| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a straightforward vectors question requiring standard techniques: (i) substitute the line equation into the plane equation to verify it lies in the plane, and (ii) find the perpendicular from P to the plane using the normal vector. Both parts are routine applications of well-practiced 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 point |
(i) $\begin{bmatrix} 2+5\lambda \\ -5-2\lambda \\ 7+3\lambda \end{bmatrix}$ substd. into $x + 4y + z + 11 = 0$ **M1**
$\Rightarrow 2 + 5\lambda - 20 - 8\lambda + 7 + 3\lambda + 11 = 0$ (for all $\lambda$) Shown correct **A1** [2]
ALTs: 2 points of $l$ shown to be in $\Pi$ **or** $\Pi$'s nml. perpr. to $l$ and 1 point of $l$ in $\Pi$
(ii) $\overrightarrow{PR} = \begin{bmatrix} 1+5\lambda \\ -7-2\lambda \\ 7+3\lambda-k \end{bmatrix}$ Setting their $\overrightarrow{PR} = m\begin{bmatrix} 1 \\ 4 \\ 1 \end{bmatrix}$ **B1 M1**
Solving for $\lambda$ in **i** & **j** components: $4(1+5\lambda) = -7-2\lambda \Rightarrow \lambda = -\frac{1}{2}$ **M1 A1**
Substituting back into **i** & **k** components to find $k$: $1 + 5\left(-\frac{1}{2}\right) = 7 + 3\left(-\frac{1}{2}\right) - k \Rightarrow k = 7$ **M1 A1** [6]
**ALTERNATIVE**
$\begin{bmatrix} 1 \\ -7 \\ 7-k \end{bmatrix} \cdot \left(\begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix} \times \begin{bmatrix} 1 \\ 4 \\ 1 \end{bmatrix}\right) = 0 \Rightarrow \begin{bmatrix} 1 \\ -7 \\ 7-k \end{bmatrix} \cdot \begin{bmatrix} -14 \\ -2 \\ 22 \end{bmatrix} = 0 \Rightarrow -14 + 14 + 22(7-k) = 0$ **M2 M1 A1 A1**
$\Rightarrow k = 7$ **A1**8 (i) Show that the line $l$ with vector equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ - 5 \\ 7 \end{array} \right) + \lambda \left( \begin{array} { r } 5 \\ - 2 \\ 3 \end{array} \right)$ lies in the plane $\Pi$ with cartesian equation $x + 4 y + z + 11 = 0$.\\
(ii) The plane $\Pi$ is horizontal, and the point $P ( 1,2 , k )$ is above it. Given that the point in $\Pi$ which is directly beneath $P$ is on the line $l$, determine the value of $k$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q8 [6]}}