| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2005 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Equal length conditions |
| Difficulty | Moderate -0.8 This is a straightforward vectors question testing basic scalar product application and distance formula. Part (i) is pure substitution into the scalar product formula with given values, requiring only routine calculation. Part (ii) involves setting up |PQ|² = 36 and solving a simple quadratic equation. Both parts are standard textbook exercises with no problem-solving insight required, making this easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\begin{pmatrix}-2\\3\\1\end{pmatrix}.\begin{pmatrix}2\\1\\q\end{pmatrix}\) with \(q=3, = -4+3+3 = 2\) | M1 | Use of \(a_1a_2+b_1b_2+c_1c_2\). |
| \(= \sqrt{14}.\sqrt{14}\cos\theta = 2, \cos\theta = \frac{1}{7}\) | M1 A1 | Dot product linked with moduli and cos. co |
| (ii) \(\vec{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix}4\\-2\\q-1\end{pmatrix}\) | M1 | Allow for p-q or q-p |
| \(16 + 4 + (q-1)^2 = 36 \rightarrow q = 5\) or \(q = -3\) | M1 A1 A1 | Use of modulus and Pythagoras Co (for both) |
$\vec{OP} = \begin{pmatrix}-2\\3\\1\end{pmatrix}$ and $\vec{OQ} = \begin{pmatrix}2\\1\\q\end{pmatrix}$
(i) $\begin{pmatrix}-2\\3\\1\end{pmatrix}.\begin{pmatrix}2\\1\\q\end{pmatrix}$ with $q=3, = -4+3+3 = 2$ | M1 | Use of $a_1a_2+b_1b_2+c_1c_2$.
$= \sqrt{14}.\sqrt{14}\cos\theta = 2, \cos\theta = \frac{1}{7}$ | M1 A1 | Dot product linked with moduli and cos. co | [3]
(ii) $\vec{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix}4\\-2\\q-1\end{pmatrix}$ | M1 | Allow for **p-q** or **q-p**
$16 + 4 + (q-1)^2 = 36 \rightarrow q = 5$ or $q = -3$ | M1 A1 A1 | Use of modulus and Pythagoras Co (for both) | [4]4 Relative to an origin $O$, the position vectors of points $P$ and $Q$ are given by
$$\overrightarrow { O P } = \left( \begin{array} { r }
- 2 \\
3 \\
1
\end{array} \right) \quad \text { and } \quad \overrightarrow { O Q } = \left( \begin{array} { l }
2 \\
1 \\
q
\end{array} \right)$$
where $q$ is a constant.\\
(i) In the case where $q = 3$, use a scalar product to show that $\cos P O Q = \frac { 1 } { 7 }$.\\
(ii) Find the values of $q$ for which the length of $\overrightarrow { P Q }$ is 6 units.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{933cdfe1-27bb-450d-8b9a-b494916242cb-3_647_741_845_699}
\end{center}
The diagram shows the cross-section of a hollow cone and a circular cylinder. The cone has radius 6 cm and height 12 cm , and the cylinder has radius $r \mathrm {~cm}$ and height $h \mathrm {~cm}$. The cylinder just fits inside the cone with all of its upper edge touching the surface of the cone.\\
\hfill \mbox{\textit{CAIE P1 2005 Q4 [7]}}