| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a Further Maths question combining 2D perpendicular distance (routine) with 3D skew line distance using the scalar triple product formula. Part (a) is straightforward, but parts (b)-(c) require knowledge of the skew lines formula and geometric interpretation, which is beyond standard A-level and involves multi-step vector calculations. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
\begin{enumerate}[label=(\alph*)]
\item Find the shortest distance between the point $(-6, 4)$ and the line $y = -0.75x + 7$. [2]
\end{enumerate}
Two lines, $l_1$ and $l_2$, are given by
$$l_1: \mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -4 \end{pmatrix} \text{ and } l_2: \mathbf{r} = \begin{pmatrix} 11 \\ -1 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the shortest distance between $l_1$ and $l_2$. [3]
\item Hence determine the geometrical arrangement of $l_1$ and $l_2$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q2 [7]}}