| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (i)(a) solving simultaneous equations to find line intersection (standard), (i)(b) finding a plane equation via cross product of direction vectors (routine FP1), and (ii) using directional cosines with constraints to find two lines (requires careful algebraic manipulation and understanding that cos²α + cos²β + cos²γ = 1). Part (ii) elevates this above routine exercises, requiring problem-solving with directional cosines and solving a system with two solutions, but remains within expected FP1 scope. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Forms and solves two equations: \(2+3\lambda=13+\mu\), \(-3+4\lambda=5-2\mu\), \(1-\lambda=8+5\mu\) giving \(\lambda=3, \mu=-2\) | M1 | 3.1a |
| Point: \(\begin{pmatrix}11\\9\\-2\end{pmatrix}\) | A1 | 1.1b; accept as coordinates or vector |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 3 & 4 & -1\\ 1 & -2 & 5\end{vmatrix} = \mathbf{i}(20-2)-\mathbf{j}(15+1)+\mathbf{k}(-6-4)\) | M1 | 3.1a; two correct components implies method |
| \(\pm(18\mathbf{i}-16\mathbf{j}-10\mathbf{k})\) | A1 | 1.1b |
| \(\mathbf{r}\bullet\begin{pmatrix}18\\-16\\-10\end{pmatrix} = \begin{pmatrix}11\\9\\-2\end{pmatrix}\bullet\begin{pmatrix}18\\-16\\-10\end{pmatrix} = \ldots\) | M1 | 1.1b; condone minor miscopies |
| \(18x - 16y - 10z = 74\), equivalently \(k(9x-8y-5z=37)\) | A1 | 2.5; accept any multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\cos^2 60° + \cos^2 45° + \cos^2\theta = 1\): \(\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2\theta = 1\) | M1 | 3.1a; allow with \(n\) or another letter for \(\cos\theta\) |
| \(\cos\theta = \pm\frac{1}{2}\) | A1 | 1.1b |
| \(\frac{x}{\cos 60°} = \frac{y}{\cos 45°} = \frac{z}{\text{their } \cos\theta}\) | M1 | 1.1b |
| \(2x = \sqrt{2}y = 2z\) and \(2x = \sqrt{2}y = -2z\) | A1 | 2.5; need not be simplified; accept "and" or "or" |
# Question 9(i)(a) - Line Intersection
| Working/Answer | Mark | Guidance |
|---|---|---|
| Forms and solves two equations: $2+3\lambda=13+\mu$, $-3+4\lambda=5-2\mu$, $1-\lambda=8+5\mu$ giving $\lambda=3, \mu=-2$ | M1 | 3.1a |
| Point: $\begin{pmatrix}11\\9\\-2\end{pmatrix}$ | A1 | 1.1b; accept as coordinates or vector |
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# Question 9(i)(b) - Normal to Plane
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 3 & 4 & -1\\ 1 & -2 & 5\end{vmatrix} = \mathbf{i}(20-2)-\mathbf{j}(15+1)+\mathbf{k}(-6-4)$ | M1 | 3.1a; two correct components implies method |
| $\pm(18\mathbf{i}-16\mathbf{j}-10\mathbf{k})$ | A1 | 1.1b |
| $\mathbf{r}\bullet\begin{pmatrix}18\\-16\\-10\end{pmatrix} = \begin{pmatrix}11\\9\\-2\end{pmatrix}\bullet\begin{pmatrix}18\\-16\\-10\end{pmatrix} = \ldots$ | M1 | 1.1b; condone minor miscopies |
| $18x - 16y - 10z = 74$, equivalently $k(9x-8y-5z=37)$ | A1 | 2.5; accept any multiple |
---
# Question 9(ii) - Direction Cosines and Line Equations
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\cos^2 60° + \cos^2 45° + \cos^2\theta = 1$: $\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2\theta = 1$ | M1 | 3.1a; allow with $n$ or another letter for $\cos\theta$ |
| $\cos\theta = \pm\frac{1}{2}$ | A1 | 1.1b |
| $\frac{x}{\cos 60°} = \frac{y}{\cos 45°} = \frac{z}{\text{their } \cos\theta}$ | M1 | 1.1b |
| $2x = \sqrt{2}y = 2z$ and $2x = \sqrt{2}y = -2z$ | A1 | 2.5; need not be simplified; accept "and" or "or" |\begin{enumerate}
\item (i) The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ - 3 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { r } 3 \\ 4 \\ - 1 \end{array} \right)$
\end{enumerate}
The line $l _ { 2 }$ has equation $\mathbf { r } = \left( \begin{array} { c } 13 \\ 5 \\ 8 \end{array} \right) + \mu \left( \begin{array} { r } 1 \\ - 2 \\ 5 \end{array} \right)$\\
where $\lambda$ and $\mu$ are scalar parameters.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $P$.\\
(a) Determine the coordinates of $P$.
Given that the plane $\Pi$ contains both $l _ { 1 }$ and $l _ { 2 }$\\
(b) determine a Cartesian equation for $\Pi$.\\
(ii) Determine a Cartesian equation for each of the two lines that
\begin{itemize}
\item pass through $( 0,0,0 )$
\item make an angle of $60 ^ { \circ }$ with the $x$-axis
\item make an angle of $45 ^ { \circ }$ with the $y$-axis
\end{itemize}
\hfill \mbox{\textit{Edexcel FP1 2024 Q9 [10]}}