| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Intersection of line with plane |
| Difficulty | Standard +0.8 This FP3 question requires finding a line-plane intersection (standard technique) and then constructing a plane from a line and perpendicularity condition (requiring cross product of direction vectors and point substitution). While methodical, it demands careful vector manipulation and multiple steps, placing it moderately above average difficulty for A-level but routine for Further Maths students. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
A line $l$ has equation $\frac{x - 6}{-4} = \frac{y + 7}{8} = \frac{z + 10}{7}$ and a plane $p$ has equation $3x - 4y - 2z = 8$.
\begin{enumerate}[label=(\roman*)]
\item Find the point of intersection of $l$ and $p$. [3]
\item Find the equation of the plane which contains $l$ and is perpendicular to $p$, giving your answer in the form $ax + by + cz = d$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 Q3 [8]}}