| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.8 This is a straightforward application of standard vector techniques: (i) uses the perpendicularity condition (dot product = 0) to solve a quadratic equation, and (ii) uses collinearity (vectors are scalar multiples) to find p, then calculates magnitude. Both parts are routine textbook exercises requiring only direct application of well-practiced methods with no problem-solving insight needed. |
| 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 multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(-2p^2 + 16p - 24 + 2p^2 - 6p + 2\) | M1 | Good attempt at scalar product |
| Set scalar product \(= 0\) and attempt solution | DM1 | |
| \(p = 2.2\) | A1 | |
| [3] | ||
| (ii) \(4 - 2p = 2(p-6)\) or \(p = 2(2p-6)\) | M1 | |
| \(p = 4 \rightarrow \overrightarrow{OA} = \begin{pmatrix}-2\\2\\1\end{pmatrix}\), \(\overrightarrow{OB} = \begin{pmatrix}-4\\4\\2\end{pmatrix}\) | A1 | At least one of OA and OB correct |
| \(\ | \overrightarrow{OA}\ | = \sqrt{(-2)^2 + 2^2 + 1^2} = 3\) |
| [4] |
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $-2p^2 + 16p - 24 + 2p^2 - 6p + 2$ | M1 | Good attempt at scalar product |
| Set scalar product $= 0$ and attempt solution | DM1 | |
| $p = 2.2$ | A1 | |
| **[3]** | | |
| **(ii)** $4 - 2p = 2(p-6)$ or $p = 2(2p-6)$ | M1 | |
| $p = 4 \rightarrow \overrightarrow{OA} = \begin{pmatrix}-2\\2\\1\end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}-4\\4\\2\end{pmatrix}$ | A1 | At least one of **OA** and **OB** correct |
| $\|\overrightarrow{OA}\| = \sqrt{(-2)^2 + 2^2 + 1^2} = 3$ | M1A1 | For M1 accept a numerical $p$ |
| **[4]** | | |
---5 Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { c }
p - 6 \\
2 p - 6 \\
1
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { c }
4 - 2 p \\
p \\
2
\end{array} \right)$$
where $p$ is a constant.\\
(i) For the case where $O A$ is perpendicular to $O B$, find the value of $p$.\\
(ii) For the case where $O A B$ is a straight line, find the vectors $\overrightarrow { O A }$ and $\overrightarrow { O B }$. Find also the length of the line $O A$.
\hfill \mbox{\textit{CAIE P1 2015 Q5 [7]}}