| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Standard +0.8 This is a standard Further Maths question on finding line of intersection of two planes (requiring cross product of normals and finding a point on the line) followed by finding where a third plane intersects this line. While it requires multiple techniques and careful algebraic manipulation, it follows a well-established procedure taught in FP3 with no novel insight required. The multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ -1 & 2 & 4 \end{vmatrix} = \begin{pmatrix} -2 \\ -11 \\ 5 \end{pmatrix}\) | M1A1 | M1: Attempt cross product of normal vectors (if method unclear, 2 components must be correct); A1: Correct vector |
| \(x=0: (0, \frac{1}{2}, \frac{3}{2})\), \(y=0: (-\frac{1}{11}, 0, \frac{19}{11})\), \(z=0: (\frac{3}{5}, \frac{19}{5}, 0)\) | M1A1 | M1: Attempt point on line (\(x\), \(y\) and \(z\)); A1: Correct coordinates |
| \(\mathbf{r} = \frac{1}{2}\mathbf{j} + \frac{3}{2}\mathbf{k} + \lambda(2\mathbf{i}+11\mathbf{j}-5\mathbf{k})\) | ddM1A1 | M1: Their point \(+ \lambda\) their direction (dependent on both previous method marks); A1: Correct equation (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2\lambda - \left(\frac{1}{2}+11\lambda\right) + 2\left(\frac{3}{2}-5\lambda\right) = 31\) | M1 | Substitutes into the third plane and solves for \(\lambda\) |
| \(\lambda = \frac{-3}{2}\) | ||
| Planes intersect at \((-3, -16, 9)\) | M1A1 | M1: Substitutes into their line; A1: Correct coordinates |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ -1 & 2 & 4 \end{vmatrix} = \begin{pmatrix} -2 \\ -11 \\ 5 \end{pmatrix}$ | M1A1 | M1: Attempt cross product of normal vectors (if method unclear, 2 components must be correct); A1: Correct vector |
| $x=0: (0, \frac{1}{2}, \frac{3}{2})$, $y=0: (-\frac{1}{11}, 0, \frac{19}{11})$, $z=0: (\frac{3}{5}, \frac{19}{5}, 0)$ | M1A1 | M1: Attempt point on line ($x$, $y$ and $z$); A1: Correct coordinates |
| $\mathbf{r} = \frac{1}{2}\mathbf{j} + \frac{3}{2}\mathbf{k} + \lambda(2\mathbf{i}+11\mathbf{j}-5\mathbf{k})$ | ddM1A1 | M1: Their point $+ \lambda$ their direction (dependent on both previous method marks); A1: Correct equation (oe) |
**(6 marks)**
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2\lambda - \left(\frac{1}{2}+11\lambda\right) + 2\left(\frac{3}{2}-5\lambda\right) = 31$ | M1 | Substitutes into the third plane and solves for $\lambda$ |
| $\lambda = \frac{-3}{2}$ | | |
| Planes intersect at $(-3, -16, 9)$ | M1A1 | M1: Substitutes into their line; A1: Correct coordinates |
**(3 marks) — Total: 9**8. The plane $\Pi _ { 1 }$ has vector equation $\mathbf { r }$. $\left( \begin{array} { l } 2 \\ 1 \\ 3 \end{array} \right) = 5$
The plane $\Pi _ { 2 }$ has vector equation $\mathbf { r } . \left( \begin{array} { r } - 1 \\ 2 \\ 4 \end{array} \right) = 7$
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$ where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors and $\lambda$ is a scalar parameter.
The plane $\Pi _ { 3 }$ has cartesian equation
$$x - y + 2 z = 31$$
\item Using your answer to part (a), or otherwise, find the coordinates of the point of intersection of the planes $\Pi _ { 1 } , \Pi _ { 2 }$ and $\Pi _ { 3 }$\\
\includegraphics[max width=\textwidth, alt={}, center]{393fd7be-c8f5-4b83-a5c7-2de04987a039-16_104_77_2469_1804}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2014 Q8 [9]}}