| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Standard +0.8 This is a substantial multi-part Further Maths question requiring: (a) finding a plane equation via cross product of vectors AB and AC, (b) finding intersection of a perpendicular line with another plane and computing distance, (c) using the tetrahedron volume formula involving scalar triple product. While the techniques are standard for FM students, the question requires careful execution across multiple steps, coordinate geometry in 3D, and the volume calculation in part (c) involves non-trivial algebraic manipulation. This is moderately challenging for Further Maths but not exceptionally difficult. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm\overrightarrow{AB} = \pm\begin{pmatrix}-4\\-1\\1\end{pmatrix}\), \(\pm\overrightarrow{AC} = \pm\begin{pmatrix}-5\\2\\0\end{pmatrix}\), \(\pm\overrightarrow{BC} = \pm\begin{pmatrix}-1\\3\\-1\end{pmatrix}\) | M1 | Correct method to obtain two relevant vectors using subtraction. Labelling can be ignored |
| \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}-4\\-1\\1\end{pmatrix}\times\begin{pmatrix}-5\\2\\0\end{pmatrix} = \begin{pmatrix}-2\\-5\\-13\end{pmatrix}\) | dM1 | Correct method for vector product of two relevant vectors (if method not shown, two correct components required) |
| \(\begin{pmatrix}3\\2\\2\end{pmatrix}\cdot\begin{pmatrix}2\\5\\13\end{pmatrix} = 6+10+26 = 42\) | ddM1 | Attempts scalar product between normal vector and any position vector of \(A\), \(B\) or \(C\) |
| \(2x + 5y + 13z = 42\) (or e.g. \(-2x-5y-13z+42=0\)) | A1 | Any correct Cartesian equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Line \(DE\): \(\mathbf{r} = \begin{pmatrix}-1\\1\\-2\end{pmatrix} \pm \lambda\begin{pmatrix}2\\5\\13\end{pmatrix}\) | M1 | Obtains parametric form for line \(DE\) with their normal (or recalculated normal) seen or implied. Allow one slip only |
| \(14(2\lambda-1)-(5\lambda+1)-17(13\lambda-2)=-66 \Rightarrow \lambda = \ldots\) | M1 | Substitutes parametric form into equation of \(\Pi_2\) and solves for \(\lambda\) |
| \(\lambda = \dfrac{85}{198}\) | A1 | Correct exact value of \(\lambda\) depending on method |
| \(DE = \sqrt{\left(2\times\frac{85}{198}\right)^2+\left(5\times\frac{85}{198}\right)^2+\left(13\times\frac{85}{198}\right)^2}\) | dM1 | Correct method to find numerical expression for distance \(DE\). Requires previous method mark |
| \(DE = \dfrac{85\sqrt{22}}{66}\) | A1 | Correct exact answer. Or \(p = \frac{85}{66}\) or \(1\frac{19}{66}\). Not \(DE = -\frac{85\sqrt{22}}{66}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AF}\cdot\overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}1\\1\\q-2\end{pmatrix}\cdot\begin{pmatrix}2\\5\\13\end{pmatrix} = 2+5+13q-26\) | M1 | Correct method for vector between \(F\) and \(A\), \(B\) or \(C\) and finds scalar product with normal, or attempts scalar triple product to obtain linear expression in \(q\). At least 2 correct elements for triple product |
| \(\frac{1}{6}(13q-19) = \pm 12 \Rightarrow q = \ldots\) | dM1 | Sets \(\frac{1}{6}\) of expression in \(q\) equal to \(\pm 12\) (or equivalent e.g. expression equal to \(\pm 72\)) and proceeds to value of \(q\) |
| \(q = 7,\ -\dfrac{53}{13}\) | A1 | Correct values. Allow exact equivalents e.g. \(-4\frac{1}{13}\) |
# Question 6(a): Equation of Plane
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{AB} = \pm\begin{pmatrix}-4\\-1\\1\end{pmatrix}$, $\pm\overrightarrow{AC} = \pm\begin{pmatrix}-5\\2\\0\end{pmatrix}$, $\pm\overrightarrow{BC} = \pm\begin{pmatrix}-1\\3\\-1\end{pmatrix}$ | M1 | Correct method to obtain two relevant vectors using subtraction. Labelling can be ignored |
| $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}-4\\-1\\1\end{pmatrix}\times\begin{pmatrix}-5\\2\\0\end{pmatrix} = \begin{pmatrix}-2\\-5\\-13\end{pmatrix}$ | dM1 | Correct method for vector product of two relevant vectors (if method not shown, two correct components required) |
| $\begin{pmatrix}3\\2\\2\end{pmatrix}\cdot\begin{pmatrix}2\\5\\13\end{pmatrix} = 6+10+26 = 42$ | ddM1 | Attempts scalar product between normal vector and any position vector of $A$, $B$ or $C$ |
| $2x + 5y + 13z = 42$ (or e.g. $-2x-5y-13z+42=0$) | A1 | Any correct Cartesian equation |
---
# Question 6(b): Distance DE
| Answer/Working | Mark | Guidance |
|---|---|---|
| Line $DE$: $\mathbf{r} = \begin{pmatrix}-1\\1\\-2\end{pmatrix} \pm \lambda\begin{pmatrix}2\\5\\13\end{pmatrix}$ | M1 | Obtains parametric form for line $DE$ with their normal (or recalculated normal) seen or implied. Allow one slip only |
| $14(2\lambda-1)-(5\lambda+1)-17(13\lambda-2)=-66 \Rightarrow \lambda = \ldots$ | M1 | Substitutes parametric form into equation of $\Pi_2$ and solves for $\lambda$ |
| $\lambda = \dfrac{85}{198}$ | A1 | Correct exact value of $\lambda$ depending on method |
| $DE = \sqrt{\left(2\times\frac{85}{198}\right)^2+\left(5\times\frac{85}{198}\right)^2+\left(13\times\frac{85}{198}\right)^2}$ | dM1 | Correct method to find numerical expression for distance $DE$. Requires previous method mark |
| $DE = \dfrac{85\sqrt{22}}{66}$ | A1 | Correct exact answer. Or $p = \frac{85}{66}$ or $1\frac{19}{66}$. Not $DE = -\frac{85\sqrt{22}}{66}$ |
---
# Question 6(c): Finding $q$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AF}\cdot\overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}1\\1\\q-2\end{pmatrix}\cdot\begin{pmatrix}2\\5\\13\end{pmatrix} = 2+5+13q-26$ | M1 | Correct method for vector between $F$ and $A$, $B$ or $C$ and finds scalar product with normal, or attempts scalar triple product to obtain linear expression in $q$. At least 2 correct elements for triple product |
| $\frac{1}{6}(13q-19) = \pm 12 \Rightarrow q = \ldots$ | dM1 | Sets $\frac{1}{6}$ of expression in $q$ equal to $\pm 12$ (or equivalent e.g. expression equal to $\pm 72$) and proceeds to value of $q$ |
| $q = 7,\ -\dfrac{53}{13}$ | A1 | Correct values. Allow exact equivalents e.g. $-4\frac{1}{13}$ |\begin{enumerate}
\item The points $A , B$ and $C$ have coordinates ( $3,2,2$ ), ( $- 1,1,3$ ) and ( $- 2,4,2$ ) respectively. The plane $\Pi _ { 1 }$ contains the points $A , B$ and $C$\\
(a) Determine a Cartesian equation of $\Pi _ { 1 }$
\end{enumerate}
Given that
\begin{itemize}
\item point $D$ has coordinates $( - 1,1 , - 2 )$
\item line $l$ passes through $D$ and is perpendicular to $\Pi _ { 1 }$
\item plane $\Pi _ { 2 }$ has equation $\mathbf { r } . ( 14 \mathbf { i } - \mathbf { j } - 17 \mathbf { k } ) = - 66$
\item $I$ meets $\Pi _ { 2 }$ at the point $E$\\
(b) show that $D E = p \sqrt { 22 }$ where $p$ is a rational number to be determined.
\end{itemize}
The point $F$ has coordinates ( $4,3 , q$ ) where $q$ is a constant.\\
Given that $A , B , C$ and $F$ are the vertices of a tetrahedron of volume 12\\
(c) determine the possible values of $q$
\hfill \mbox{\textit{Edexcel F3 2024 Q6 [12]}}