| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 14 |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths vectors question requiring: (i) deriving the shortest distance formula between skew lines using cross product and parametric equations, (ii) solving for when distance equals zero (intersection condition), and (iii) computing a plane equation and point-to-plane distance. The variable parameter t adds complexity, requiring trigonometric manipulation. While the techniques are standard for Further Maths, the multi-stage reasoning and algebraic manipulation place it well above average difficulty. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| \(PQ = \frac{ | (\mathbf{i}-\mathbf{j})\cdot\{(2\sin t)\mathbf{i} + \mathbf{j} + 4\mathbf{k}\} | }{\sqrt{4\sin^2 t + 17}}\) depM1A1 |
| \(PQ = \frac{ | 2\sin t - 1 | }{\sqrt{4\sin^2 t + 17}}\) A1 [5] |
**(i)** $\{\mathbf{i} - (2\sin t)\mathbf{j}\} \times \{4\mathbf{j} - \mathbf{k}\} = (2\sin t)\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ M1A1
$PQ = \frac{|(\mathbf{i}-\mathbf{j})\cdot\{(2\sin t)\mathbf{i} + \mathbf{j} + 4\mathbf{k}\}|}{\sqrt{4\sin^2 t + 17}}$ depM1A1
$PQ = \frac{|2\sin t - 1|}{\sqrt{4\sin^2 t + 17}}$ A1 **[5]**
**(ii)** $PQ = 0 \Rightarrow \sin t = 0.5$
$t = \frac{1}{6}\pi, \frac{5}{6}\pi$ (Accept 0.524 and 2.62) M1, A1A1 **[3]**
**(iii)** Obtains normal vector to plane $BPQ$
$(\sqrt{2}\mathbf{i} + \mathbf{j} + 4\mathbf{k}) \times (\mathbf{i} - \sqrt{2}\mathbf{j}) = 4\sqrt{2}\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$ M1A1
EITHER: $\overrightarrow{BA} = \mathbf{i} - \mathbf{j}$ B1
Perp from $A$ to $BPQ = \frac{(4\sqrt{2}\mathbf{i} + 4\mathbf{j} - 3\mathbf{k})\cdot(\mathbf{i}-\mathbf{j})}{\sqrt{57}}$ M1A1
$= 0.219$ A1 **[6]**
OR: Plane $BPQ$ is $4\sqrt{2}x + 4y - 3z = 4(\sqrt{2}-1)$
Uses distance formula with point $A(2, 1, 4)$ and this plane M1A1A1
Perp from $A$ to $BPQ = \frac{8\sqrt{2} + 4 - 12 + 4 - 4\sqrt{2}}{\sqrt{32+16+9}}$
$= 0.219$ A1 **(6)**10 The line $l _ { 1 }$ is parallel to the vector $4 \mathbf { j } - \mathbf { k }$ and passes through the point $A$ whose position vector is $2 \mathbf { i } + \mathbf { j } + 4 \mathbf { k }$. The variable line $l _ { 2 }$ is parallel to the vector $\mathbf { i } - ( 2 \sin t ) \mathbf { j }$, where $0 \leqslant t < 2 \pi$, and passes through the point $B$ whose position vector is $\mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k }$. The points $P$ and $Q$ are on $l _ { 1 }$ and $l _ { 2 }$ respectively, and $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$.\\
(i) Find the length of $P Q$ in terms of $t$.\\
(ii) Hence find the values of $t$ for which $l _ { 1 }$ and $l _ { 2 }$ intersect.\\
(iii) For the case $t = \frac { 1 } { 4 } \pi$, find the perpendicular distance from $A$ to the plane $B P Q$, giving your answer correct to 3 decimal places.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q10 [14]}}