| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Vector equation of a plane |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring application of matrix transformations to planes and conversion to Cartesian form. Students must transform the position vector and direction vectors, find the normal via cross product, then form the scalar product equation. While it involves multiple steps and FM content, each step follows standard procedures without requiring novel insight. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.04b Plane equations: cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2&0&3\\0&2&-1\\0&1&2\end{pmatrix}\begin{pmatrix}1+s+t\\-1+s+2t\\2&-2t\end{pmatrix}\) | M1 | Writes \(\Pi_i\) as a single vector |
| A1 | Correct statement | |
| \(= \begin{pmatrix}2+2s+2t+6-6t\\-2+2s+4t-2+2t\\-1+s+2t+4-4t\end{pmatrix}\) | M1 | Correct attempt to multiply |
| A1 | Correct vector in any form | |
| \(= \begin{pmatrix}8+2s-4t\\-4+2s+6t\\3+s-2t\end{pmatrix}\) | B1 | Correct simplified vector |
| \(\mathbf{r} = \begin{pmatrix}8\\-4\\3\end{pmatrix} + s\begin{pmatrix}2\\2\\1\end{pmatrix} + t\begin{pmatrix}-4\\6\\-2\end{pmatrix}\) | ||
| \(\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&2&1\\-4&6&-2\end{vmatrix} = -10\mathbf{i}+20\mathbf{k}\) | M1 | Attempts cross product of their direction vectors |
| A1 | Any multiple of \(-10\mathbf{i}+20\mathbf{k}\) | |
| \((8\mathbf{i}-4\mathbf{j}+3\mathbf{k})\cdot(\mathbf{i}-2\mathbf{k}) = 8-6\) | M1 | Attempt scalar product of their normal vector with their position vector |
| \(\mathbf{r}\cdot(\mathbf{i}-2\mathbf{k}) = 2\) | A1 | Correct equation (accept any correct equivalent, e.g. \(\mathbf{r}\cdot(-10\mathbf{i}+20\mathbf{k})=-20\)) |
| (9) |
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2&0&3\\0&2&-1\\0&1&2\end{pmatrix}\begin{pmatrix}1+s+t\\-1+s+2t\\2&-2t\end{pmatrix}$ | M1 | Writes $\Pi_i$ as a single vector |
| | A1 | Correct statement |
| $= \begin{pmatrix}2+2s+2t+6-6t\\-2+2s+4t-2+2t\\-1+s+2t+4-4t\end{pmatrix}$ | M1 | Correct attempt to multiply |
| | A1 | Correct vector in any form |
| $= \begin{pmatrix}8+2s-4t\\-4+2s+6t\\3+s-2t\end{pmatrix}$ | B1 | Correct simplified vector |
| $\mathbf{r} = \begin{pmatrix}8\\-4\\3\end{pmatrix} + s\begin{pmatrix}2\\2\\1\end{pmatrix} + t\begin{pmatrix}-4\\6\\-2\end{pmatrix}$ | | |
| $\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&2&1\\-4&6&-2\end{vmatrix} = -10\mathbf{i}+20\mathbf{k}$ | M1 | Attempts cross product of their direction vectors |
| | A1 | Any multiple of $-10\mathbf{i}+20\mathbf{k}$ |
| $(8\mathbf{i}-4\mathbf{j}+3\mathbf{k})\cdot(\mathbf{i}-2\mathbf{k}) = 8-6$ | M1 | Attempt scalar product of their normal vector with their position vector |
| $\mathbf{r}\cdot(\mathbf{i}-2\mathbf{k}) = 2$ | A1 | Correct equation (accept any correct equivalent, e.g. $\mathbf{r}\cdot(-10\mathbf{i}+20\mathbf{k})=-20$) |
| | **(9)** | |
---\begin{enumerate}
\item The plane $\Pi _ { 1 }$ has vector equation
\end{enumerate}
$$\mathbf { r } = \left( \begin{array} { r }
1 \\
- 1 \\
2
\end{array} \right) + s \left( \begin{array} { l }
1 \\
1 \\
0
\end{array} \right) + t \left( \begin{array} { r }
1 \\
2 \\
- 2
\end{array} \right) ,$$
where $s$ and $t$ are real parameters.
The plane $\Pi _ { 1 }$ is transformed to the plane $\Pi _ { 2 }$ by the transformation represented by the matrix $\mathbf { T }$, where
$$\mathbf { T } = \left( \begin{array} { r r r }
2 & 0 & 3 \\
0 & 2 & - 1 \\
0 & 1 & 2
\end{array} \right)$$
Find an equation of the plane $\Pi _ { 2 }$ in the form r.n=p\\
\hfill \mbox{\textit{Edexcel FP3 2013 Q4 [9]}}