| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Acute angle between two planes |
| Difficulty | Challenging +1.3 This FP3 question requires finding a plane equation from three points (standard vector method), recognizing geometric symmetry in a tetrahedron, and calculating angles between planes using normal vectors. While it involves multiple steps and 3D geometry with awkward coordinates, the techniques are all standard Further Maths material with no novel insight required. The symmetry observation in part (ii) is straightforward given the coordinate structure. Moderately challenging but well within typical FP3 scope. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
A regular tetrahedron has vertices at the points
$$A\left(0, 0, \frac{2}{\sqrt{3}}\sqrt{6}\right), \quad B\left(\frac{2}{\sqrt{3}}\sqrt{3}, 0, 0\right), \quad C\left(-\frac{1}{3}\sqrt{3}, 1, 0\right), \quad D\left(-\frac{1}{3}\sqrt{3}, -1, 0\right).$$
\begin{enumerate}[label=(\roman*)]
\item Obtain the equation of the face $ABC$ in the form
$$x + \sqrt{3}y + \left(\frac{1}{2}\sqrt{2}\right)z = \frac{2}{3}\sqrt{3}.$$ [5]
(Answers which only verify the given equation will not receive full credit.)
\item Give a geometrical reason why the equation of the face $ABD$ can be expressed as
$$x - \sqrt{3}y + \left(\frac{1}{2}\sqrt{2}\right)z = \frac{2}{3}\sqrt{3}.$$ [2]
\item Hence find the cosine of the angle between two faces of the tetrahedron. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2010 Q5 [11]}}