| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Challenging +1.2 This is a standard Further Maths vectors question requiring finding a plane equation via cross product, angle between planes using normal vectors, and shortest distance between skew lines. While it involves multiple parts and several techniques, each step follows routine procedures taught in Further Maths with no novel insight required. The calculations are moderately lengthy but straightforward, placing it slightly above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d 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 |
| \(\overrightarrow{AB} = -\mathbf{j} + 3\mathbf{k}\), \(\overrightarrow{AC} = 2\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\) | B1 | Finds direction vectors of two lines in the plane. \(\overrightarrow{BC} = 2\mathbf{i} - 3\mathbf{j} - \mathbf{k}\) |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 3 \\ 1 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 5 \\ 3 \\ 1 \end{pmatrix}\) | M1 A1 | Finds normal to plane \(ABC\). OE |
| \(5(0) + 3(-2) + 1(1) = -5 \Rightarrow 5x + 3y + z = -5\) | M1 A1 | Substitutes point. OE |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 2 \\ 0 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix}\) | M1 A1 | Finds normal to plane \(OBC\) |
| \(\begin{pmatrix} 5 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} = \sqrt{35}\sqrt{45}\cos\theta\) | M1 | Uses dot product of normal vectors |
| \(28.1°\) | A1 | \(0.49(0)\) rad |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 3 \\ t & 1 & -1 \end{vmatrix} = \begin{pmatrix} -2 \\ 3t \\ t \end{pmatrix}\) | M1 A1 | Finds common perpendicular |
| \(\frac{1}{\sqrt{4+10t^2}}\begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 3t \\ t \end{pmatrix} = \left\lvert\frac{-4-10t}{\sqrt{4+10t^2}}\right\rvert\) | M1 A1 | Uses formula for perpendicular distance |
| \(\left\lvert\frac{-4-10t}{\sqrt{4+10t^2}}\right\rvert = \sqrt{10} \Rightarrow (4+10t)^2 = 10(4+10t^2)\) | M1 | Sets equal to \(\sqrt{10}\) and solves for \(t\) |
| \(t = \frac{3}{10}\) | A1 | |
| Total: 6 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = -\mathbf{j} + 3\mathbf{k}$, $\overrightarrow{AC} = 2\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}$ | **B1** | Finds direction vectors of two lines in the plane. $\overrightarrow{BC} = 2\mathbf{i} - 3\mathbf{j} - \mathbf{k}$ |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 3 \\ 1 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 5 \\ 3 \\ 1 \end{pmatrix}$ | **M1 A1** | Finds normal to plane $ABC$. OE |
| $5(0) + 3(-2) + 1(1) = -5 \Rightarrow 5x + 3y + z = -5$ | **M1 A1** | Substitutes point. OE |
| **Total: 5** | | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 2 \\ 0 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix}$ | **M1 A1** | Finds normal to plane $OBC$ |
| $\begin{pmatrix} 5 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix} = \sqrt{35}\sqrt{45}\cos\theta$ | **M1** | Uses dot product of normal vectors |
| $28.1°$ | **A1** | $0.49(0)$ rad |
| **Total: 4** | | |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 3 \\ t & 1 & -1 \end{vmatrix} = \begin{pmatrix} -2 \\ 3t \\ t \end{pmatrix}$ | **M1 A1** | Finds common perpendicular |
| $\frac{1}{\sqrt{4+10t^2}}\begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 3t \\ t \end{pmatrix} = \left\lvert\frac{-4-10t}{\sqrt{4+10t^2}}\right\rvert$ | **M1 A1** | Uses formula for perpendicular distance |
| $\left\lvert\frac{-4-10t}{\sqrt{4+10t^2}}\right\rvert = \sqrt{10} \Rightarrow (4+10t)^2 = 10(4+10t^2)$ | **M1** | Sets equal to $\sqrt{10}$ and solves for $t$ |
| $t = \frac{3}{10}$ | **A1** | |
| **Total: 6** | | |7 The points $A , B , C$ have position vectors
$$- 2 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } , \quad - 2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } , \quad - 2 \mathbf { j } + \mathbf { k } ,$$
respectively, relative to the origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the plane $A B C$, giving your answer in the form $a x + b y + c z = d$.
\item Find the acute angle between the planes $O B C$ and $A B C$.\\
The point $D$ has position vector $t \mathbf { i } - \mathbf { j }$.
\item Given that the shortest distance between the lines $A B$ and $C D$ is $\sqrt { \mathbf { 1 0 } }$, find the value of $t$.\\
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]}}