| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 12 |
| Topic | Vectors: Lines & Planes |
| Type | Plane containing line and point/vector |
| Difficulty | Standard +0.3 This is a standard three-part vectors question testing routine techniques: (i) substituting a point from the line into the plane equation, (ii) using perpendicular distance formula from point to plane, (iii) finding a plane through a line and external point. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Shortest distance is \( | m | \left |
**(i)**
Substituting $\begin{pmatrix}1+3\lambda\\-3+4\lambda\\2+6\lambda\end{pmatrix}$ into plane equation; i.e. $\begin{pmatrix}1+3\lambda\\-3+4\lambda\\2+6\lambda\end{pmatrix}\cdot\begin{pmatrix}2\\-6\\3\end{pmatrix} = k$ **M1**
**OR** any point on line (since "given")
$k = 2 + 6\lambda + 18 - 24\lambda + 6 + 18\lambda = 26$ **A1**
**(ii)**
Working with vector $\begin{pmatrix}10+2m\\2-6m\\3m-43\end{pmatrix}$ **B1**
Substituting into the plane equation: $\begin{pmatrix}10+2m\\2+6m\\3m-43\end{pmatrix}\cdot\begin{pmatrix}2\\-6\\3\end{pmatrix} = k$ **M1**
Solving a linear equation in $m$: $20 + 4m - 12 + 36m + 9m - 129 = 26$ **M1**
$m = 3 \Rightarrow Q = (16, -16, -34)$ **A1**
Shortest distance is $|m|\left|\begin{pmatrix}2\\-6\\3\end{pmatrix}\right| = 21$ or $PQ = \sqrt{6^2+18^2+9^2} = 21$ **A1**
**(iii)**
Finding 3 points in the plane: e.g. $A(1,-3,2)$, $B(4,1,8)$, $C(10,2,-43)$ **M1**
Then 2 vectors in (// to) plane: e.g. $\overrightarrow{AB} = \begin{pmatrix}3\\4\\6\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}9\\5\\-45\end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix}6\\1\\-51\end{pmatrix}$ **M1**
**OR B1 B1** for any two vectors in the plane
Vector product of any two of these to get normal to plane: $\begin{pmatrix}10\\-9\\1\end{pmatrix}$ **M1 A1**
(any non-zero multiple)
$d = \begin{pmatrix}10\\-9\\1\end{pmatrix}\cdot(\text{any position vector}) = \begin{pmatrix}10\\-9\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-3\\2\end{pmatrix}$ e.g. $= 39$ **M1 A1**
$\Rightarrow 10x - 9y + z = 39$ CAO (aef)
**ALTERNATE SOLUTION**
$ax + by + cz = d$ contains $\begin{pmatrix}1+3\lambda\\-3+4\lambda\\2+6\lambda\end{pmatrix}$ and $\begin{pmatrix}10\\2\\-43\end{pmatrix}$ **M1 B1**
… so $a + 3a\lambda + 4b\lambda - 3b + 2c + 6c\lambda = d$ and $10a + 2b - 43c = d$ **M1**
Then $a - 3b + 2c = d$ and $3a + 4b + 6c = 0$ ($\lambda$ terms) i.e. equating terms **M1**
Eliminating (e.g.) $c$ from 1st two equations $\Rightarrow 9a + 10b = 0$ **M1**
Choosing $a = 10$, $b = -9 \Rightarrow c = 1$ and $d = 39$ i.e. $10x - 9y + z = 39$ CAO **M1 A1**
**Total: 12 marks**10 The line $L$ has equation $\mathbf { r } = \left( \begin{array} { c } 1 \\ - 3 \\ 2 \end{array} \right) + \lambda \left( \begin{array} { l } 3 \\ 4 \\ 6 \end{array} \right)$ and the plane $\Pi$ has equation $\mathbf { r } \cdot \left( \begin{array} { c } 2 \\ - 6 \\ 3 \end{array} \right) = k$.\\
(i) Given that $L$ lies in $\Pi$, determine the value of $k$.\\
(ii) Find the coordinates of the point, $Q$, in $\Pi$ which is closest to $P ( 10,2 , - 43 )$. Deduce the shortest distance from $P$ to $\Pi$.\\
(iii) Find, in the form $a x + b y + c z = d$, where $a , b , c$ and $d$ are integers, an equation for the plane which contains both $L$ and $P$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q10 [12]}}