| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: dot product for perpendicularity, magnitude formula, and unit vector calculation. All parts require direct application of formulas with minimal problem-solving insight, making it slightly easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(OA.OB = -7 + 3 - 3p + p^2\), \((p + 1)(p - 4) = 0\) | M1, DM1 | Correct method for scalar product. Equate to zero & attempt to factorise/solve. '= 0' implied by answers |
| \(p = -1\) or \(4\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(49 + (1 - p^2) + p^2 = 2[(1 + 9 + p^2)]\) | M1 | Scalar result required |
| \(p = 15\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB = -8i + 6j\) | B1 | |
| Divide \(AB\) by \( | AB | = \sqrt{(-8)^2 + 6^2} = 10\) soi |
| Unit vector \(= \frac{1}{10}(-8i + 6j)\) oe cao | A1 | \(p = 15\) used – treat as MR \(\rightarrow \begin{pmatrix}-8\\-17\\0\end{pmatrix}\) [3] |
### (i)
$OA.OB = -7 + 3 - 3p + p^2$, $(p + 1)(p - 4) = 0$ | **M1, DM1** | Correct method for scalar product. Equate to zero & attempt to factorise/solve. '= 0' implied by answers
$p = -1$ or $4$ | **A1** | [3]
### (ii)
$49 + (1 - p^2) + p^2 = 2[(1 + 9 + p^2)]$ | **M1** | Scalar result required
$p = 15$ | **A1** | [2]
### (iii)
$AB = -8i + 6j$ | **B1** |
Divide $AB$ by $|AB| = \sqrt{(-8)^2 + 6^2} = 10$ soi | **M1** |
Unit vector $= \frac{1}{10}(-8i + 6j)$ oe cao | **A1** | $p = 15$ used – treat as MR $\rightarrow \begin{pmatrix}-8\\-17\\0\end{pmatrix}$ [3]
---Three points, $O$, $A$ and $B$, are such that $\overrightarrow{OA} = \mathbf{i} + 3\mathbf{j} + p\mathbf{k}$ and $\overrightarrow{OB} = -7\mathbf{i} + (1 - p)\mathbf{j} + p\mathbf{k}$, where $p$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Find the values of $p$ for which $\overrightarrow{OA}$ is perpendicular to $\overrightarrow{OB}$. [3]
\item The magnitudes of $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are $a$ and $b$ respectively. Find the value of $p$ for which $b^2 = 2a^2$. [2]
\item Find the unit vector in the direction of $\overrightarrow{AB}$ when $p = -8$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q7 [8]}}