5p + 4 = 0$ | **A1** |
$p = 1$ or $4$ | **A1** [4] |
### Part (ii):
$\mathbf{XP} = -4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \rightarrow |\mathbf{XP}| = \sqrt{16+16+4}$ | **M1** | Expect 6
Unit vector $= 1/6(-4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})$ oe | **A1** [2] |
### Part (iii):
$\mathbf{AG} = -4\mathbf{i} + 15\mathbf{j} + 2\mathbf{k}$ | **B1** |
$\mathbf{XQ} = \lambda\mathbf{AG}$ soi | **M1** |
$\lambda = 2/3 \rightarrow \mathbf{XQ} = -\dfrac{8}{3}\mathbf{i} + 10\mathbf{j} + \dfrac{4}{3}\mathbf{k}$ | **A1** [3] |
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{9f17f7b8-b54d-467d-be26-21c599ce6ca2-4_724_1488_257_330}
The diagram shows a cuboid $O A B C D E F G$ with a horizontal base $O A B C$ in which $O A = 4 \mathrm {~cm}$ and $A B = 15 \mathrm {~cm}$. The height $O D$ of the cuboid is 2 cm . The point $X$ on $A B$ is such that $A X = 5 \mathrm {~cm}$ and the point $P$ on $D G$ is such that $D P = p \mathrm {~cm}$, where $p$ is a constant. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O C$ and $O D$ respectively.\\
(i) Find the possible values of $p$ such that angle $O P X = 90 ^ { \circ }$.\\
(ii) For the case where $p = 9$, find the unit vector in the direction of $\overrightarrow { X P }$.\\
(iii) A point $Q$ lies on the face $C B F G$ and is such that $X Q$ is parallel to $A G$. Find $\overrightarrow { X Q }$.
\hfill \mbox{\textit{CAIE P1 2016 Q9 [9]}}