| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Moderate -0.8 This is a straightforward vectors question requiring only basic operations: finding a midpoint using the midpoint formula, calculating magnitude using Pythagoras, and applying the vector equation BC=AD to find D. All steps are routine recall with no problem-solving or geometric insight needed. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
2 The points $A$, $B$ and $C$ have position vectors $3 \mathbf { i } - 4 \mathbf { j } + 2 \mathbf { k } , - \mathbf { i } + 6 \mathbf { k }$ and $7 \mathbf { i } - 4 \mathbf { j } - 2 \mathbf { k }$ respectively. M is the midpoint of BC .
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of $\overrightarrow { O M }$ is equal to $\sqrt { 17 }$.
Point D is such that $\overrightarrow { B C } = \overrightarrow { A D }$.
\item Show that position vector of the point D is $11 \mathbf { i } - 8 \mathbf { j } - 6 \mathbf { k }$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 Q2 [5]}}