| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angles between vectors |
| Difficulty | Moderate -0.5 This is a straightforward application of standard vector formulas (dot product, magnitude, and angle) with no conceptual challenges. While it's a multi-part question requiring several calculations, each step is routine and follows directly from memorized formulas. It's slightly easier than average because it's purely computational with no problem-solving or geometric insight required. |
| Spec | 1.10c Magnitude and direction: of vectors4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} \cdot \mathbf{b} = 3\times2 + 4\times-1 + -2\times-5 = 12\) | B1 | Obtains 12 |
| Answer | Marks | Guidance |
|---|---|---|
| \( | \mathbf{a} | = \sqrt{3^2 + 4^2 + (-2)^2} = \sqrt{29}\) |
| \( | \mathbf{b} | = \sqrt{2^2 + (-1)^2 + (-5)^2} = \sqrt{30}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\theta = \dfrac{12}{\sqrt{29}\times\sqrt{30}}\), \(\theta = 65.9936\ldots\) | M1 | Writes correct equation in \(\theta\); use of \(\mathbf{a}\times\mathbf{b}\) must at least proceed to correct calculation of \( |
| \(\theta = 66°\) (nearest degree) | A1 | Obtains 66 |
## Question 5:
### Part 5(a):
| $\mathbf{a} \cdot \mathbf{b} = 3\times2 + 4\times-1 + -2\times-5 = 12$ | B1 | Obtains 12 |
### Part 5(b):
| $|\mathbf{a}| = \sqrt{3^2 + 4^2 + (-2)^2} = \sqrt{29}$ | M1 | Obtains correct expression for at least one of $|\mathbf{a}|$ or $|\mathbf{b}|$; condone missing brackets on negative values |
| $|\mathbf{b}| = \sqrt{2^2 + (-1)^2 + (-5)^2} = \sqrt{30}$ | A1 | Obtains $\sqrt{29}$ and $\sqrt{30}$; condone AWRT 5.39 and 5.48 |
### Part 5(c):
| $\cos\theta = \dfrac{12}{\sqrt{29}\times\sqrt{30}}$, $\theta = 65.9936\ldots$ | M1 | Writes correct equation in $\theta$; use of $\mathbf{a}\times\mathbf{b}$ must at least proceed to correct calculation of $|\mathbf{a}\times\mathbf{b}|$ leading to equation in $\theta$; FT their $\mathbf{a}\cdot\mathbf{b}$ and $|\mathbf{a}|$, $|\mathbf{b}|$ |
| $\theta = 66°$ (nearest degree) | A1 | Obtains 66 |
---5 The vectors $\mathbf { a }$ and $\mathbf { b }$ are given by
$$\mathbf { a } = 3 \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } \quad \text { and } \quad \mathbf { b } = 2 \mathbf { i } - \mathbf { j } - 5 \mathbf { k }$$
5
\begin{enumerate}[label=(\alph*)]
\item Calculate a.b
5
\item $\quad$ Calculate $| \mathbf { a } |$ and $| \mathbf { b } |$\\
$| \mathbf { a } | =$ $\_\_\_\_$\\
5
\item Calculate the acute angle between $\mathbf { a }$ and $\mathbf { b }$\\
Give your answer to the nearest degree.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q5 [5]}}