| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Plane containing line and point/vector |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring finding a plane's equation from a line and parallel vector (using cross product for the normal), then finding the angle between a line and plane. Both parts follow routine procedures taught in Further Pure, with straightforward calculations and no novel problem-solving required. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -4 \\ 2 & 5 & -4 \end{vmatrix} = \begin{pmatrix} 24 \\ -4 \\ 7 \end{pmatrix}\) | M1 A1 | Finds vector perpendicular to the plane |
| \(24(1) - 4(3) + 7(-1) = d \Rightarrow 24x - 4y + 7z = 5\) | M1 A1 | Uses point in the plane |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 24 \\ -4 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -5 \\ -2 \end{pmatrix} = \sqrt{641}\sqrt{54}\cos\alpha \Rightarrow \cos\alpha = \dfrac{126}{\sqrt{641}\sqrt{54}}\) | M1A1FT | Uses dot product of \(5\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}\) and their normal |
| Acute angle between \(l_2\) and \(\Pi\) is \(90 - \alpha = 42.6°\) | A1 | 0.744 radians |
| Total: 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -4 \\ 2 & 5 & -4 \end{vmatrix} = \begin{pmatrix} 24 \\ -4 \\ 7 \end{pmatrix}$ | **M1 A1** | Finds vector perpendicular to the plane |
| $24(1) - 4(3) + 7(-1) = d \Rightarrow 24x - 4y + 7z = 5$ | **M1 A1** | Uses point in the plane |
| **Total: 4** | | |
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## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 24 \\ -4 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -5 \\ -2 \end{pmatrix} = \sqrt{641}\sqrt{54}\cos\alpha \Rightarrow \cos\alpha = \dfrac{126}{\sqrt{641}\sqrt{54}}$ | **M1A1FT** | Uses dot product of $5\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}$ and their normal |
| Acute angle between $l_2$ and $\Pi$ is $90 - \alpha = 42.6°$ | **A1** | 0.744 radians |
| **Total: 3** | | |2 The line $l _ { 1 }$ has equation $\mathbf { r } = \mathbf { i } + 3 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } - \mathbf { j } - 4 \mathbf { k } )$.\\
The plane $\Pi$ contains $l _ { 1 }$ and is parallel to the vector $2 \mathbf { i } + 5 \mathbf { j } - 4 \mathbf { k }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of $\Pi$, giving your answer in the form $a x + b y + c z = d$.\\
\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-05_2723_33_99_22}
The line $l _ { 2 }$ is parallel to the vector $5 \mathbf { i } - 5 \mathbf { j } - 2 \mathbf { k }$.
\item Find the acute angle between $l _ { 2 }$ and $\Pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q2 [7]}}