| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 1 \\ 3 & 2 & -1 \end{vmatrix} = \begin{pmatrix}-4\\2\\-8\end{pmatrix} \sim \begin{pmatrix}2\\-1\\4\end{pmatrix}\) | M1 A1 | Finds vector perpendicular to \(\Pi\) |
| \(2(3) - (-2) + 4(1) = 12\) | M1 | Substitutes point on \(\Pi\) |
| \(2x - y + 4z = 12\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0\\1\\1\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\4\end{pmatrix} = \sqrt{2}\sqrt{21}\cos\alpha\) leading to \(\cos\alpha = \dfrac{3}{\sqrt{42}}\) | M1 A1 FT | Uses dot product of \(\mathbf{j}+\mathbf{k}\) and their normal. May be implied by angle 62.4° |
| Acute angle between \(l\) and \(\Pi\) is \(90° - \alpha = 27.6°\) | A1 | 0.48 radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{OF} = \overrightarrow{OP} + t\begin{pmatrix}2\\-1\\4\end{pmatrix} = \begin{pmatrix}2+2t\\3-t\\1+4t\end{pmatrix}\) | M1 A1 | Forms \(\overrightarrow{OF}\) using parameter |
| \(2(1+5) = 12\) i.e. \(21t + 5 = 12\) | M1 | Substitutes into equation of plane |
| \(\overrightarrow{OF} = \dfrac{1}{3}\begin{pmatrix}8\\8\\7\end{pmatrix}\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 1 \\ 3 & 2 & -1 \end{vmatrix} = \begin{pmatrix}-4\\2\\-8\end{pmatrix} \sim \begin{pmatrix}2\\-1\\4\end{pmatrix}$ | M1 A1 | Finds vector perpendicular to $\Pi$ |
| $2(3) - (-2) + 4(1) = 12$ | M1 | Substitutes point on $\Pi$ |
| $2x - y + 4z = 12$ | A1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0\\1\\1\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\4\end{pmatrix} = \sqrt{2}\sqrt{21}\cos\alpha$ leading to $\cos\alpha = \dfrac{3}{\sqrt{42}}$ | M1 A1 FT | Uses dot product of $\mathbf{j}+\mathbf{k}$ and their normal. May be implied by angle 62.4° |
| Acute angle between $l$ and $\Pi$ is $90° - \alpha = 27.6°$ | A1 | 0.48 radians |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OF} = \overrightarrow{OP} + t\begin{pmatrix}2\\-1\\4\end{pmatrix} = \begin{pmatrix}2+2t\\3-t\\1+4t\end{pmatrix}$ | M1 A1 | Forms $\overrightarrow{OF}$ using parameter |
| $2(1+5) = 12$ i.e. $21t + 5 = 12$ | M1 | Substitutes into equation of plane |
| $\overrightarrow{OF} = \dfrac{1}{3}\begin{pmatrix}8\\8\\7\end{pmatrix}$ | A1 | |
---4 The plane $\Pi$ contains the lines $\mathbf { r } = 3 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } + \lambda ( - \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )$ and $\mathbf { r } = 4 \mathbf { i } + 4 \mathbf { j } + 2 \mathbf { k } + \mu ( 3 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } )$.
\begin{enumerate}[label=(\alph*)]
\item Find a Cartesian equation of $\Pi$, giving your answer in the form $a x + b y + c z = d$.\\
The line $l$ passes through the point $P$ with position vector $2 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }$ and is parallel to the vector $\mathbf { j } + \mathbf { k }$.
\item Find the acute angle between $I$ and $\Pi$.
\item Find the position vector of the foot of the perpendicular from $P$ to $\Pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q4 [11]}}