| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane through three points |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine application of cross product to find a plane normal, then using the plane equation formula and distance formula. Part (c) on common perpendicular is slightly less routine but still a standard technique. All parts follow textbook methods with no novel insight required, making it slightly easier than average for Further Maths content. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = -2\mathbf{i} + \mathbf{j} + 4\mathbf{k}\), \(\overrightarrow{AC} = -3\mathbf{i} + 3\mathbf{k}\), \(\overrightarrow{BC} = -\mathbf{i} - \mathbf{j} - \mathbf{k}\) | B1 | Finds direction vectors of two lines in the plane |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 4 \\ -3 & 0 & 3 \end{vmatrix} = \begin{pmatrix} 3 \\ -6 \\ 3 \end{pmatrix} \sim \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\) | M1 A1FT | Finds normal to the plane \(ABC\) |
| \(1(1) - 2(1) + 1(0) = -1 \Rightarrow x - 2y + z = -1\) | M1 A1 | Substitutes point. CAO |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{\sqrt{1^2 + 2^2 + 1^2}} = \dfrac{1}{\sqrt{6}} = 0.408\) | M1 A1FT | Divides by magnitude of normal vector. FT *their* (a) |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{OP} = \begin{pmatrix} -2\lambda \\ \lambda \\ 3\lambda \end{pmatrix}\), \(\overrightarrow{OQ} = \begin{pmatrix} 1-2\mu \\ 1+\mu \\ 4\mu \end{pmatrix} \Rightarrow \overrightarrow{PQ} = \begin{pmatrix} 1-2\mu+2\lambda \\ 1+\mu-\lambda \\ 4\mu-3\lambda \end{pmatrix}\) | M1 A1 | Finds \(\overrightarrow{PQ}\), where \(P\) is a point on \(OC\) and \(Q\) is a point on \(AB\) |
| \(\begin{pmatrix} 1-2\mu+2\lambda \\ 1+\mu-\lambda \\ 4\mu-3\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} = 0\) | M1* | Uses that dot product of \(\overrightarrow{PQ}\) with line direction is zero |
| \(17\mu - 14\lambda = 1\) | dM1 | Deduces one equation |
| \(\begin{pmatrix} 1-2\mu+2\lambda \\ 1+\mu-\lambda \\ 4\mu-3\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} = 0 \Rightarrow 21\mu - 17\lambda = 1\) | dM1 | Deduces second equation |
| \(\lambda = -\dfrac{4}{5} \Rightarrow \overrightarrow{OP} = -\dfrac{4}{5}\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}\) | dM1 A1 | Solves for \(\lambda\) or \(\mu\) and substitutes into \(\overrightarrow{OP}\) |
| \(\mathbf{r} = -\dfrac{4}{5}\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + k\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\) | A1 | OE |
| Total: 8 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = -2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$, $\overrightarrow{AC} = -3\mathbf{i} + 3\mathbf{k}$, $\overrightarrow{BC} = -\mathbf{i} - \mathbf{j} - \mathbf{k}$ | B1 | Finds direction vectors of **two** lines in the plane |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 4 \\ -3 & 0 & 3 \end{vmatrix} = \begin{pmatrix} 3 \\ -6 \\ 3 \end{pmatrix} \sim \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ | M1 A1FT | Finds normal to the plane $ABC$ |
| $1(1) - 2(1) + 1(0) = -1 \Rightarrow x - 2y + z = -1$ | M1 A1 | Substitutes point. CAO |
| **Total: 5** | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{\sqrt{1^2 + 2^2 + 1^2}} = \dfrac{1}{\sqrt{6}} = 0.408$ | M1 A1FT | Divides by magnitude of normal vector. FT *their* **(a)** |
| **Total: 2** | | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OP} = \begin{pmatrix} -2\lambda \\ \lambda \\ 3\lambda \end{pmatrix}$, $\overrightarrow{OQ} = \begin{pmatrix} 1-2\mu \\ 1+\mu \\ 4\mu \end{pmatrix} \Rightarrow \overrightarrow{PQ} = \begin{pmatrix} 1-2\mu+2\lambda \\ 1+\mu-\lambda \\ 4\mu-3\lambda \end{pmatrix}$ | M1 A1 | Finds $\overrightarrow{PQ}$, where $P$ is a point on $OC$ and $Q$ is a point on $AB$ |
| $\begin{pmatrix} 1-2\mu+2\lambda \\ 1+\mu-\lambda \\ 4\mu-3\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} = 0$ | M1* | Uses that dot product of $\overrightarrow{PQ}$ with line direction is zero |
| $17\mu - 14\lambda = 1$ | dM1 | Deduces one equation |
| $\begin{pmatrix} 1-2\mu+2\lambda \\ 1+\mu-\lambda \\ 4\mu-3\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} = 0 \Rightarrow 21\mu - 17\lambda = 1$ | dM1 | Deduces second equation |
| $\lambda = -\dfrac{4}{5} \Rightarrow \overrightarrow{OP} = -\dfrac{4}{5}\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}$ | dM1 A1 | Solves for $\lambda$ or $\mu$ and substitutes into $\overrightarrow{OP}$ |
| $\mathbf{r} = -\dfrac{4}{5}\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + k\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ | A1 | OE |
| **Total: 8** | | |6 The points $A , B , C$ have position vectors
$$\mathbf { i } + \mathbf { j } , \quad - \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } , \quad - 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } ,$$
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 perpendicular distance from $O$ to the plane $A B C$.
\item Find a vector equation of the common perpendicular to the lines $O C$ and $A B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q6 [15]}}