| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Vector between two points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing basic concepts: midpoint formula (routine), magnitude calculation (solving a simple quadratic), and parallel vectors for trapezium (standard property). All parts use direct application of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks |
|---|---|
| \(k = 3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-4)^2 + (2-k)^2 = 13\) | M1 | oe e.g. allow consistent use of square roots; must be using subtraction in brackets; may be implied by one correct value for \(k\) |
| \(k = 0\) | A1 | |
| \(k = 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{4-2}{7-1} = \dfrac{k-5}{4-3}\) oe | M1 | or \(\dfrac{5-4}{3-7} = \dfrac{k-2}{4-1}\) oe or \(\dfrac{5-2}{3-1} = \dfrac{4-k}{7-4}\) oe; consistent application of gradients (allow one sign error) |
| \(k = \dfrac{16}{3}\) or \(k = -\dfrac{1}{2}\) or \(k = \dfrac{5}{4}\) | A1 |
# Question 5:
## Part (a):
$k = 3$ | **B1** |
## Part (b):
$(1-4)^2 + (2-k)^2 = 13$ | **M1** | oe e.g. allow consistent use of square roots; must be using subtraction in brackets; may be implied by one correct value for $k$
$k = 0$ | **A1** |
$k = 4$ | **A1** |
## Part (c):
$\dfrac{4-2}{7-1} = \dfrac{k-5}{4-3}$ oe | **M1** | or $\dfrac{5-4}{3-7} = \dfrac{k-2}{4-1}$ oe or $\dfrac{5-2}{3-1} = \dfrac{4-k}{7-4}$ oe; consistent application of gradients (allow one sign error)
$k = \dfrac{16}{3}$ or $k = -\dfrac{1}{2}$ or $k = \dfrac{5}{4}$ | **A1** |
---5 Points $A , B , C$ and $D$ have position vectors $\mathbf { a } = \binom { 1 } { 2 } , \mathbf { b } = \binom { 3 } { 5 } , \mathbf { c } = \binom { 7 } { 4 }$ and $\mathbf { d } = \binom { 4 } { k }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$ for which $D$ is the midpoint of $A C$.
\item Find the two values of $k$ for which $| \overrightarrow { A D } | = \sqrt { 13 }$.
\item Find one value of $k$ for which the four points form a trapezium.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q5 [6]}}