| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2023 |
| Session | February |
| Marks | 8 |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question testing routine techniques: finding a normal vector via cross product, converting to Cartesian form, and using the distance formula from point to plane. Part (c) requires solving a quadratic but follows a predictable method with no novel insight needed. Slightly easier than average due to straightforward application of learned procedures. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04f Line-plane intersection: find point |
The plane $\Pi$ has equation
$$\mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$$
where $\lambda$ and $\mu$ are scalar parameters.
\begin{enumerate}[label=(\alph*)]
\item Show that vector $2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$ is perpendicular to $\Pi$.
[2]
\item Hence find a Cartesian equation of $\Pi$.
[2]
\end{enumerate}
The line $l$ has equation
$$\mathbf{r} = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 6 \\ -3 \end{pmatrix}$$
where $t$ is a scalar parameter.
The point $A$ lies on $l$.
Given that the shortest distance between $A$ and $\Pi$ is $2\sqrt{29}$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item determine the possible coordinates of $A$.
[4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q4 [8]}}