| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 0 \\ 0 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \sim \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) | M1 A1 | Finds common perpendicular |
| \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{\sqrt{1^2 + 3^2 + 2^2}} = \frac{1}{\sqrt{14}}\) | B1 | Divides by magnitude of the normal to \(\Pi\), 0.267 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix} = \sqrt{3}\sqrt{14}\cos\alpha\) leading to \(\cos\alpha = \frac{6}{\sqrt{3}\sqrt{14}}\) | M1 A1 FT | Takes dot product of normal vectors |
| \(22.2°\) | A1 | Accept 0.388 radians. Mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -7 \\ -2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} -20 \\ 12 \\ -4 \end{pmatrix} \sim \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\) | M1 A1 | Finds direction of common perpendicular |
| \(\overrightarrow{OP} = \lambda\begin{pmatrix} 2 \\ 1 \\ -7 \end{pmatrix}\), \(\overrightarrow{OQ} = \begin{pmatrix} 2-2\mu \\ 2-3\mu \\ \mu \end{pmatrix}\) leading to \(\overrightarrow{PQ} = \begin{pmatrix} 2-2\mu-2\lambda \\ 2-3\mu-\lambda \\ \mu+7\lambda \end{pmatrix}\) | M1 A1 | Finds \(\overrightarrow{PQ}\) |
| \(\begin{pmatrix} 2-2\mu-2\lambda \\ 2-3\mu-\lambda \\ \mu+7\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix} = 0\) or \(= k\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\) | M1 | Uses dot product of \(\overrightarrow{PQ}\) with line direction is zero, or \(\overrightarrow{PQ}\) is multiple of common perpendicular (parameter \(k\) not 1) |
| \(14\mu + 14\lambda = 10\) | A1 | Deduces one equation |
| \(\begin{pmatrix} 2-2\mu-2\lambda \\ 2-3\mu-\lambda \\ \mu+7\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -7 \end{pmatrix} = 0 \Rightarrow 14\mu + 54\lambda = 6\) | A1 | Deduces second equation |
| \(\lambda = -\frac{1}{10}\) leading to \(\overrightarrow{OP} = -\frac{1}{10}\begin{pmatrix} 2 \\ 1 \\ -7 \end{pmatrix}\) | M1 A1 | Solves for \(\lambda\) and substitutes into \(\overrightarrow{OP}\) |
| \(\mathbf{r} = \begin{pmatrix} -0.2 \\ -0.1 \\ 0.7 \end{pmatrix} + k\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\) | B1 FT | FT using their common perpendicular |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 0 \\ 0 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \sim \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$ | M1 A1 | Finds common perpendicular |
| $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0$ | A1 | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{\sqrt{1^2 + 3^2 + 2^2}} = \frac{1}{\sqrt{14}}$ | B1 | Divides by magnitude of the normal to $\Pi$, 0.267 |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix} = \sqrt{3}\sqrt{14}\cos\alpha$ leading to $\cos\alpha = \frac{6}{\sqrt{3}\sqrt{14}}$ | M1 A1 FT | Takes dot product of normal vectors |
| $22.2°$ | A1 | Accept 0.388 radians. Mark final answer |
---
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -7 \\ -2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} -20 \\ 12 \\ -4 \end{pmatrix} \sim \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}$ | M1 A1 | Finds direction of common perpendicular |
| $\overrightarrow{OP} = \lambda\begin{pmatrix} 2 \\ 1 \\ -7 \end{pmatrix}$, $\overrightarrow{OQ} = \begin{pmatrix} 2-2\mu \\ 2-3\mu \\ \mu \end{pmatrix}$ leading to $\overrightarrow{PQ} = \begin{pmatrix} 2-2\mu-2\lambda \\ 2-3\mu-\lambda \\ \mu+7\lambda \end{pmatrix}$ | M1 A1 | Finds $\overrightarrow{PQ}$ |
| $\begin{pmatrix} 2-2\mu-2\lambda \\ 2-3\mu-\lambda \\ \mu+7\lambda \end{pmatrix} \cdot \begin{pmatrix} -2 \\ -3 \\ 1 \end{pmatrix} = 0$ or $= k\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}$ | M1 | Uses dot product of $\overrightarrow{PQ}$ with line direction is zero, or $\overrightarrow{PQ}$ is multiple of common perpendicular (parameter $k$ not 1) |
| $14\mu + 14\lambda = 10$ | A1 | Deduces one equation |
| $\begin{pmatrix} 2-2\mu-2\lambda \\ 2-3\mu-\lambda \\ \mu+7\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -7 \end{pmatrix} = 0 \Rightarrow 14\mu + 54\lambda = 6$ | A1 | Deduces second equation |
| $\lambda = -\frac{1}{10}$ leading to $\overrightarrow{OP} = -\frac{1}{10}\begin{pmatrix} 2 \\ 1 \\ -7 \end{pmatrix}$ | M1 A1 | Solves for $\lambda$ and substitutes into $\overrightarrow{OP}$ |
| $\mathbf{r} = \begin{pmatrix} -0.2 \\ -0.1 \\ 0.7 \end{pmatrix} + k\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}$ | B1 FT | FT using their common perpendicular |7 The points $A , B , C$ have position vectors
$$2 \mathbf { i } + 2 \mathbf { j } , \quad - \mathbf { j } + \mathbf { k } \quad \text { and } \quad 2 \mathbf { i } + \mathbf { j } - 7 \mathbf { k }$$
respectively, relative to the origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the plane $O A B$, giving your answer in the form $\mathbf { r } . \mathbf { n } = p$.\\
The plane $\Pi$ has equation $\mathrm { x } - 3 \mathrm { y } - 2 \mathrm { z } = 1$.
\item Find the perpendicular distance of $\Pi$ from the origin.
\item Find the acute angle between the planes $O A B$ and $\Pi$.
\item Find an equation for the common perpendicular to the lines $O C$ and $A B$.\\
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 2021 Q7 [17]}}