| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Vector between two points |
| Difficulty | Easy -1.3 This is a straightforward two-part question testing basic vector definitions: finding a displacement vector by subtraction and calculating its magnitude using Pythagoras. Both are direct applications of standard formulas with no problem-solving required, making it easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\) | M1 | Subtraction either way; implied by one correct component \(\pm 9\mathbf{i} \pm 3\mathbf{j}\) |
| \(\overrightarrow{AB} = -9\mathbf{i} + 3\mathbf{j}\) | A1 | Accept \(\begin{pmatrix}-9\\3\end{pmatrix}\) but not \(\begin{pmatrix}-9\mathbf{i}\\3\mathbf{j}\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \( | AB | = \sqrt{(-9)^2 + (3)^2}\) |
| \( | AB | = 3\sqrt{10}\) |
# Question 3(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ | M1 | Subtraction either way; implied by one correct component $\pm 9\mathbf{i} \pm 3\mathbf{j}$ |
| $\overrightarrow{AB} = -9\mathbf{i} + 3\mathbf{j}$ | A1 | Accept $\begin{pmatrix}-9\\3\end{pmatrix}$ but not $\begin{pmatrix}-9\mathbf{i}\\3\mathbf{j}\end{pmatrix}$ |
# Question 3(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $|AB| = \sqrt{(-9)^2 + (3)^2}$ | M1 | Correct use of Pythagoras using answer to (a) |
| $|AB| = 3\sqrt{10}$ | A1ft | Follow through from (a); must be simplified |
---\begin{enumerate}
\item Given that the point $A$ has position vector $4 \mathbf { i } - 5 \mathbf { j }$ and the point $B$ has position vector $- 5 \mathbf { i } - 2 \mathbf { j }$, (a) find the vector $\overrightarrow { A B }$,\\
(b) find $| \overrightarrow { A B } |$.
\end{enumerate}
Give your answer as a simplified surd.
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q3 [4]}}