| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2025 |
| Session | February |
| Marks | 9 |
| Topic | Vectors: Lines & Planes |
| Type | Angle between two planes |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question covering routine techniques: finding a normal vector via cross product to get Cartesian form, substituting parametric line equations into a plane, and using the scalar product formula for angle between planes. All three parts are textbook exercises requiring methodical application of learned procedures with no novel problem-solving or geometric insight needed. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
The plane $\Pi_1$ has equation
$$\mathbf{r} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} + \lambda (\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) + \mu(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})$$
where $\lambda$ and $\mu$ are scalar parameters.
\begin{enumerate}[label=(\alph*)]
\item Find a Cartesian equation for $\Pi_1$ [4]
\end{enumerate}
The line $l$ has equation
$$\frac{x-1}{5} = \frac{y-3}{-3} = \frac{z+2}{4}$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the coordinates of the point of intersection of $l$ with $\Pi_1$ [3]
\end{enumerate}
The plane $\Pi_2$ has equation
$$\mathbf{r}.(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = 5$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, to the nearest degree, the acute angle between $\Pi_1$ and $\Pi_2$ [2]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q4 [9]}}