| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 10 |
| 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: collinearity condition (proportional vectors), perpendicularity (dot product = 0), and parallelogram properties. All parts are routine applications of basic vector concepts with no novel insight required, making it slightly easier than average for A-level. |
| 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 |
|---|---|---|
| (i) \(p = 2\); Unit vector \(\frac{1}{\sqrt{6}}\begin{pmatrix}1\\1\end{pmatrix}\) or \(\frac{1}{\sqrt{24}}\begin{pmatrix}2\\2\end{pmatrix}\) oe | B1, M1, A1, \(\checkmark\) | ft for their \(p\) [3] |
| (ii) \(\overrightarrow{AB} = \begin{pmatrix}2\\1\\p\end{pmatrix} - \begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}4-p\\1\\p-1\end{pmatrix}\); \(\overrightarrow{OA} \cdot \overrightarrow{AB} = 0 \Rightarrow 4p - p^2 + 1 + p - 1 = 0\); \(5p - p^2 = 0 \Rightarrow p = 0\) or 5 | M1, A1, M1, A1, \(\checkmark\) | ft from their AB; cao [5] |
| (iii) \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}2\\3\end{pmatrix} - \begin{pmatrix}1\\1\end{pmatrix}\) oe \(= \begin{pmatrix}1\\2\end{pmatrix}\) | M1, A1 | [2] |
(i) $p = 2$; Unit vector $\frac{1}{\sqrt{6}}\begin{pmatrix}1\\1\end{pmatrix}$ or $\frac{1}{\sqrt{24}}\begin{pmatrix}2\\2\end{pmatrix}$ oe | B1, M1, A1, $\checkmark$ | ft for their $p$ [3]
(ii) $\overrightarrow{AB} = \begin{pmatrix}2\\1\\p\end{pmatrix} - \begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}4-p\\1\\p-1\end{pmatrix}$; $\overrightarrow{OA} \cdot \overrightarrow{AB} = 0 \Rightarrow 4p - p^2 + 1 + p - 1 = 0$; $5p - p^2 = 0 \Rightarrow p = 0$ or 5 | M1, A1, M1, A1, $\checkmark$ | ft from their AB; cao [5]
(iii) $\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}2\\3\end{pmatrix} - \begin{pmatrix}1\\1\end{pmatrix}$ oe $= \begin{pmatrix}1\\2\end{pmatrix}$ | M1, A1 | [2]9 The position vectors of points $A$ and $B$ relative to an origin $O$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { c }
p \\
1 \\
1
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { l }
4 \\
2 \\
p
\end{array} \right)$$
where $p$ is a constant.\\
(i) In the case where $O A B$ is a straight line, state the value of $p$ and find the unit vector in the direction of $\overrightarrow { O A }$.\\
(ii) In the case where $O A$ is perpendicular to $A B$, find the possible values of $p$.\\
(iii) In the case where $p = 3$, the point $C$ is such that $O A B C$ is a parallelogram. Find the position vector of $C$.
\hfill \mbox{\textit{CAIE P1 2012 Q9 [10]}}