| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2019 |
| Session | Specimen |
| Marks | 8 |
| 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: (a) substituting a point from the line into the plane equation, (b) using perpendicular distance formula from point to plane, (c) 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 forms4.04j Shortest distance: between a point and a plane |
Finding 3 points in the plane: e.g. $A(1, -3, 2)$, $B(4, 1, 8)$, $C(10, 2, -43)$ [M1]
**OR** B1 for one vector in the plane
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 for another vector in the plane
Vector product of any two of these to get normal to plane: $\begin{pmatrix}10\\-9\\1\end{pmatrix}$ (any non-zero multiple) [M1A1]
$d = \begin{pmatrix}10\\-9\\1\end{pmatrix} \bullet \text{(any position vector)} = \begin{pmatrix}10\\-9\\1\end{pmatrix} \bullet \begin{pmatrix}1\\-3\\2\end{pmatrix}$ e.g. $= 39$ [M1]
$\Rightarrow 10x - 9y + z = 39$ CAO (aef) [A1]
**OR ALTERNATE SOLUTION** [M1B1]
$ax + by + cz = d$ contains $\begin{pmatrix}1+3\lambda\\-3+4\lambda\\2+6\lambda\end{pmatrix}$ and $\begin{pmatrix}10\\2\\-43\end{pmatrix}$
... so $a + 3a\lambda + 4b\lambda - 3b + 2c + 6c\lambda = d$ and $10a + 2b - 43c = d$
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 [M1A1]
**Available marks: 6**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 $\mathbf { r } \cdot \left( \begin{array} { c } 2 \\ - 6 \\ 3 \end{array} \right) = k$.
\begin{enumerate}[label=(\alph*)]
\item Given that $L$ lies in $\Pi$, determine the value of $k$.
\item Find the coordinates of the point, $Q$, in $\Pi$ which is closest to $P ( 10,2 , - 43 )$. Deduce the shortest distance from $P$ to $\Pi$.
\item 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 $P$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2019 Q10 [8]}}