| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Moderate -0.5 This is a straightforward vectors question requiring standard techniques: finding a direction vector by subtraction and writing a line equation in vector form. Part (a) is routine bookwork with no problem-solving element. While it's from Further Maths P4, this particular part involves only basic vector manipulation that's below average difficulty even for FM students. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts \(\pm\left(\begin{pmatrix}6\\0\\7\end{pmatrix}-\begin{pmatrix}6\\6\\2\end{pmatrix}\right) = \pm\begin{pmatrix}0\\-6\\5\end{pmatrix}\) | M1 | |
| \(\mathbf{r} = \begin{pmatrix}6\\6\\2\end{pmatrix}+\lambda\begin{pmatrix}0\\-6\\5\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}6\\0\\7\end{pmatrix}+\lambda\begin{pmatrix}0\\6\\-5\end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Meet if \(\begin{pmatrix}6\\6\\2\end{pmatrix}+\lambda\begin{pmatrix}0\\-6\\5\end{pmatrix} = \begin{pmatrix}3\\1\\4\end{pmatrix}+\mu\begin{pmatrix}1\\5\\9\end{pmatrix}\) giving \(\begin{cases}6=3+\mu\\6-6\lambda=1+5\mu\\2+5\lambda=4+9\mu\end{cases}\) | M1 | |
| From first equation \(\mu = 3\) | A1 | Alt: solves equations 2 and 3 to give \(\mu=\frac{13}{79}\) or \(\lambda=\frac{55}{79}\) |
| Then need \(\lambda = \frac{5-5\times3}{6} = -\frac{5}{3}\) from 2nd equation; \(\lambda = \frac{2+9\times3}{5} = \frac{29}{5}\) from 3rd equation | M1 | Alt: checks \(\mu\) in first equation |
| Values of \(\lambda\) do not agree and hence lines do not meet. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\overrightarrow{OC} = \begin{pmatrix}3\\1\\4\end{pmatrix}-\begin{pmatrix}1\\5\\9\end{pmatrix} = \begin{pmatrix}2\\-4\\-5\end{pmatrix}\) | M1 | |
| \(\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA} = \begin{pmatrix}2\\-4\\-5\end{pmatrix}-\begin{pmatrix}6\\6\\2\end{pmatrix} = \begin{pmatrix}-4\\-10\\-7\end{pmatrix}\) Accept \(\pm\) | dM1 | Depends on first mark |
| \(\frac{\pm\overrightarrow{AC}\cdot(\mathbf{i}+5\mathbf{j}+9\mathbf{k})}{\left | \overrightarrow{AC}\right | \left |
| \(\theta = \arccos\left | \frac{-117}{\sqrt{165}\sqrt{107}}\right | = 28.3°\) or \(\theta = 90°-\arcsin\left |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts difference between the two given vectors | M1 | Implied by 2 out of 3 correct coordinates if no method shown. Can be either way round. |
| Any correct equation e.g. \(\mathbf{r} = \begin{pmatrix}6\\6\\2\end{pmatrix} + \lambda\begin{pmatrix}0\\-6\\5\end{pmatrix}\) | A1 | Must have "\(\mathbf{r} = \ldots\)". Any point on line valid as starting point; any multiple of direction vector acceptable. Allow in i, j, k form but not e.g. \(\mathbf{r} = \begin{pmatrix}6\mathbf{i}\\6\mathbf{j}\\2\mathbf{k}\end{pmatrix} + \lambda\begin{pmatrix}0\\-6\mathbf{j}\\5\mathbf{k}\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equates lines and extracts at least one equation in \(\lambda\) and \(\mu\) | M1 | |
| Correct value for one parameter e.g. \(\lambda = -\frac{5}{3}\) or \(\mu = 3\) | A1 | Depends on which equations solved. Must follow M1. |
| Substitutes into both other equations to compare results | M1 | |
| Correct values for \(\lambda\) found with conclusion that lines do not meet | A1 | All work must be correct with suitable conclusion. When checking, e.g. \(2 - \frac{25}{3} \neq 4 + 27\) so they do not meet — calculations need not be fully carried out as long as result is obvious. If either side evaluated incorrectly, withhold final mark. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(\mu = -1\) into \(l_2\) to find coordinates of \(C\) or vector \(\overrightarrow{OC}\) | M1 | |
| Uses \(\overrightarrow{OC}\) to find \(\pm\overrightarrow{AC}\) | dM1 | Look for attempt at difference of vectors with at least two correct coordinates |
| Attempts \(\dfrac{\pm\overrightarrow{AC} \cdot (\mathbf{i}+5\mathbf{j}+9\mathbf{k})}{\left | \overrightarrow{AC}\right | \left |
| Correct unsimplified expression e.g. \(\pm\left(\dfrac{-4\times1 + -10\times5 + -7\times9}{\sqrt{16+100+49}\sqrt{1+25+81}}\right)\), \(\pm\dfrac{117}{\sqrt{17655}}\) | A1 | |
| \(\approx 28.3°\) | A1 | Accept awrt 28.3 |
## Question 8(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts $\pm\left(\begin{pmatrix}6\\0\\7\end{pmatrix}-\begin{pmatrix}6\\6\\2\end{pmatrix}\right) = \pm\begin{pmatrix}0\\-6\\5\end{pmatrix}$ | M1 | |
| $\mathbf{r} = \begin{pmatrix}6\\6\\2\end{pmatrix}+\lambda\begin{pmatrix}0\\-6\\5\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}6\\0\\7\end{pmatrix}+\lambda\begin{pmatrix}0\\6\\-5\end{pmatrix}$ | A1 | |
---
## Question 8(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Meet if $\begin{pmatrix}6\\6\\2\end{pmatrix}+\lambda\begin{pmatrix}0\\-6\\5\end{pmatrix} = \begin{pmatrix}3\\1\\4\end{pmatrix}+\mu\begin{pmatrix}1\\5\\9\end{pmatrix}$ giving $\begin{cases}6=3+\mu\\6-6\lambda=1+5\mu\\2+5\lambda=4+9\mu\end{cases}$ | M1 | |
| From first equation $\mu = 3$ | A1 | Alt: solves equations 2 and 3 to give $\mu=\frac{13}{79}$ or $\lambda=\frac{55}{79}$ |
| Then need $\lambda = \frac{5-5\times3}{6} = -\frac{5}{3}$ from 2nd equation; $\lambda = \frac{2+9\times3}{5} = \frac{29}{5}$ from 3rd equation | M1 | Alt: checks $\mu$ in first equation |
| Values of $\lambda$ do not agree and hence lines do not meet. | A1 | |
---
## Question 8(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OC} = \begin{pmatrix}3\\1\\4\end{pmatrix}-\begin{pmatrix}1\\5\\9\end{pmatrix} = \begin{pmatrix}2\\-4\\-5\end{pmatrix}$ | M1 | |
| $\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA} = \begin{pmatrix}2\\-4\\-5\end{pmatrix}-\begin{pmatrix}6\\6\\2\end{pmatrix} = \begin{pmatrix}-4\\-10\\-7\end{pmatrix}$ Accept $\pm$ | dM1 | Depends on first mark |
| $\frac{\pm\overrightarrow{AC}\cdot(\mathbf{i}+5\mathbf{j}+9\mathbf{k})}{\left|\overrightarrow{AC}\right|\left|\mathbf{i}+5\mathbf{j}+9\mathbf{k}\right|} = \pm\frac{-4\times1+-10\times5+-7\times9}{\sqrt{16+100+49}\sqrt{1+25+81}}$ | ddM1 A1 | Depends on both previous marks |
| $\theta = \arccos\left|\frac{-117}{\sqrt{165}\sqrt{107}}\right| = 28.3°$ or $\theta = 90°-\arcsin\left|\frac{-117}{\sqrt{165}\sqrt{107}}\right| = 28.3°$ | A1 | |
# Mark Scheme Extraction
---
## Question (a) [Vectors - Line Equation]
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts difference between the two given vectors | M1 | Implied by 2 out of 3 correct coordinates if no method shown. Can be either way round. |
| Any correct equation e.g. $\mathbf{r} = \begin{pmatrix}6\\6\\2\end{pmatrix} + \lambda\begin{pmatrix}0\\-6\\5\end{pmatrix}$ | A1 | Must have "$\mathbf{r} = \ldots$". Any point on line valid as starting point; any multiple of direction vector acceptable. Allow in **i, j, k** form but not e.g. $\mathbf{r} = \begin{pmatrix}6\mathbf{i}\\6\mathbf{j}\\2\mathbf{k}\end{pmatrix} + \lambda\begin{pmatrix}0\\-6\mathbf{j}\\5\mathbf{k}\end{pmatrix}$ |
---
## Question (b) [Vectors - Lines Do Not Meet]
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equates lines and extracts at least one equation in $\lambda$ **and** $\mu$ | M1 | |
| Correct value for one parameter e.g. $\lambda = -\frac{5}{3}$ or $\mu = 3$ | A1 | Depends on which equations solved. Must follow M1. |
| Substitutes into both other equations to compare results | M1 | |
| Correct values for $\lambda$ found with conclusion that lines do not meet | A1 | All work must be correct with suitable conclusion. When checking, e.g. $2 - \frac{25}{3} \neq 4 + 27$ so they do not meet — calculations need not be fully carried out as long as result is obvious. If either side evaluated incorrectly, withhold final mark. |
> **Note:** Maximum marks in (b) if line in (a) is incorrect is **1010**
---
## Question (c) [Vectors - Angle]
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $\mu = -1$ into $l_2$ to find coordinates of $C$ or vector $\overrightarrow{OC}$ | M1 | |
| Uses $\overrightarrow{OC}$ to find $\pm\overrightarrow{AC}$ | dM1 | Look for attempt at difference of vectors with at least two correct coordinates |
| Attempts $\dfrac{\pm\overrightarrow{AC} \cdot (\mathbf{i}+5\mathbf{j}+9\mathbf{k})}{\left|\overrightarrow{AC}\right|\left|\mathbf{i}+5\mathbf{j}+9\mathbf{k}\right|}$ with their $\overrightarrow{AC}$ | ddM1 | Requires attempt at scalar product of $\overrightarrow{AC}$ with $\mathbf{i}+5\mathbf{j}+9\mathbf{k}$ in numerator with at least 2 components correct; correct attempts at product of magnitudes in denominator |
| Correct unsimplified expression e.g. $\pm\left(\dfrac{-4\times1 + -10\times5 + -7\times9}{\sqrt{16+100+49}\sqrt{1+25+81}}\right)$, $\pm\dfrac{117}{\sqrt{17655}}$ | A1 | |
| $\approx 28.3°$ | A1 | Accept awrt 28.3 |
---8. With respect to a fixed origin $O$ the points $A$ and $B$ have position vectors
$$\left( \begin{array} { l }
6 \\
6 \\
2
\end{array} \right) \text { and } \left( \begin{array} { l }
6 \\
0 \\
7
\end{array} \right)$$
respectively.
The line $l _ { 1 }$ passes through the points $A$ and $B$.\\
(a) Write down an equation for $l _ { 1 }$
Give your answer in the form $\mathbf { r } = \mathbf { p } + \lambda \mathbf { q }$, where $\lambda$ is a scalar parameter.
The line $l _ { 2 }$ has equation
$$\mathbf { r } = \left( \begin{array} { l }
3 \\
1 \\
4
\end{array} \right) + \mu \left( \begin{array} { l }
1 \\
5 \\
9
\end{array} \right)$$
where $\mu$ is a scalar parameter.\\
(b) Show that $l _ { 1 }$ and $l _ { 2 }$ do not meet.
The point $C$ is on $l _ { 2 }$ where $\mu = - 1$\\
(c) Find the acute angle between $A C$ and $l _ { 2 }$
Give your answer in degrees to one decimal place.\\
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\section*{Question 8 continued}
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\section*{Question 8 continued}
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\section*{Question 8 continued}
\begin{enumerate}
\item (a) Find the derivative with respect to $y$ of
\end{enumerate}
$$\frac { 1 } { ( 1 + 2 \ln y ) ^ { 2 } }$$
(b) Hence find a general solution to the differential equation
$$3 \operatorname { cosec } ( 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 1 + 2 \ln y ) ^ { 3 } \quad y > 0 \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$
(c) Show that the particular solution of this differential equation for which $y = 1$ at $x = \frac { \pi } { 6 }$ is given by
$$y = \mathrm { e } ^ { A \sec x - \frac { 1 } { 2 } }$$
where $A$ is an irrational number to be found.\\
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\section*{Question 9 continued}
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\section*{Question 9 continued}
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\includegraphics[max width=\textwidth, alt={}, center]{fe07afad-9cfc-48c0-84f1-5717f81977d4-32_2649_1894_109_173}
\hfill \mbox{\textit{Edexcel P4 2022 Q8 [11]}}