| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring finding a plane from two lines (using cross product of direction vectors), then standard applications (intersection, perpendicular distance, angle). While it involves several steps and FM content, each technique is routine and methodical with no novel insight required. The computational load and FM context place it above average difficulty but well within standard textbook territory. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&-1\\1&-3&4\end{vmatrix} = \mathbf{i}-9\mathbf{j}-7\mathbf{k}\) | M1A1 | Finds normal to plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\Pi:\ x-9y-7z = \text{constant}\) | ||
| Sub \((1,-1,2) \Rightarrow\) constant \(= -4\) | ||
| \(\Pi:\ x-9y-7z=-4\) | M1A1 | 4 marks |
| \(l:\ x=6+2\lambda,\ y=-2+\lambda,\ z=1-4\lambda\) | General point on line | |
| Sub in \(\Pi \Rightarrow 6+2\lambda+18-9\lambda-7+28\lambda=-4\) | M1 | |
| \(\Rightarrow \lambda=-1\) | A1 | |
| Position vector of intersection is \(4\mathbf{i}-3\mathbf{j}+5\mathbf{k}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(\left\ | \frac{6+18-7+4}{\sqrt{1+81+49}}\right | \) or \(\frac{(2\mathbf{i}+\mathbf{j}-4\mathbf{k})\cdot(\mathbf{i}-9\mathbf{j}-7\mathbf{k})}{\sqrt{1+81+49}}\) |
| \(= \frac{21}{\sqrt{131}}\ (=1.83)\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2\mathbf{i}+\mathbf{j}-4\mathbf{k})\cdot(\mathbf{i}-9\mathbf{j}-7\mathbf{k})=21\) | M1 | Scalar product |
| \(= \sqrt{4+1+16}\sqrt{1+81+49}\sin\theta\) | A1 | |
| \(\sin\theta = \sqrt{\frac{21}{131}} \Rightarrow \theta = 23.6°\) or \(0.412\) rad | A1 | 3 marks, total [13] |
# Question 9:
## Vectors — Plane and Line
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&-1\\1&-3&4\end{vmatrix} = \mathbf{i}-9\mathbf{j}-7\mathbf{k}$ | M1A1 | Finds normal to plane |
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Pi:\ x-9y-7z = \text{constant}$ | | |
| Sub $(1,-1,2) \Rightarrow$ constant $= -4$ | | |
| $\Pi:\ x-9y-7z=-4$ | M1A1 | 4 marks |
| $l:\ x=6+2\lambda,\ y=-2+\lambda,\ z=1-4\lambda$ | | General point on line |
| Sub in $\Pi \Rightarrow 6+2\lambda+18-9\lambda-7+28\lambda=-4$ | M1 | |
| $\Rightarrow \lambda=-1$ | A1 | |
| Position vector of intersection is $4\mathbf{i}-3\mathbf{j}+5\mathbf{k}$ | A1 | 3 marks |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $\left\|\frac{6+18-7+4}{\sqrt{1+81+49}}\right|$ or $\frac{(2\mathbf{i}+\mathbf{j}-4\mathbf{k})\cdot(\mathbf{i}-9\mathbf{j}-7\mathbf{k})}{\sqrt{1+81+49}}$ | M1A1 | Distance of point from plane |
| $= \frac{21}{\sqrt{131}}\ (=1.83)$ | A1 | 3 marks |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2\mathbf{i}+\mathbf{j}-4\mathbf{k})\cdot(\mathbf{i}-9\mathbf{j}-7\mathbf{k})=21$ | M1 | Scalar product |
| $= \sqrt{4+1+16}\sqrt{1+81+49}\sin\theta$ | A1 | |
| $\sin\theta = \sqrt{\frac{21}{131}} \Rightarrow \theta = 23.6°$ or $0.412$ rad | A1 | 3 marks, total [13] |
---9 Find a cartesian equation of the plane $\Pi$ containing the lines
$$\mathbf { r } = 3 \mathbf { i } + \mathbf { k } + s ( 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = 3 \mathbf { i } - 7 \mathbf { j } + 10 \mathbf { k } + t ( \mathbf { i } - 3 \mathbf { j } + 4 \mathbf { k } )$$
The line $l$ passes through the point $P$ with position vector $6 \mathbf { i } - 2 \mathbf { j } + \mathbf { k }$ and is parallel to the vector $2 \mathbf { i } + \mathbf { j } - 4 \mathbf { k }$. Find\\
(i) the position vector of the point where $l$ meets $\Pi$,\\
(ii) the perpendicular distance from $P$ to $\Pi$,\\
(iii) the acute angle between $l$ and $\Pi$.
\hfill \mbox{\textit{CAIE FP1 2011 Q9 [13]}}