| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Challenging +1.2 This is a standard Further Maths vectors question requiring multiple techniques (cross product for normal, scalar product form, parametric line-plane intersection, perpendicular distance formula) but follows a predictable structure with clear signposting. Part (b) being a 'show that' reduces difficulty. The perpendicular distance calculation in (c) is routine application of the formula once the setup is complete. Slightly above average due to the multi-step nature and Further Maths content, but no novel insight required. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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 of direction vectors |
| \(\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} = 5\) | M1 A1 | Dot product with point on plane |
| \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5\) | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation of \(l\): \(\mathbf{r} = \begin{pmatrix} 6 \\ 13 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}\) | M1 | Line through given point with normal direction |
| At intersection: \(\begin{pmatrix} 6+t \\ 13+4t \\ 5+2t \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5\) | M1 | Substituting into plane equation |
| \(\Rightarrow 6+t+4(13+4t)+2(5+2t)=5 \Rightarrow t=-3\) | M1 | Solving for \(t\) |
| N is \((3, 1, -1)\) ✱ | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{PN} \cdot \overrightarrow{PR} = (-3\mathbf{i}-12\mathbf{j}-6\mathbf{k})\cdot(-5\mathbf{i}-13\mathbf{j}-3\mathbf{k}) = 189\) | M1 A1ft | ft from N |
| \(\sqrt{9+144+36}\sqrt{25+169+9}\cos NPR = 189\) | A1 | |
| \(NX = NP\sin NPR = \sqrt{189}\sin NPR = 3.61\) | M1 A1 | (5 marks total, 14 overall) |
## Question 7:
### Part (a):
| Answer/Working | 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 of direction vectors |
| $\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} = 5$ | M1 A1 | Dot product with point on plane |
| $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5$ | | **(5 marks)** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation of $l$: $\mathbf{r} = \begin{pmatrix} 6 \\ 13 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}$ | M1 | Line through given point with normal direction |
| At intersection: $\begin{pmatrix} 6+t \\ 13+4t \\ 5+2t \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5$ | M1 | Substituting into plane equation |
| $\Rightarrow 6+t+4(13+4t)+2(5+2t)=5 \Rightarrow t=-3$ | M1 | Solving for $t$ |
| N is $(3, 1, -1)$ ✱ | A1 | **(4 marks)** |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PN} \cdot \overrightarrow{PR} = (-3\mathbf{i}-12\mathbf{j}-6\mathbf{k})\cdot(-5\mathbf{i}-13\mathbf{j}-3\mathbf{k}) = 189$ | M1 A1ft | ft from N |
| $\sqrt{9+144+36}\sqrt{25+169+9}\cos NPR = 189$ | A1 | |
| $NX = NP\sin NPR = \sqrt{189}\sin NPR = 3.61$ | M1 A1 | **(5 marks total, 14 overall)** |
---7. The plane $\Pi$ has vector equation
$$\mathbf { r } = 3 \mathbf { i } + \mathbf { k } + \lambda ( - 4 \mathbf { i } + \mathbf { j } ) + \mu ( 6 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } )$$
\begin{enumerate}[label=(\alph*)]
\item Find an equation of $\Pi$ in the form $\mathbf { r } \cdot \mathbf { n } = p$, 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 $N$.
\item Show that the coordinates of $N$ are $( 3,1 , - 1 )$.
The point $R$ lies on $\Pi$ and has coordinates $( 1,0,2 )$.
\item Find the perpendicular distance from $N$ to the line $P R$. Give your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2010 Q7 [14]}}