| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 7 |
| Topic | Vectors 3D & Lines |
| Type | Angle between two vectors/lines (direct) |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard vector techniques: finding distance using magnitude of difference vectors, then using scalar product formula to find an angle. Both are routine A-level procedures with no conceptual challenges, making it slightly easier than average but not trivial since it requires careful calculation with 3D vectors. |
| Spec | 1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Attempt \(\cos\theta = \frac{\overrightarrow{AO}\cdot\overrightarrow{AB}}{ | \overrightarrow{AO} |
(i) Find $\mathbf{a} - \mathbf{b}$ or $\mathbf{b} - \mathbf{a}$ [M1]
Use correct method to find the magnitude of any vector [M1]
$\sqrt{154}$ or equivalent [A1]
(ii) Attempt $\cos\theta = \frac{\overrightarrow{AO}\cdot\overrightarrow{AB}}{|\overrightarrow{AO}||\overrightarrow{AB}|}$ [M1]
Obtain 70 anywhere [B1]
Obtain $\frac{70}{\sqrt{45}\sqrt{154}}$ [A1]
Obtain $32.8°$ [A1]9 The points $A$ and $B$ have position vectors $\mathbf { a }$ and $\mathbf { b }$ relative to an origin $O$, where $\mathbf { a } = 5 \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k }$ and $\mathbf { b } = - 7 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }$.\\
(i) Find the length of $A B$.\\
(ii) Use a scalar product to find angle $O A B$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2016 Q9 [7]}}