| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a standard Further Maths question on skew lines requiring the cross product formula for shortest distance and a plane-line angle calculation. While it involves multiple steps and vector techniques beyond A-level Pure, it follows a well-established method that Further Maths students practice regularly, making it moderately above average difficulty. |
| Spec | 4.04d Angles: between planes and between line and plane4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}2\\4\\5\end{pmatrix}\), \(\overrightarrow{CD} = \begin{pmatrix}1\\1\\0\end{pmatrix}\) | B1 | Find the directions of the lines |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\2 & 4 & 5\\1 & 1 & 0\end{vmatrix} = \begin{pmatrix}-5\\5\\-2\end{pmatrix}\) | M1 A1 | Finds direction of common perpendicular. Allow any non-zero scalar multiple |
| \(\frac{\begin{pmatrix}1\\5\\1\end{pmatrix}\cdot\begin{pmatrix}-5\\5\\-2\end{pmatrix}}{\sqrt{5^2+5^2+2^2}} = \frac{18}{\sqrt{54}} = \sqrt{6} = 2.45\) | M1 A1 | Uses formula for shortest distance |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\2 & 4 & 5\\2 & 6 & 1\end{vmatrix} = \begin{pmatrix}-26\\8\\4\end{pmatrix} \sim \begin{pmatrix}-13\\4\\2\end{pmatrix}\) | M1 A1 | Finds normal to the plane |
| \(\frac{\begin{pmatrix}1\\1\\0\end{pmatrix}\cdot\begin{pmatrix}-13\\4\\2\end{pmatrix}}{\sqrt{1^2+1^2+0^2}\sqrt{13^2+4^2+2^2}} = -\frac{9}{\sqrt{378}} = -\frac{\sqrt{42}}{14}\) | M1 A1 | FT Uses dot product of their normal with direction of line |
| \(\cos^{-1}\left(-\frac{\sqrt{42}}{14}\right) - 90 = 27.6°\) (or \(0.481\) rad) | A1 | |
| Total: 5 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}2\\4\\5\end{pmatrix}$, $\overrightarrow{CD} = \begin{pmatrix}1\\1\\0\end{pmatrix}$ | B1 | Find the directions of the lines |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\2 & 4 & 5\\1 & 1 & 0\end{vmatrix} = \begin{pmatrix}-5\\5\\-2\end{pmatrix}$ | M1 A1 | Finds direction of common perpendicular. Allow any non-zero scalar multiple |
| $\frac{\begin{pmatrix}1\\5\\1\end{pmatrix}\cdot\begin{pmatrix}-5\\5\\-2\end{pmatrix}}{\sqrt{5^2+5^2+2^2}} = \frac{18}{\sqrt{54}} = \sqrt{6} = 2.45$ | M1 A1 | Uses formula for shortest distance |
| **Total: 5** | | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\2 & 4 & 5\\2 & 6 & 1\end{vmatrix} = \begin{pmatrix}-26\\8\\4\end{pmatrix} \sim \begin{pmatrix}-13\\4\\2\end{pmatrix}$ | M1 A1 | Finds normal to the plane |
| $\frac{\begin{pmatrix}1\\1\\0\end{pmatrix}\cdot\begin{pmatrix}-13\\4\\2\end{pmatrix}}{\sqrt{1^2+1^2+0^2}\sqrt{13^2+4^2+2^2}} = -\frac{9}{\sqrt{378}} = -\frac{\sqrt{42}}{14}$ | M1 A1 | FT Uses dot product of their normal with direction of line |
| $\cos^{-1}\left(-\frac{\sqrt{42}}{14}\right) - 90 = 27.6°$ (or $0.481$ rad) | A1 | |
| **Total: 5** | | |7 The line $l _ { 1 }$ passes through the points $A ( - 3,1,4 )$ and $B ( - 1,5,9 )$. The line $l _ { 2 }$ passes through the points $C ( - 2,6,5 )$ and $D ( - 1,7,5 )$.\\
(i) Find the shortest distance between the lines $l _ { 1 }$ and $l _ { 2 }$.\\
(ii) Find the acute angle between the line $l _ { 2 }$ and the plane containing $A , B$ and $D$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q7 [10]}}