| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding a normal vector via cross product, converting to Cartesian form, substituting a line equation into a plane, and using standard angle formulas. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 2 & 3 & 0 \end{vmatrix} = \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\) | M1 A1 | Finds vector perpendicular to \(\Pi\) |
| \(-3(-2) + 2(3) + 3(3) = 21\) | M1 | Substitutes point on \(\Pi\) |
| \(-3x + 2y + 3z = 21\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 2 \\ -3 \\ 5+t \end{pmatrix}\) | B1 | Forms general point on line (given as a single vector) |
| \(-3(2) + 2(-3) + 3(5+t) = 21\) leading to \(t = 6\) | M1 | Substitutes into the equation for \(\Pi\) |
| \(\begin{pmatrix} 2 \\ -3 \\ 11 \end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \sqrt{1}\sqrt{22}\cos\alpha\) leading to \(\cos\alpha = \frac{3}{\sqrt{22}}\) | M1 A1 FT | Uses dot product of k and their normal |
| Acute angle between \(l\) and \(\Pi\) is \(90 - \alpha = 39.8°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 0 \\ 0 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \\ 2 \end{pmatrix}\) | B1 | Finds direction vector from \(P\) to plane |
| \(\frac{1}{\sqrt{22}}\begin{pmatrix} -2 \\ 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \frac{18}{\sqrt{22}} = 3.84\) | M1 A1 | Uses dot product of their direction and normal vectors |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 2 & 3 & 0 \end{vmatrix} = \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}$ | M1 A1 | Finds vector perpendicular to $\Pi$ |
| $-3(-2) + 2(3) + 3(3) = 21$ | M1 | Substitutes point on $\Pi$ |
| $-3x + 2y + 3z = 21$ | A1 | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 2 \\ -3 \\ 5+t \end{pmatrix}$ | B1 | Forms general point on line (given as a single vector) |
| $-3(2) + 2(-3) + 3(5+t) = 21$ leading to $t = 6$ | M1 | Substitutes into the equation for $\Pi$ |
| $\begin{pmatrix} 2 \\ -3 \\ 11 \end{pmatrix}$ | A1 | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \sqrt{1}\sqrt{22}\cos\alpha$ leading to $\cos\alpha = \frac{3}{\sqrt{22}}$ | M1 A1 FT | Uses dot product of **k** and their normal |
| Acute angle between $l$ and $\Pi$ is $90 - \alpha = 39.8°$ | A1 | |
## Question 5(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 0 \\ 0 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \\ 2 \end{pmatrix}$ | B1 | Finds direction vector from $P$ to plane |
| $\frac{1}{\sqrt{22}}\begin{pmatrix} -2 \\ 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \frac{18}{\sqrt{22}} = 3.84$ | M1 A1 | Uses dot product of their direction and normal vectors |5 The plane $\Pi$ has equation $\mathbf { r } = - 2 \mathbf { i } + 3 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } + \mathbf { k } ) + \mu ( 2 \mathbf { i } + 3 \mathbf { j } )$.
\begin{enumerate}[label=(\alph*)]
\item Find a Cartesian equation of $\Pi$, giving your answer in the form $a x + b y + c = d$.\\
The line $l$ passes through the point $P$ with position vector $2 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k }$ and is parallel to the vector $\mathbf { k }$.
\item Find the position vector of the point where $l$ meets $\Pi$.
\item Find the acute angle between $l$ and $\Pi$.
\item Find the perpendicular distance from $P$ to $\Pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q5 [13]}}