| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Ratio division of line segment |
| Difficulty | Moderate -0.8 This question tests basic vector operations (scalar multiplication, addition/subtraction), distance calculation using Pythagoras, and understanding position vectors on a line. Part (a) is routine computation, part (b)(i) requires simple ratio division on a line segment, and part (b)(ii) tests conceptual understanding that coefficients must sum to 1 for collinearity. All are standard textbook exercises with no problem-solving insight required, making this easier than average but not trivial due to the multi-part nature. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(4u – 3v = 20i – 27j\) | B1 | AO1 |
| B1 | AO1 | |
| (ii) A correct method for finding the length of \(UV\) | M1 | AO1 |
| Length of \(UV = 10\) | A1 | AO1 |
| (b) (i) Position vector of \(C = \frac{1}{10}a + \frac{9}{10}b\) or \(C = \frac{9}{10}a + \frac{1}{10}b\) | M1 | AO3 |
| Position vector \(C = -\frac{1}{10}a + \frac{9}{10}b\) | A1 | AO3 |
| (ii) The position vector of any point on the road will be of the form \(\lambda a + (1 – \lambda)b\) for some value of \(\lambda\) | B1 | AO2 |
**(a)** (i) $4u – 3v = 20i – 27j$ | B1 | AO1
| B1 | AO1
(ii) A correct method for finding the length of $UV$ | M1 | AO1
Length of $UV = 10$ | A1 | AO1
**(b)** (i) Position vector of $C = \frac{1}{10}a + \frac{9}{10}b$ or $C = \frac{9}{10}a + \frac{1}{10}b$ | M1 | AO3
Position vector $C = -\frac{1}{10}a + \frac{9}{10}b$ | A1 | AO3
(ii) The position vector of any point on the road will be of the form $\lambda a + (1 – \lambda)b$ for some value of $\lambda$ | B1 | AO2
**Total: [7]**\begin{enumerate}[label=(\alph*)]
\item The vectors $\mathbf{u}$ and $\mathbf{v}$ are defined by $\mathbf{u} = 2\mathbf{i} - 3\mathbf{j}$, $\mathbf{v} = -4\mathbf{i} + 5\mathbf{j}$.
\begin{enumerate}[label=(\roman*)]
\item Find the vector $4\mathbf{u} - 3\mathbf{v}$.
\item The vectors $\mathbf{u}$ and $\mathbf{v}$ are the position vectors of the points $U$ and $V$, respectively. Find the length of the line $UV$. [4]
\end{enumerate}
\item Two villages $A$ and $B$ are 40 km apart on a long straight road passing through a desert. The position vectors of $A$ and $B$ are denoted by $\mathbf{a}$ and $\mathbf{b}$, respectively.
\begin{enumerate}[label=(\roman*)]
\item Village $C$ lies on the road between $A$ and $B$ at a distance 4 km from $B$. Find the position vector of $C$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
\item Village $D$ has position vector $\frac{2}{9}\mathbf{a} + \frac{5}{9}\mathbf{b}$. Explain why village $D$ cannot possibly be on the straight road passing through $A$ and $B$. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 1 Q18 [7]}}