| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.3 This is a straightforward application of standard vector techniques: part (i) uses the dot product condition for perpendicularity (OA·OB = 0) leading to a simple quadratic, and part (ii) requires finding a vector and normalizing it. Both are routine procedures with no conceptual challenges, making this slightly easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(OA.OB = -3p^2 - 4 + p^4\) soi \((p^2 + 1)(p^2 - 4) = 0\) oe e.g. with substitution \(p = \pm 2\) and no other real solutions | M1 M1 A1 [3] | Put = 0 (soi) and attempt to solve |
| (ii) \(\overrightarrow{BA} = \begin{pmatrix} 9 \\ 4 \\ 9 \end{pmatrix} - \begin{pmatrix} -3 \\ -1 \\ 9 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}\) | M1 | Reversed subtraction can score M1M1A0 |
| \(\overrightarrow{BA} = \sqrt{12^2 + 5^2} = 13\) and division by their 13 | M1 | |
| Unit vector \(= \frac{1}{13}\begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}\) cao | A1 [3] |
(i) $OA.OB = -3p^2 - 4 + p^4$ soi $(p^2 + 1)(p^2 - 4) = 0$ oe e.g. with substitution $p = \pm 2$ and no other real solutions | M1 M1 A1 [3] | Put = 0 (soi) and attempt to solve
(ii) $\overrightarrow{BA} = \begin{pmatrix} 9 \\ 4 \\ 9 \end{pmatrix} - \begin{pmatrix} -3 \\ -1 \\ 9 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}$ | M1 | Reversed subtraction can score M1M1A0
$\overrightarrow{BA} = \sqrt{12^2 + 5^2} = 13$ and division by their 13 | M1 |
Unit vector $= \frac{1}{13}\begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}$ cao | A1 [3] |8 Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { c }
3 p \\
4 \\
p ^ { 2 }
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { c }
- p \\
- 1 \\
p ^ { 2 }
\end{array} \right)$$
(i) Find the values of $p$ for which angle $A O B$ is $90 ^ { \circ }$.\\
(ii) For the case where $p = 3$, find the unit vector in the direction of $\overrightarrow { B A }$.
\hfill \mbox{\textit{CAIE P1 2014 Q8 [6]}}