| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | June |
| Marks | 7 |
| 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 vector question requiring coordinate setup from a diagram, basic vector arithmetic (finding position vectors), and applying the scalar product formula to find an angle. While it involves 3D geometry, the steps are routine: establish coordinates, subtract to find vectors, use dot product formula. Slightly easier than average due to the simple geometry (cylinder with perpendicular axes) and standard technique application. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(MC = 4i + 4j - 6k\) | B1, B2\* | Correct only. One off for each error in \(i\), \(j\) and \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(a \cdot b = | a | |
| Angle \(= 109.7°\) (allow \(109.6\)) | Correct only. No penalty for use of column vectors |
**(i)** $MO = 4i - 6k$
$MC = 4i + 4j - 6k$ | B1, B2\* | Correct only. One off for each error in $i$, $j$ and $k$
**(ii)** $MO.MC = 16 + 0 - 36 = -20$
$= \sqrt{(4^2 + 6^2)(4^2 + 4^2 + 6^2)} \cos\theta$
Use of $a \cdot b = |a||b|\cos\theta$ Use of Modulus | M1, M1, M1, A1 | Use of $a \cdot b = a_1b_1 + a_2b_2 + a_3b_3$; Use of $|a||b|\cos\theta$ |
Angle $= 109.7°$ (allow $109.6$) | | Correct only. No penalty for use of column vectors5\\
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The diagram shows a solid cylinder standing on a horizontal circular base, centre $O$ and radius 4 units. The line $B A$ is a diameter and the radius $O C$ is at $90 ^ { \circ }$ to $O A$. Points $O ^ { \prime } , A ^ { \prime } , B ^ { \prime }$ and $C ^ { \prime }$ lie on the upper surface of the cylinder such that $O O ^ { \prime } , A A ^ { \prime } , B B ^ { \prime }$ and $C C ^ { \prime }$ are all vertical and of length 12 units. The mid-point of $B B ^ { \prime }$ is $M$.
Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O C$ and $O O ^ { \prime }$ respectively.\\
(i) Express each of the vectors $\overrightarrow { M O }$ and $\overrightarrow { M C ^ { \prime } }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(ii) Hence find the angle $O M C ^ { \prime }$.
\hfill \mbox{\textit{CAIE P1 2002 Q5 [7]}}