| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Acute angle between two planes |
| Difficulty | Challenging +1.2 This is a solid FP3 3D coordinate geometry question requiring multiple techniques (finding point from perpendicularity condition, distance formula, plane equations, angle between planes via normal vectors) but follows standard methods throughout. The perpendicularity condition AB ⊥ plane BCD means AB is parallel to the normal vector, giving a straightforward parametric approach to find B. Part (ii) requires finding the normal to plane ACD and using the scalar product formula—all textbook procedures. While multi-step and requiring careful calculation, it demands no novel insight beyond applying learned techniques systematically. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane |
A tetrahedron $ABCD$ is such that $AB$ is perpendicular to the base $BCD$. The coordinates of the points $A, C$ and $D$ are $(-1, -7, 2), (5, 0, 3)$ and $(-1, 3, 3)$ respectively, and the equation of the plane $BCD$ is $x + 2y - 2z = -1$.
\begin{enumerate}[label=(\roman*)]
\item Find, in either order, the coordinates of $B$ and the length of $AB$. [5]
\item Find the acute angle between the planes $ACD$ and $BCD$. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 Q6 [11]}}