| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 8 |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Moderate -0.3 Part (i) requires finding two midpoints using standard formula and calculating distance—routine vector operations. Part (ii) involves finding direction vector of AB and using parallel line condition, which is straightforward application of vector concepts. Both parts are standard textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1.5 \\ 0.5 \\ 4 \end{pmatrix}\) \(\begin{pmatrix} -1 \\ -0.5 \\ 4 \end{pmatrix}\) | M1, A1 | Correct method for one midpoint |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.5^2 + 1^2 (+ 0^2)\) | M1 | ft their midpoints; \(\sqrt{\;}\) not necessary for M1 |
| Distance \(= \frac{\sqrt{29}}{2}\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}\) \(\overrightarrow{CD} = \begin{pmatrix} -4-x \\ 6 \\ 3-z \end{pmatrix}\) | M1 | |
| \(\overrightarrow{CD} = -2\overrightarrow{AB}\) | M1 | For scale factor -2 |
| Answer | Marks |
|---|---|
| \(3 - z = -4 \Rightarrow z = 7\) | A1, A1 |
| [4] |
## 5(i)
Position vectors of midpoints $AB$ & $BC$ are
$\begin{pmatrix} 1.5 \\ 0.5 \\ 4 \end{pmatrix}$ $\begin{pmatrix} -1 \\ -0.5 \\ 4 \end{pmatrix}$ | M1, A1 | Correct method for one midpoint
Both midpoints correct
$2.5^2 + 1^2 (+ 0^2)$ | M1 | ft their midpoints; $\sqrt{\;}$ not necessary for M1
Distance $= \frac{\sqrt{29}}{2}$ | A1 |
| [4]
## 5(ii)
$\overrightarrow{AB} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$ $\overrightarrow{CD} = \begin{pmatrix} -4-x \\ 6 \\ 3-z \end{pmatrix}$ | M1 |
$\overrightarrow{CD} = -2\overrightarrow{AB}$ | M1 | For scale factor -2
$-x - 4 = -2 \Rightarrow x = -2$
$3 - z = -4 \Rightarrow z = 7$ | A1, A1 |
| [4]
---5 Points $A , B$ and $C$ have position vectors $\left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right) , \left( \begin{array} { c } 2 \\ - 1 \\ 5 \end{array} \right)$ and $\left( \begin{array} { c } - 4 \\ 0 \\ 3 \end{array} \right)$ respectively.\\
(i) Find the exact distance between the midpoint of $A B$ and the midpoint of $B C$.
Point $D$ has position vector $\left( \begin{array} { c } x \\ - 6 \\ z \end{array} \right)$ and the line $C D$ is parallel to the line $A B$.\\
(ii) Find all the possible pairs of $x$ and $z$.
\hfill \mbox{\textit{OCR H240/02 2018 Q5 [8]}}