| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane containing line and point/parallel to vector |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths vectors question requiring cross products to find a plane normal, distance formula, and perpendicular distance between skew lines. While it involves several steps and Further Maths content, each part follows standard procedures: (a) cross product of direction vectors for normal, (b) point-to-plane distance formula, (c) setting up perpendicularity conditions. The techniques are well-practiced in Further Maths courses with no novel insight required, making it moderately above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 0 & 2 \\ 0 & 1 & 1 \end{vmatrix} = \begin{pmatrix}-2\\-5\\5\end{pmatrix}\) | M1 A1 | |
| \(-5(-5) = 25\) | M1 | |
| \(-2x - 5y + 5z = 25\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0\\-5\\0\end{pmatrix} - \begin{pmatrix}4\\2\\-2\end{pmatrix} = \begin{pmatrix}-4\\-7\\2\end{pmatrix}\) | M1 | |
| \(\frac{1}{\sqrt{54}}\begin{vmatrix}\begin{pmatrix}-4\\-7\\2\end{pmatrix} \cdot \begin{pmatrix}-2\\-5\\5\end{pmatrix}\end{vmatrix} = \frac{53}{\sqrt{54}} = 7.21\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{OP} = \begin{pmatrix}5\lambda\\-5\\2\lambda\end{pmatrix}\), \(\overrightarrow{OQ} = \begin{pmatrix}4\\2+\mu\\-2+\mu\end{pmatrix} \Rightarrow \overrightarrow{PQ} = \begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix}\) | M1 A1 | |
| \(\begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix} \cdot \begin{pmatrix}5\\0\\2\end{pmatrix} = 0\) or \(\begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix} = k\begin{pmatrix}-2\\-5\\5\end{pmatrix}\) | M1 | |
| \(-29\lambda + 2\mu = -16\) | A1 | |
| \(\begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix} \cdot \begin{pmatrix}0\\1\\1\end{pmatrix} = 0 \Rightarrow -2\lambda + 2\mu = -5\) | A1 | |
| \(\lambda = \frac{11}{27} \Rightarrow \overrightarrow{OP} = \frac{1}{27}\begin{pmatrix}55\\-135\\22\end{pmatrix} = \frac{55}{27}\mathbf{i} - 5\mathbf{j} + \frac{22}{27}\mathbf{k}\) | M1 A1 | |
| \(\mathbf{r} = \frac{1}{27}\begin{pmatrix}55\\-135\\22\end{pmatrix} + k\begin{pmatrix}-2\\-5\\5\end{pmatrix}\) | B1 FT | Follow through their common perpendicular |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 0 & 2 \\ 0 & 1 & 1 \end{vmatrix} = \begin{pmatrix}-2\\-5\\5\end{pmatrix}$ | **M1 A1** | |
| $-5(-5) = 25$ | **M1** | |
| $-2x - 5y + 5z = 25$ | **A1** | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0\\-5\\0\end{pmatrix} - \begin{pmatrix}4\\2\\-2\end{pmatrix} = \begin{pmatrix}-4\\-7\\2\end{pmatrix}$ | **M1** | |
| $\frac{1}{\sqrt{54}}\begin{vmatrix}\begin{pmatrix}-4\\-7\\2\end{pmatrix} \cdot \begin{pmatrix}-2\\-5\\5\end{pmatrix}\end{vmatrix} = \frac{53}{\sqrt{54}} = 7.21$ | **M1 A1** | |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OP} = \begin{pmatrix}5\lambda\\-5\\2\lambda\end{pmatrix}$, $\overrightarrow{OQ} = \begin{pmatrix}4\\2+\mu\\-2+\mu\end{pmatrix} \Rightarrow \overrightarrow{PQ} = \begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix}$ | **M1 A1** | |
| $\begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix} \cdot \begin{pmatrix}5\\0\\2\end{pmatrix} = 0$ or $\begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix} = k\begin{pmatrix}-2\\-5\\5\end{pmatrix}$ | **M1** | |
| $-29\lambda + 2\mu = -16$ | **A1** | |
| $\begin{pmatrix}4-5\lambda\\7+\mu\\-2+\mu-2\lambda\end{pmatrix} \cdot \begin{pmatrix}0\\1\\1\end{pmatrix} = 0 \Rightarrow -2\lambda + 2\mu = -5$ | **A1** | |
| $\lambda = \frac{11}{27} \Rightarrow \overrightarrow{OP} = \frac{1}{27}\begin{pmatrix}55\\-135\\22\end{pmatrix} = \frac{55}{27}\mathbf{i} - 5\mathbf{j} + \frac{22}{27}\mathbf{k}$ | **M1 A1** | |
| $\mathbf{r} = \frac{1}{27}\begin{pmatrix}55\\-135\\22\end{pmatrix} + k\begin{pmatrix}-2\\-5\\5\end{pmatrix}$ | **B1 FT** | Follow through their common perpendicular |7 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = - 5 \mathbf { j } + \lambda ( 5 \mathbf { i } + 2 \mathbf { k } )$ and $\mathbf { r } = 4 \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k } + \mu ( \mathbf { j } + \mathbf { k } )$ respectively. The plane $\Pi$ contains $l _ { 1 }$ and is parallel to $l _ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of $\Pi$, giving your answer in the form $a x + b y + c z = d$.
\item Find the distance between $l _ { 2 }$ and $\Pi$.\\
The point $P$ on $l _ { 1 }$ and the point $Q$ on $l _ { 2 }$ are such that $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$.
\item Show that $P$ has position vector $\frac { 55 } { 27 } \mathbf { i } - 5 \mathbf { j } + \frac { 22 } { 27 } \mathbf { k }$ and state a vector equation for $P Q$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q7 [15]}}