| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| 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 vector question testing the skew lines distance formula, finding a plane containing one line parallel to another, and using intersection properties. Part (a) is direct application of the formula d = |(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|/|\mathbf{b}_1 \times \mathbf{b}_2|. Part (b) requires finding a normal via cross product of the two direction vectors. Part (c) uses the given line of intersection to find the normal to Π₂ via cross product with Π₁'s normal, then uses the point (1,1,1). While multi-step, each part follows standard procedures taught in Further Maths with no novel insight required, making it moderately above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-3\\4\\0\end{pmatrix} - \begin{pmatrix}1\\4\\-1\end{pmatrix} = \begin{pmatrix}-4\\0\\1\end{pmatrix}\) | B1 | OE. Finds direction of one line to another. |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\0 & 1 & -2\\1 & 2 & 1\end{vmatrix} = \begin{pmatrix}5\\-2\\-1\end{pmatrix}\) | M1 A1 | Find common perpendicular. |
| \(\frac{1}{\sqrt{30}}\left | \begin{pmatrix}-4\\0\\1\end{pmatrix}\cdot\begin{pmatrix}5\\-2\\-1\end{pmatrix}\right | = \frac{21}{\sqrt{30}} (= 3.83)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5(1) - 2(4) - 1(-1) = -2 \Rightarrow 5x - 2y - z = -2\) | M1 | Uses point in the plane. |
| A1 FT | Follow through their normal in part 5(a). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\mathbf{i} - \mathbf{j} + \mathbf{k}\) | B1 | Finding direction between the two given points |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 1 \\ 1 & 4 & -3 \end{vmatrix} = \begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix}\) | M1 | Find common perpendicular. Allow use of their normal from part (b) |
| \(x + 2y + 3z = 6\) | A1 | OE |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-3\\4\\0\end{pmatrix} - \begin{pmatrix}1\\4\\-1\end{pmatrix} = \begin{pmatrix}-4\\0\\1\end{pmatrix}$ | **B1** | OE. Finds direction of one line to another. |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\0 & 1 & -2\\1 & 2 & 1\end{vmatrix} = \begin{pmatrix}5\\-2\\-1\end{pmatrix}$ | **M1 A1** | Find common perpendicular. |
| $\frac{1}{\sqrt{30}}\left|\begin{pmatrix}-4\\0\\1\end{pmatrix}\cdot\begin{pmatrix}5\\-2\\-1\end{pmatrix}\right| = \frac{21}{\sqrt{30}} (= 3.83)$ | **M1 A1** | Uses formula for shortest distance. |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5(1) - 2(4) - 1(-1) = -2 \Rightarrow 5x - 2y - z = -2$ | **M1** | Uses point in the plane. |
| | **A1 FT** | Follow through their normal in part 5(a). |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\mathbf{i} - \mathbf{j} + \mathbf{k}$ | B1 | Finding direction between the two given points |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 1 \\ 1 & 4 & -3 \end{vmatrix} = \begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix}$ | M1 | Find common perpendicular. Allow use of their normal from part (b) |
| $x + 2y + 3z = 6$ | A1 | OE |
---5 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = \mathbf { i } + 4 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { j } - 2 \mathbf { k } )$ and $\mathbf { r } = - 3 \mathbf { i } + 4 \mathbf { j } + \mu ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.\\
The plane $\Pi _ { 1 }$ contains $l _ { 1 }$ and is parallel to $l _ { 2 }$.
\item Obtain an equation of $\Pi _ { 1 }$ in the form $p x + q y + r z = s$.\\
\includegraphics[max width=\textwidth, alt={}, center]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-10_2715_40_144_2007}
\item The point $( 1,1,1 )$ lies on the plane $\Pi _ { 2 }$.
It is given that the line of intersection of the planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ passes through the point ( $0,0,2$ ) and is parallel to the vector $\mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }$.
Obtain an equation of $\Pi _ { 2 }$ in the form $a x + b y + c z = d$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q5 [10]}}