| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.8 This is a multi-part Further Maths question requiring: (a) finding shortest distance between skew lines using the scalar triple product formula and solving for parameter m, (b) point-to-line distance using cross product, (c) angle between planes using normal vectors. While it involves several standard Further Maths techniques, the execution is relatively straightforward once the formulas are known, though the multi-step nature and computational complexity elevate it above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04d Angles: between planes and between line and plane4.04e Line intersections: parallel, skew, or intersecting4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\4 & -1 & 1\\0 & 1 & m-1\end{vmatrix} = -m\mathbf{i} - 4(m-1)\mathbf{j} + 4\mathbf{k}\) | M1A1 | 2 marks |
| \(\sqrt{(m^2 + 16(1-2m+m^2)+16)} = 3\) | M1A1 | 2 marks |
| \(19m^2 - 40m + 4 = 0\) | M1A1 | 2 marks |
| \((19m-2)(m-2) = 0\), \(m=2\) since \(m\) is an integer | A1 | 1 mark |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{CA} = \begin{pmatrix}0\\1\\-4\end{pmatrix}\) and \(\overrightarrow{AD} = \begin{pmatrix}-1\\1\\5\end{pmatrix}\) | B1 | 1 mark |
| \(\frac{1}{\sqrt{17}}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\0 & 1 & -4\\1 & 0 & -4\end{vmatrix} = \frac{1}{\sqrt{17}}\begin{pmatrix}-4\\1\\-1\end{pmatrix}\) | M1 | 1 mark |
| \(\frac{1}{\sqrt{17}}\sqrt{4^2+1^2+2} = \sqrt{\frac{18}{17}}\ (=1.03)\) | A1 | 1 mark |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{BC} = 3\mathbf{i} - \mathbf{j} + 5\mathbf{k}\) | B1 | 1 mark |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\3 & -1 & 5\\0 & 1 & 1\end{vmatrix} = -6\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}\) | M1 | 1 mark |
| \(\cos\theta = \frac{(4\mathbf{i}-\mathbf{j}+\mathbf{k})\cdot(2\mathbf{i}+\mathbf{j}-\mathbf{k})}{\sqrt{16+1+1+1}\sqrt{4+1+1+1}} = \frac{6}{\sqrt{18}\sqrt{6}} = \frac{1}{\sqrt{3}}\) | M1 | 1 mark |
| Angle between planes \(= \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\) | A1 | 1 mark |
| 4 |
## Question 6(a):
| $\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\4 & -1 & 1\\0 & 1 & m-1\end{vmatrix} = -m\mathbf{i} - 4(m-1)\mathbf{j} + 4\mathbf{k}$ | M1A1 | 2 marks | Finds direction of common perpendicular |
| $\sqrt{(m^2 + 16(1-2m+m^2)+16)} = 3$ | M1A1 | 2 marks | Uses result for shortest distance between lines |
| $19m^2 - 40m + 4 = 0$ | M1A1 | 2 marks | AG — Dependent on method seen for solving quadratic equation |
| $(19m-2)(m-2) = 0$, $m=2$ since $m$ is an integer | A1 | 1 mark | |
| | | **7** |
## Question 6(b):
| $\overrightarrow{CA} = \begin{pmatrix}0\\1\\-4\end{pmatrix}$ and $\overrightarrow{AD} = \begin{pmatrix}-1\\1\\5\end{pmatrix}$ | B1 | 1 mark | Finds relevant vectors |
| $\frac{1}{\sqrt{17}}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\0 & 1 & -4\\1 & 0 & -4\end{vmatrix} = \frac{1}{\sqrt{17}}\begin{pmatrix}-4\\1\\-1\end{pmatrix}$ | M1 | 1 mark | Uses cross-product; unit vector parallel to $\overrightarrow{CA}$ is required |
| $\frac{1}{\sqrt{17}}\sqrt{4^2+1^2+2} = \sqrt{\frac{18}{17}}\ (=1.03)$ | A1 | 1 mark | |
| | | **3** |
## Question 6(c):
| $\overrightarrow{BC} = 3\mathbf{i} - \mathbf{j} + 5\mathbf{k}$ | B1 | 1 mark | Finds second vector in $BCD$ |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\3 & -1 & 5\\0 & 1 & 1\end{vmatrix} = -6\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}$ | M1 | 1 mark | Finds normal vector to $BCD$ |
| $\cos\theta = \frac{(4\mathbf{i}-\mathbf{j}+\mathbf{k})\cdot(2\mathbf{i}+\mathbf{j}-\mathbf{k})}{\sqrt{16+1+1+1}\sqrt{4+1+1+1}} = \frac{6}{\sqrt{18}\sqrt{6}} = \frac{1}{\sqrt{3}}$ | M1 | 1 mark | Finds angle between normal vectors |
| Angle between planes $= \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$ | A1 | 1 mark | AG |
| | | **4** |
---6 The position vectors of the points $A , B , C , D$ are
$$2 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k } , \quad - 2 \mathbf { i } + 5 \mathbf { j } - 4 \mathbf { k } , \quad \mathbf { i } + 4 \mathbf { j } + \mathbf { k } , \quad \mathbf { i } + 5 \mathbf { j } + m \mathbf { k } ,$$
respectively, where $m$ is an integer. It is given that the shortest distance between the line through $A$ and $B$ and the line through $C$ and $D$ is 3 .\\
(a) Show that the only possible value of $m$ is 2 .\\
(b) Find the shortest distance of $D$ from the line through $A$ and $C$.\\
(c) Show that the acute angle between the planes $A C D$ and $B C D$ is $\cos ^ { - 1 } \left( \frac { 1 } { \sqrt { 3 } } \right)$.\\
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q6 [14]}}