| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Unit vector in given direction |
| Difficulty | Moderate -0.8 This is a straightforward vectors question requiring basic recall of position vector relationships and unit vector calculation. Part (a) involves simple vector subtraction to find x and y (routine algebra), and part (b) requires dividing by magnitude—both standard AS-level procedures with no problem-solving insight needed. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1, y = 3\) | M1, A1, A1 | Attempts \(\overline{AB} = \overline{OB} - \overline{OA}\); Equating the coefficients of \(i\) and \(j\) and attempts to find the value of \(x\) or \(y\); Both correct values of \(x\) and \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Unit vector: \(-\frac{3}{5}i + \frac{4}{5}j\) | M1, A1 | Makes an attempt to use Pythagoras theorem to find \( |
### Part a:
$x = 1, y = 3$ | M1, A1, A1 | Attempts $\overline{AB} = \overline{OB} - \overline{OA}$; Equating the coefficients of $i$ and $j$ and attempts to find the value of $x$ or $y$; Both correct values of $x$ and $y$
### Part b:
Unit vector: $-\frac{3}{5}i + \frac{4}{5}j$ | M1, A1 | Makes an attempt to use Pythagoras theorem to find $|\overline{AB}|$; Correct answer onlyGiven that the point $A$ has position vector $x\mathbf{i} - \mathbf{j}$, the point $B$ has position vector $-2\mathbf{i} + y\mathbf{j}$ and $\overrightarrow{AB} = -3\mathbf{i} + 4\mathbf{j}$, find
\begin{enumerate}[label=(\alph*)]
\item the values of $x$ and $y$ [3]
\item a unit vector in the direction of $\overrightarrow{AB}$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q3 [5]}}