| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 4 |
| 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 vector, then using scalar product formula to find an angle. Both are routine A-level procedures with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt \(\cos\theta = \frac{\overrightarrow{AO}\cdot\overrightarrow{AB}}{ | \overrightarrow{AO} |
**(a)**
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** [3]
**(b)**
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** [4]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 $\mathbf { b } = - 7 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }$.
\begin{enumerate}[label=(\alph*)]
\item Find the length of $A B$.
\item Use a scalar product to find angle $O A B$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2019 Q9 [4]}}