| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Common perpendicular to two skew lines |
| Difficulty | Standard +0.8 This is a Further Pure 3 question on skew lines requiring the cross product of direction vectors and the scalar triple product formula for shortest distance. While the techniques are standard for FP3, the question demands careful vector manipulation and formula application beyond typical A-level content, placing it moderately above average difficulty. |
| Spec | 4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
Two skew lines have equations
$$\frac{x}{2} = \frac{y + 3}{1} = \frac{z - 6}{3} \quad \text{and} \quad \frac{x - 5}{3} = \frac{y + 1}{1} = \frac{z - 7}{5}.$$
\begin{enumerate}[label=(\roman*)]
\item Find the direction of the common perpendicular to the lines. [2]
\item Find the shortest distance between the lines. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 Q3 [6]}}