| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.2 This is a standard Further Maths vectors question testing the formula for shortest distance between skew lines, finding a plane equation from a line and parallel vector, and angle between line and plane. All parts use routine techniques with clear methods, though part (a) requires careful application of the cross product formula. The multi-part structure and Further Maths content places it above average difficulty, but it's a textbook-style question without requiring novel insight. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d 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 |
| \(\begin{pmatrix}3\\-5\\-6\end{pmatrix} - \begin{pmatrix}3\\0\\3\end{pmatrix} = \begin{pmatrix}0\\-5\\-9\end{pmatrix}\) | B1 | |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 4 & 4\\0 & 5 & 6\end{vmatrix} = \begin{pmatrix}4\\-6\\5\end{pmatrix}\) | M1 A1 | |
| \(\frac{1}{\sqrt{77}}\begin{pmatrix}0\\-5\\-9\end{pmatrix} \cdot \begin{pmatrix}4\\-6\\5\end{pmatrix} = \frac{15}{\sqrt{77}} = 1.71\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 4 & 4\\1 & 0 & 1\end{vmatrix} = \begin{pmatrix}4\\3\\-4\end{pmatrix}\) | M1 A1 | |
| \(4(3)+3(0)-4(3)=0 \Rightarrow 4x+3y-4z=0\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0\\5\\6\end{pmatrix}\cdot\begin{pmatrix}4\\3\\-4\end{pmatrix} = \sqrt{61}\sqrt{41}\cos\alpha \Rightarrow \cos\alpha = \frac{-9}{\sqrt{61}\sqrt{41}}\) | M1 A1FT | A1 FT their normal |
| Acute angle between \(l_2\) and \(\Pi\) is \(\alpha - 90 = 10.4°\) | A1 |
## Question 5:
### Part 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}3\\-5\\-6\end{pmatrix} - \begin{pmatrix}3\\0\\3\end{pmatrix} = \begin{pmatrix}0\\-5\\-9\end{pmatrix}$ | B1 | |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 4 & 4\\0 & 5 & 6\end{vmatrix} = \begin{pmatrix}4\\-6\\5\end{pmatrix}$ | M1 A1 | |
| $\frac{1}{\sqrt{77}}\begin{pmatrix}0\\-5\\-9\end{pmatrix} \cdot \begin{pmatrix}4\\-6\\5\end{pmatrix} = \frac{15}{\sqrt{77}} = 1.71$ | M1 A1 | |
### Part 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 4 & 4\\1 & 0 & 1\end{vmatrix} = \begin{pmatrix}4\\3\\-4\end{pmatrix}$ | M1 A1 | |
| $4(3)+3(0)-4(3)=0 \Rightarrow 4x+3y-4z=0$ | M1 A1 | |
### Part 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0\\5\\6\end{pmatrix}\cdot\begin{pmatrix}4\\3\\-4\end{pmatrix} = \sqrt{61}\sqrt{41}\cos\alpha \Rightarrow \cos\alpha = \frac{-9}{\sqrt{61}\sqrt{41}}$ | M1 A1FT | A1 FT their normal |
| Acute angle between $l_2$ and $\Pi$ is $\alpha - 90 = 10.4°$ | A1 | |
---5 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = 3 \mathbf { i } + 3 \mathbf { k } + \lambda ( \mathbf { i } + 4 \mathbf { j } + 4 \mathbf { k } )$ and $\mathbf { r } = 3 \mathbf { i } - 5 \mathbf { j } - 6 \mathbf { k } + \mu ( 5 \mathbf { j } + 6 \mathbf { k } )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.\\
The plane $\Pi$ contains $l _ { 1 }$ and is parallel to the vector $\mathbf { i } + \mathbf { k }$.
\item Find the equation of $\Pi$, giving your answer in the form $a x + b y + c z = d$.
\item Find the acute angle between $l _ { 2 }$ and $\Pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q5 [12]}}