| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.8 This is a Further Maths question requiring conversion of line equation to parametric form, substitution into plane equation, then finding a plane through a line perpendicular to another plane (requiring cross product of direction vectors). Multi-step with several techniques, but follows standard procedures without requiring novel insight. |
| Spec | 4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point |
3 A line $l$ has equation $\frac { x - 6 } { - 4 } = \frac { y + 7 } { 8 } = \frac { z + 10 } { 7 }$ and a plane $p$ has equation $3 x - 4 y - 2 z = 8$.\\
(i) Find the point of intersection of $l$ and $p$.\\
(ii) Find the equation of the plane which contains $l$ and is perpendicular to $p$, giving your answer in the form $a x + b y + c z = d$.
\hfill \mbox{\textit{OCR FP3 2009 Q3 [8]}}