| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Standard +0.8 This is a Further Maths question requiring finding the line of intersection of two planes (involving cross product of normals and solving simultaneous equations), then finding the distance between two skew lines parallel to this intersection. While the techniques are standard for FM students, the multi-step nature and requirement to work with 3D geometry across two parts elevates it above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| VIIV STHI NI JINM ION OC | VIAV SIHI NI JMAM/ION OC | VIAV SIHL NI JIIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{n} = \begin{pmatrix}1\\-5\\3\end{pmatrix} \times \begin{pmatrix}3\\-2\\2\end{pmatrix} = \begin{pmatrix}-10+6\\-(2-9)\\-2+15\end{pmatrix}\) | M1 | Attempt vector product between normal vectors |
| \(= \begin{pmatrix}-4\\7\\13\end{pmatrix}\) | A1 | Correct vector |
| \(x=0 \Rightarrow -5y+3z=11,\; -2y+2z=7 \Rightarrow y=-\frac{1}{4}, z=\frac{13}{4}\) or \(y=0 \Rightarrow x+3z=11,\; 3x+2z=7 \Rightarrow x=-\frac{1}{7}, z=\frac{26}{7}\) or \(z=0 \Rightarrow x-5y=11,\; 3x-2y=7 \Rightarrow x=1, y=-2\) | M1 | Correct strategy to find a point on \(l\) |
| Correct position vector of point on \(l\) | A1 | Correct position vector of point on \(l\) |
| \(\mathbf{r} = \mathbf{i} - 2\mathbf{j} + \lambda(-4\mathbf{i}+7\mathbf{j}+13\mathbf{k})\) | A1ft | Correct equation; follow through their position and direction vectors but must be "\(\mathbf{r} =\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=11+5y-3z\); substituting: \(\Rightarrow y - \frac{7z}{13} = -\frac{26}{13}\) | M1 | Eliminate one variable |
| \(x=11+5\!\left(-\frac{26}{13}+\frac{7z}{13}\right) \Rightarrow z = \frac{13-13x}{4}\) | A1 | Obtain 2 correct expressions for one of the variables |
| \(\dfrac{x-1}{-\dfrac{4}{13}} = \dfrac{y+2}{\dfrac{7}{13}} = z\) | M1A1 | M1 Obtain a Cartesian equation for \(l\); A1 Correct equation |
| \(\mathbf{r} = (\mathbf{i}-2\mathbf{j}) + \lambda\!\left(-\frac{4}{13}\mathbf{i}+\frac{7}{13}\mathbf{j}+\mathbf{k}\right)\) oe | A1ft | Deduce a vector equation for \(l\); follow through their Cartesian equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}3\\2\\1\end{pmatrix}-\begin{pmatrix}2\\0\\3\end{pmatrix}=\begin{pmatrix}1\\2\\-2\end{pmatrix}\) | B1 | Correct vector joining \(P\) to \(Q\) |
| \(\begin{pmatrix}-4\\7\\13\end{pmatrix}\times\begin{pmatrix}1\\2\\-2\end{pmatrix}=\begin{pmatrix}-40\\5\\-15\end{pmatrix}\) | M1 | Attempt vector product between direction of \(l\) and \(\mathbf{i}+2\mathbf{j}-2\mathbf{k}\) |
| Correct vector \((-40, 5, -15)\) | A1 | Correct vector |
| \(\sin\theta = \dfrac{ | -40\mathbf{i}+5\mathbf{j}-15\mathbf{k} | }{ |
| \(d = \dfrac{ | -40\mathbf{i}+5\mathbf{j}-15\mathbf{k} | }{ |
| \(d = \dfrac{5\sqrt{481}}{39}\) | A1 | cao; allow equivalent exact forms e.g. \(d=\dfrac{5\sqrt{74}}{\sqrt{234}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r}_m = \begin{pmatrix}2\\0\\3\end{pmatrix}+\lambda\begin{pmatrix}-4/7\\1\\13/7\end{pmatrix}\) or \(\mathbf{r}_n = \begin{pmatrix}3\\2\\1\end{pmatrix}+\mu\begin{pmatrix}-4/7\\1\\13/7\end{pmatrix}\) | B1ft | Vector equation for either line with their direction vector from (a) |
| \(\overrightarrow{NP}=\begin{pmatrix}-1+\frac{4}{7}\mu\\-2-\mu\\2-\frac{13}{7}\mu\end{pmatrix}\) | Uses either \(P\) and parametric form of a point on \(n\), OR \(Q\) and parametric form of a point on \(m\) | |
| \(\begin{pmatrix}-1+\frac{4}{7}\mu\\-2-\mu\\2-\frac{13}{7}\mu\end{pmatrix}\cdot\begin{pmatrix}-4/7\\1\\13/7\end{pmatrix}=0\) | M1A1 | M1: Forms scalar product of \(NP\) and direction vector of \(l\) equated to zero; A1: Correct vectors |
| \(\Rightarrow \mu = \dfrac{56}{117}\) | M1 | Solves |
| \(d = \sqrt{\left(-\frac{85}{117}\right)^2+\left(-\frac{290}{117}\right)^2+\left(\frac{10}{9}\right)^2} = \dfrac{5\sqrt{481}}{39}\) | A1 | Obtains the correct distance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct vector \(PQ\) | B1 | |
| \(\begin{pmatrix}1\\2\\-2\end{pmatrix}\cdot\begin{pmatrix}-4\\7\\13\end{pmatrix} = \begin{vmatrix}1\\2\\-2\end{vmatrix}\begin{vmatrix}-4\\7\\13\end{vmatrix}\cos\theta\) | M1 | Forms scalar product and attempts to evaluate LHS |
| \(\cos\theta = \dfrac{-16}{3\sqrt{234}}\) | A1 | Correct value for \(\cos\theta\) exact or decimal |
| \(d= | PQ | \sin\theta = 3\sqrt{1-\left(\dfrac{-16}{3\sqrt{234}}\right)^2} = \dfrac{5\sqrt{74}}{\sqrt{234}}\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{pmatrix}1\\-5\\3\end{pmatrix} \times \begin{pmatrix}3\\-2\\2\end{pmatrix} = \begin{pmatrix}-10+6\\-(2-9)\\-2+15\end{pmatrix}$ | M1 | Attempt vector product between normal vectors |
| $= \begin{pmatrix}-4\\7\\13\end{pmatrix}$ | A1 | Correct vector |
| $x=0 \Rightarrow -5y+3z=11,\; -2y+2z=7 \Rightarrow y=-\frac{1}{4}, z=\frac{13}{4}$ **or** $y=0 \Rightarrow x+3z=11,\; 3x+2z=7 \Rightarrow x=-\frac{1}{7}, z=\frac{26}{7}$ **or** $z=0 \Rightarrow x-5y=11,\; 3x-2y=7 \Rightarrow x=1, y=-2$ | M1 | Correct strategy to find a point on $l$ |
| Correct position vector of point on $l$ | A1 | Correct position vector of point on $l$ |
| $\mathbf{r} = \mathbf{i} - 2\mathbf{j} + \lambda(-4\mathbf{i}+7\mathbf{j}+13\mathbf{k})$ | A1ft | Correct equation; follow through their position and direction vectors but must be "$\mathbf{r} =$" |
**ALT:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=11+5y-3z$; substituting: $\Rightarrow y - \frac{7z}{13} = -\frac{26}{13}$ | M1 | Eliminate one variable |
| $x=11+5\!\left(-\frac{26}{13}+\frac{7z}{13}\right) \Rightarrow z = \frac{13-13x}{4}$ | A1 | Obtain 2 correct expressions for one of the variables |
| $\dfrac{x-1}{-\dfrac{4}{13}} = \dfrac{y+2}{\dfrac{7}{13}} = z$ | M1A1 | M1 Obtain a Cartesian equation for $l$; A1 Correct equation |
| $\mathbf{r} = (\mathbf{i}-2\mathbf{j}) + \lambda\!\left(-\frac{4}{13}\mathbf{i}+\frac{7}{13}\mathbf{j}+\mathbf{k}\right)$ oe | A1ft | Deduce a vector equation for $l$; follow through their Cartesian equation |
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3\\2\\1\end{pmatrix}-\begin{pmatrix}2\\0\\3\end{pmatrix}=\begin{pmatrix}1\\2\\-2\end{pmatrix}$ | B1 | Correct vector joining $P$ to $Q$ |
| $\begin{pmatrix}-4\\7\\13\end{pmatrix}\times\begin{pmatrix}1\\2\\-2\end{pmatrix}=\begin{pmatrix}-40\\5\\-15\end{pmatrix}$ | M1 | Attempt vector product between direction of $l$ and $\mathbf{i}+2\mathbf{j}-2\mathbf{k}$ |
| Correct vector $(-40, 5, -15)$ | A1 | Correct vector |
| $\sin\theta = \dfrac{|-40\mathbf{i}+5\mathbf{j}-15\mathbf{k}|}{|-4\mathbf{i}+7\mathbf{j}+13\mathbf{k}||\mathbf{i}+2\mathbf{j}-2\mathbf{k}|}$, $d=|\overrightarrow{PQ}|\sin\theta$ | | Angle between $PQ$ and line $n$ |
| $d = \dfrac{|-40\mathbf{i}+5\mathbf{j}-15\mathbf{k}|}{|-4\mathbf{i}+7\mathbf{j}+13\mathbf{k}|} = \dfrac{1}{\sqrt{234}}\sqrt{40^2+5^2+15^2}$ | M1 | Fully correct method for the distance |
| $d = \dfrac{5\sqrt{481}}{39}$ | A1 | cao; allow equivalent exact forms e.g. $d=\dfrac{5\sqrt{74}}{\sqrt{234}}$ |
**ALT 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r}_m = \begin{pmatrix}2\\0\\3\end{pmatrix}+\lambda\begin{pmatrix}-4/7\\1\\13/7\end{pmatrix}$ or $\mathbf{r}_n = \begin{pmatrix}3\\2\\1\end{pmatrix}+\mu\begin{pmatrix}-4/7\\1\\13/7\end{pmatrix}$ | B1ft | Vector equation for either line with their direction vector from (a) |
| $\overrightarrow{NP}=\begin{pmatrix}-1+\frac{4}{7}\mu\\-2-\mu\\2-\frac{13}{7}\mu\end{pmatrix}$ | | Uses either $P$ and parametric form of a point on $n$, OR $Q$ and parametric form of a point on $m$ |
| $\begin{pmatrix}-1+\frac{4}{7}\mu\\-2-\mu\\2-\frac{13}{7}\mu\end{pmatrix}\cdot\begin{pmatrix}-4/7\\1\\13/7\end{pmatrix}=0$ | M1A1 | M1: Forms scalar product of $NP$ and direction vector of $l$ equated to zero; A1: Correct vectors |
| $\Rightarrow \mu = \dfrac{56}{117}$ | M1 | Solves |
| $d = \sqrt{\left(-\frac{85}{117}\right)^2+\left(-\frac{290}{117}\right)^2+\left(\frac{10}{9}\right)^2} = \dfrac{5\sqrt{481}}{39}$ | A1 | Obtains the correct distance |
**ALT 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct vector $PQ$ | B1 | |
| $\begin{pmatrix}1\\2\\-2\end{pmatrix}\cdot\begin{pmatrix}-4\\7\\13\end{pmatrix} = \begin{vmatrix}1\\2\\-2\end{vmatrix}\begin{vmatrix}-4\\7\\13\end{vmatrix}\cos\theta$ | M1 | Forms scalar product and attempts to evaluate LHS |
| $\cos\theta = \dfrac{-16}{3\sqrt{234}}$ | A1 | Correct value for $\cos\theta$ exact or decimal |
| $d=|PQ|\sin\theta = 3\sqrt{1-\left(\dfrac{-16}{3\sqrt{234}}\right)^2} = \dfrac{5\sqrt{74}}{\sqrt{234}}$ | M1A1 | M1: Correct method for distance; A1: Correct **exact** distance |8. The plane $\Pi _ { 1 }$ has equation
$$x - 5 y + 3 z = 11$$
The plane $\Pi _ { 2 }$ has equation
$$3 x - 2 y + 2 z = 7$$
The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ intersect in the line $l$.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for $l$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$ where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors and $\lambda$ is a scalar parameter.
The point $P ( 2,0,3 )$ lies on $\Pi _ { 1 }$
The line $m$, which passes through $P$, is parallel to $l$.
The point $Q ( 3,2,1 )$ lies on $\Pi _ { 2 }$\\
The line $n$, which passes through $Q$, is also parallel to $l$.
\item Find, in exact simplified form, the shortest distance between $m$ and $n$.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIIV STHI NI JINM ION OC & VIAV SIHI NI JMAM/ION OC & VIAV SIHL NI JIIYM ION OO \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2020 Q8 [10]}}