| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| 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 two-part question testing basic vector concepts: (i) uses the perpendicularity condition (dot product = 0) with simple arithmetic, and (ii) requires finding a vector, calculating its magnitude, then scaling—all standard procedures with no problem-solving insight needed. Easier than average A-level content. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
**Question 3**
**(i)**
M1: Use of $x_1x_2 + y_1y_2 + z_1z_2$
$\text{OA} \cdot \text{OB} = 10 + 7 + 2p$
DM1: Set equal to $0$
$= 0 \rightarrow p = -8\frac{1}{2}$
A1: Correct value (accept negative)
**[3]**
**(ii)**
B1: $\text{AB} = -3i + 6j + 2k$
M1: For modulus
Modulus $= \sqrt{9 + 36 + 4}$
M1: Scales by $\times 28 \div \text{modulus}$
$\rightarrow -12i + 24j + 8k$
A1: Correct answer (could leave as $4 \times \ldots$)
**[4]**3 Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by
$$\overrightarrow { O A } = 5 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 2 \mathbf { i } + 7 \mathbf { j } + p \mathbf { k }$$
where $p$ is a constant.\\
(i) Find the value of $p$ for which angle $A O B$ is $90 ^ { \circ }$.\\
(ii) In the case where $p = 4$, find the vector which has magnitude 28 and is in the same direction as $\overrightarrow { A B }$.
\hfill \mbox{\textit{CAIE P1 2011 Q3 [7]}}