| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| 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, calculating angle between line and plane using dot product, and finding perpendicular distance. All parts follow textbook methods with no novel insight required, though it requires careful execution across multiple steps. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\mathbf{n} = (2\mathbf{j}-\mathbf{k}) \times (3\mathbf{i}+2\mathbf{j}+2\mathbf{k}) = 6\mathbf{i}-3\mathbf{j}-6\mathbf{k}\) o.a.e. (e.g. \(2\mathbf{i}-\mathbf{j}-2\mathbf{k}\)) | M1 A1 | (2 marks) |
| Answer | Marks |
|---|---|
| Answer/Working: Line \(l\) has direction \(2\mathbf{i}-2\mathbf{j}-\mathbf{k}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha = 90 - \beta = 63\) degrees to nearest degree | M1 A1lft, A1 awrt | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Plane \(P\) has equation \(\mathbf{r} \cdot (2\mathbf{i}-\mathbf{j}-2\mathbf{k}) = 1\) | M1 A1 | |
| Perpendicular distance is \(\frac{1-(-7)}{\sqrt{9}} = \frac{8}{3}\) | M1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Parallel plane through A has equation \(\mathbf{r} \cdot \frac{2\mathbf{i}-\mathbf{j}-2\mathbf{k}}{3} = -\frac{7}{3}\) | M1 A1 | |
| Plane P has equation \(\mathbf{r} \cdot \frac{2\mathbf{i}-\mathbf{j}-2\mathbf{k}}{3} = \frac{1}{3}\) | M1 | |
| So O lies between the two and perpendicular distance is \(\frac{1}{3} + \frac{7}{3} = \frac{8}{3}\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Distance A to \((3,1,2) = \sqrt{2^2+2^2+1^2} = 3\) | M1A1 | |
| Perpendicular distance is \('3' \sin\alpha = 3 \times \frac{8}{9} = \frac{8}{3}\) | M1A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Finding Cartesian equation of plane P: \(2x - y - 2z - 1 = 0\) | M1 A1 | |
| \(d = \frac{ | n_1\alpha + n_2\beta + n_3\gamma + d | }{\sqrt{n_1^2+n_2^2+n_3^2}} = \frac{ |
## Part (a)
**Answer/Working:** $\mathbf{n} = (2\mathbf{j}-\mathbf{k}) \times (3\mathbf{i}+2\mathbf{j}+2\mathbf{k}) = 6\mathbf{i}-3\mathbf{j}-6\mathbf{k}$ o.a.e. (e.g. $2\mathbf{i}-\mathbf{j}-2\mathbf{k}$) | **M1 A1** | (2 marks)
## Part (b)
**Answer/Working:** Line $l$ has direction $2\mathbf{i}-2\mathbf{j}-\mathbf{k}$ | **B1**
Angle between line $l$ and normal is given by $(\cos\beta \text{ or }\sin\alpha) = \frac{4+2+2}{\sqrt{9}\sqrt{9}} = \frac{8}{9}$
$\alpha = 90 - \beta = 63$ degrees to nearest degree | **M1 A1lft, A1 awrt** | (4 marks)
## Part (c) - Alt 1
**Answer/Working:** Plane $P$ has equation $\mathbf{r} \cdot (2\mathbf{i}-\mathbf{j}-2\mathbf{k}) = 1$ | **M1 A1**
Perpendicular distance is $\frac{1-(-7)}{\sqrt{9}} = \frac{8}{3}$ | **M1 A1** | (4 marks)
## Part (c) - Alt 2
**Answer/Working:** Parallel plane through A has equation $\mathbf{r} \cdot \frac{2\mathbf{i}-\mathbf{j}-2\mathbf{k}}{3} = -\frac{7}{3}$ | **M1 A1**
Plane P has equation $\mathbf{r} \cdot \frac{2\mathbf{i}-\mathbf{j}-2\mathbf{k}}{3} = \frac{1}{3}$ | **M1**
So O lies between the two and perpendicular distance is $\frac{1}{3} + \frac{7}{3} = \frac{8}{3}$ | **A1** | (4 marks)
## Part (c) - Alt 3
**Answer/Working:** Distance A to $(3,1,2) = \sqrt{2^2+2^2+1^2} = 3$ | **M1A1**
Perpendicular distance is $'3' \sin\alpha = 3 \times \frac{8}{9} = \frac{8}{3}$ | **M1A1** | (4 marks)
## Part (c) - Alt 4
**Answer/Working:** Finding Cartesian equation of plane P: $2x - y - 2z - 1 = 0$ | **M1 A1**
$d = \frac{|n_1\alpha + n_2\beta + n_3\gamma + d|}{\sqrt{n_1^2+n_2^2+n_3^2}} = \frac{|2(1)-1(3)-2(3)-1|}{\sqrt{2^2+1^2+2^2}} = \frac{8}{3}$ | **M1 A1** | (4 marks)
---The plane $P$ has equation
$$\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$$
\begin{enumerate}[label=(\alph*)]
\item Find a vector perpendicular to the plane $P$.
[2]
The line $l$ passes through the point $A(1, 3, 3)$ and meets $P$ at $(3, 1, 2)$.
The acute angle between the plane $P$ and the line $l$ is $\alpha$.
\item Find $\alpha$ to the nearest degree.
[4]
\item Find the perpendicular distance from $A$ to the plane $P$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2011 Q6 [10]}}