| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Topic | Vectors: Cross Product & Distances |
| Type | Perpendicular distance from point to plane |
| Difficulty | Standard +0.3 This is a standard Further Maths question testing routine 3D geometry formulas: point-to-plane distance (direct formula application) and skew line distance (standard vector product method). Both parts require straightforward substitution into well-known formulas with minimal problem-solving, making it slightly easier than average even for Further Maths content. |
| Spec | 4.04i Shortest distance: between a point and a line4.04j Shortest distance: between a point and a plane |
\begin{enumerate}[label=(\alph*)]
\item A plane $\Pi$ has the equation $\mathbf{r} \cdot \begin{pmatrix} 3 \\ 6 \\ -2 \end{pmatrix} = 15$. $C$ is the point $(4, -5, 1)$.
Find the shortest distance between $\Pi$ and $C$. [3]
\item Lines $l_1$ and $l_2$ have the following equations.
$l_1: \mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 4 \\ -2 \end{pmatrix}$
$l_2: \mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$
Find, in exact form, the distance between $l_1$ and $l_2$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q1 [8]}}