| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.8 This is a multi-part Further Maths vectors question requiring: (a) finding a normal vector via cross product and converting plane equations, (b) finding line-plane intersection (shown result), and (c) computing perpendicular distance from point to line using vector methods. While systematic, it demands multiple techniques, careful vector manipulation, and is from F3 (Further Pure 3), making it moderately challenging but still procedural for Further Maths students. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 0 \\ 6 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}\) | M1, A2(1,0) | Cross product calculation |
| \(\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \bullet \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} = 5\) | M1A1 | Dot product with point to find constant |
| \(\mathbf{r} \bullet \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5\) | Final plane equation (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Equation of \(l\): \(\mathbf{r} = \begin{pmatrix} 6 \\ 13 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}\) | M1 | |
| At intersection: \(\begin{pmatrix} 6+t \\ 13+4t \\ 5+2t \end{pmatrix} \bullet \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5\) | M1 | |
| \(\Rightarrow 6+t+4(13+4t)+2(5+2t)=5 \Rightarrow t=-3\) | M1 | |
| N is \((3, 1, -1)\) ✱ | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{PN} \bullet \overrightarrow{PR} = (-3\mathbf{i}-12\mathbf{j}-6\mathbf{k})\bullet(-5\mathbf{i}-13\mathbf{j}-3\mathbf{k}) = 189\) | M1 A1ft | |
| \(\sqrt{9+144+36}\sqrt{25+169+9}\cos NPR = 189\) | A1 | |
| \(NX = NP\sin NPR = \sqrt{189}\sin NPR = 3.61\) | M1A1 | (5 marks) |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 0 \\ 6 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}$ | M1, A2(1,0) | Cross product calculation |
| $\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \bullet \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} = 5$ | M1A1 | Dot product with point to find constant |
| $\mathbf{r} \bullet \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5$ | | Final plane equation **(5 marks)** |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Equation of $l$: $\mathbf{r} = \begin{pmatrix} 6 \\ 13 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}$ | M1 | |
| At intersection: $\begin{pmatrix} 6+t \\ 13+4t \\ 5+2t \end{pmatrix} \bullet \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5$ | M1 | |
| $\Rightarrow 6+t+4(13+4t)+2(5+2t)=5 \Rightarrow t=-3$ | M1 | |
| N is $(3, 1, -1)$ ✱ | A1 | **(4 marks)** |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PN} \bullet \overrightarrow{PR} = (-3\mathbf{i}-12\mathbf{j}-6\mathbf{k})\bullet(-5\mathbf{i}-13\mathbf{j}-3\mathbf{k}) = 189$ | M1 A1ft | |
| $\sqrt{9+144+36}\sqrt{25+169+9}\cos NPR = 189$ | A1 | |
| $NX = NP\sin NPR = \sqrt{189}\sin NPR = 3.61$ | M1A1 | **(5 marks)** |
---\begin{enumerate}
\item The plane $\Pi$ has vector equation
\end{enumerate}
$$\mathbf { r } = 3 \mathbf { i } + \mathbf { k } + \lambda ( - 4 \mathbf { i } + \mathbf { j } ) + \mu ( 6 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } )$$
(a) Find an equation of $\Pi$ in the form $\mathbf { r } . \mathbf { n } = p$, where $\mathbf { n }$ is a vector perpendicular to $\Pi$ and $p$ is a constant.
The point $P$ has coordinates $( 6,13,5 )$. The line $l$ passes through $P$ and is perpendicular to $\Pi$. The line $l$ intersects $\Pi$ at the point $N$.\\
(b) Show that the coordinates of $N$ are $( 3,1 , - 1 )$.
The point $R$ lies on $\Pi$ and has coordinates $( 1,0,2 )$.\\
(c) Find the perpendicular distance from $N$ to the line $P R$. Give your answer to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel F3 Q7 [14]}}