| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (a) proving lines are skew by showing they don't intersect and aren't parallel, (b) applying the skew lines distance formula involving cross product and dot product, and (c) finding a plane equation given constraints. While the techniques are standard for FP3, the question requires careful vector manipulation across multiple parts and is more demanding than typical A-level pure maths questions due to the Further Maths content. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\3\\-1\end{pmatrix} \neq k\begin{pmatrix}1\\1\\3\end{pmatrix}\) | B1 | Shows lines are not parallel. If they say "different direction vectors", the direction vectors must be identified. |
| i: \(1+2\lambda = -1+\mu\) (1) | ||
| j: \(3\lambda = 4+\mu\) (2) | ||
| k: \(2-\lambda = 1+3\mu\) (3) | ||
| Solving any pair: e.g. (1) and (2) yields \(\lambda=6, \mu=14\); (1) and (3) yields \(\lambda=-\frac{5}{7}, \mu=\frac{4}{7}\); (2) and (3) yields \(\lambda=\frac{13}{10}, \mu=-\frac{1}{10}\) | M1 | Attempts to solve a pair of equations to find at least one of either \(\lambda=\ldots\) or \(\mu=\ldots\) |
| Checking third equation shows contradiction, e.g. Checking (3): \(-4\neq43\); Checking (2): \(-\frac{15}{7}\neq\frac{32}{7}\); Checking (1): \(3.6\neq-1.1\) | M1 | Attempts to show a contradiction |
| So the lines are not parallel and do not intersect so the lines are skew | A1 | All complete and with no errors and conclusion. If they have already stated "not parallel" there is no need to repeat this. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}1\\1\\3\end{pmatrix}=\pm\begin{pmatrix}10\\-7\\-1\end{pmatrix}\) | M1 | Attempt cross product of direction vectors. If no method is shown, 2 components should be correct. |
| Correct vector \(\begin{pmatrix}10\\-7\\-1\end{pmatrix}\) | A1 | Correct vector |
| \(\pm\left(\begin{pmatrix}1\\0\\2\end{pmatrix}-\begin{pmatrix}-1\\4\\1\end{pmatrix}\right)\bullet\pm\begin{pmatrix}10\\-7\\-1\end{pmatrix}=\pm(20+28-1)=\pm47\) | M1 | Attempt scalar product between the difference of the position vectors and their normal vector. |
| \(d=\left\ | \frac{\pm47}{\sqrt{10^2+7^2+1^2}}\right\ | =\frac{47}{\sqrt{150}}\) |
| Any equivalent or awrt \(3.84\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}1\\1\\3\end{pmatrix}=\begin{pmatrix}10\\-7\\-1\end{pmatrix}\) | M1 | Attempt cross product of direction vectors |
| Correct vector | A1 | |
| \(\begin{pmatrix}10\\-7\\-1\end{pmatrix}\bullet\begin{pmatrix}1\\0\\2\end{pmatrix}=8\), \(\begin{pmatrix}10\\-7\\-1\end{pmatrix}\bullet\begin{pmatrix}-1\\4\\1\end{pmatrix}=-39\) | M1 | Attempt equation of both planes |
| \(d=\frac{8}{\sqrt{10^2+7^2+1^2}}-\frac{-39}{\sqrt{10^2+7^2+1^2}}=\frac{47}{\sqrt{150}}\) | M1 | Correct completion |
| Any equivalent e.g. \(\frac{47\sqrt{6}}{30}\) or awrt \(3.84\), but must be positive | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| General chord: \(\begin{pmatrix}2+2\lambda-\mu\\-4+3\lambda-\mu\\1-\lambda-3\mu\end{pmatrix}\); set dot products with both direction vectors \(=0\) giving two equations in two unknowns | M1 | Finds a general chord between the 2 lines and attempts scalar product with directions, sets \(=0\) |
| \(2\lambda-11\mu=-1\); \(14\lambda-2\mu=9\) | ||
| \(\lambda=\frac{101}{150},\ \mu=\frac{16}{75}\) | A1 | Correct values |
| Finds ends of chord or substitutes into chord vector: \(\begin{pmatrix}2+2\lambda-\mu\\-4+3\lambda-\mu\\1-\lambda-3\mu\end{pmatrix}=\begin{pmatrix}\frac{47}{15}\\-\frac{329}{150}\\-\frac{47}{150}\end{pmatrix}\) | M1 | Uses values to find ends of chord or substitutes into chord vector |
| \(d=\sqrt{\left(\frac{47}{15}\right)^2+\left(\frac{329}{150}\right)^2+\left(\frac{47}{150}\right)^2}=\frac{47\sqrt{6}}{30}\) | M1 | Correct completion by finding distance between the two points |
| Any equivalent e.g. \(\frac{47\sqrt{6}}{30}\) or awrt \(3.84\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}1\\1\\3\end{pmatrix}=\pm\begin{pmatrix}10\\-7\\-1\end{pmatrix}\) | M1 | Attempt cross product of direction vectors |
| Correct vector | A1 | |
| Sets chord \(=k\begin{pmatrix}10\\-7\\-1\end{pmatrix}\), giving 3 equations in 3 unknowns, solves to find \(k=\frac{47}{150}\) | M1 | Finds common chord parallel to normal vector |
| \(d=\sqrt{\left(\frac{47}{15}\right)^2+\left(\frac{329}{150}\right)^2+\left(\frac{47}{150}\right)^2}=\frac{47\sqrt{6}}{30}\) | M1 | Correct completion by finding length of vector |
| Any equivalent e.g. \(\frac{47\sqrt{6}}{30}\) or awrt \(3.84\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}3\\8\\13\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}2\\8\\11\end{pmatrix}\) | M1 | Attempt another non-parallel vector in \(\Pi\) |
| \(\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}2\\8\\11\end{pmatrix}=\begin{pmatrix}41\\-24\\10\end{pmatrix}\) | dM1 | Attempt cross product of two non-parallel vectors in the plane. If method not shown, at least 2 components correct. Dependent on first M mark. |
| \(\begin{pmatrix}41\\-24\\10\end{pmatrix}\bullet\begin{pmatrix}1\\0\\2\end{pmatrix}=\ldots\) or \(\begin{pmatrix}41\\-24\\10\end{pmatrix}\bullet\begin{pmatrix}3\\8\\13\end{pmatrix}=\ldots\) | ddM1 | Attempt scalar product with a point in the plane. Dependent on both previous method marks. |
| \(41x-24y+10z=61\) | A1 | Any multiple but must be a Cartesian equation. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}3\\8\\13\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}2\\8\\11\end{pmatrix}\) | M1 | Attempt another vector in \(\Pi\) |
| \(r=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\-1\end{pmatrix}+\mu\begin{pmatrix}2\\8\\11\end{pmatrix}\) i.e. \(x=1+2\lambda+2\mu\) (1), \(y=3\lambda+8\mu\) (2), \(z=2-\lambda+11\mu\) (3) | dM1 | Forms vector equation of plane. Dependent on first M mark. |
| \((1)+2(3): x+2z=5+24\mu\); \((2)+3(3): 3z+y=6+41\mu\) | ddM1 | Eliminates \(\lambda\) or \(\mu\). Dependent on both previous method marks. |
| \(\frac{3z+y-6}{41}=\frac{x+2z-5}{24}\) | A1 | Any correct equation but must be a correct Cartesian equation. |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\3\\-1\end{pmatrix} \neq k\begin{pmatrix}1\\1\\3\end{pmatrix}$ | B1 | Shows lines are not parallel. If they say "different direction vectors", the direction vectors must be identified. |
| **i:** $1+2\lambda = -1+\mu$ (1) | | |
| **j:** $3\lambda = 4+\mu$ (2) | | |
| **k:** $2-\lambda = 1+3\mu$ (3) | | |
| Solving any pair: e.g. (1) and (2) yields $\lambda=6, \mu=14$; (1) and (3) yields $\lambda=-\frac{5}{7}, \mu=\frac{4}{7}$; (2) and (3) yields $\lambda=\frac{13}{10}, \mu=-\frac{1}{10}$ | M1 | Attempts to solve a pair of equations to find at least one of either $\lambda=\ldots$ or $\mu=\ldots$ |
| Checking third equation shows contradiction, e.g. Checking (3): $-4\neq43$; Checking (2): $-\frac{15}{7}\neq\frac{32}{7}$; Checking (1): $3.6\neq-1.1$ | M1 | Attempts to show a contradiction |
| So the lines are not parallel and do not intersect so the lines are skew | A1 | All complete and with no errors and conclusion. If they have already stated "not parallel" there is no need to repeat this. |
**(4 marks)**
---
## Part (b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}1\\1\\3\end{pmatrix}=\pm\begin{pmatrix}10\\-7\\-1\end{pmatrix}$ | M1 | Attempt cross product of direction vectors. If no method is shown, 2 components should be correct. |
| Correct vector $\begin{pmatrix}10\\-7\\-1\end{pmatrix}$ | A1 | Correct vector |
| $\pm\left(\begin{pmatrix}1\\0\\2\end{pmatrix}-\begin{pmatrix}-1\\4\\1\end{pmatrix}\right)\bullet\pm\begin{pmatrix}10\\-7\\-1\end{pmatrix}=\pm(20+28-1)=\pm47$ | M1 | Attempt scalar product between the difference of the position vectors and their normal vector. |
| $d=\left\|\frac{\pm47}{\sqrt{10^2+7^2+1^2}}\right\|=\frac{47}{\sqrt{150}}$ | M1 | Correct completion. Divides scalar product by modulus of their vector product. |
| Any equivalent or awrt $3.84$ | A1 | |
**(5 marks)**
---
## Part (b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}1\\1\\3\end{pmatrix}=\begin{pmatrix}10\\-7\\-1\end{pmatrix}$ | M1 | Attempt cross product of direction vectors |
| Correct vector | A1 | |
| $\begin{pmatrix}10\\-7\\-1\end{pmatrix}\bullet\begin{pmatrix}1\\0\\2\end{pmatrix}=8$, $\begin{pmatrix}10\\-7\\-1\end{pmatrix}\bullet\begin{pmatrix}-1\\4\\1\end{pmatrix}=-39$ | M1 | Attempt equation of both planes |
| $d=\frac{8}{\sqrt{10^2+7^2+1^2}}-\frac{-39}{\sqrt{10^2+7^2+1^2}}=\frac{47}{\sqrt{150}}$ | M1 | Correct completion |
| Any equivalent e.g. $\frac{47\sqrt{6}}{30}$ or awrt $3.84$, but must be positive | A1 | |
**(5 marks)**
---
## Part (b) Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| General chord: $\begin{pmatrix}2+2\lambda-\mu\\-4+3\lambda-\mu\\1-\lambda-3\mu\end{pmatrix}$; set dot products with both direction vectors $=0$ giving two equations in two unknowns | M1 | Finds a general chord between the 2 lines and attempts scalar product with directions, sets $=0$ |
| $2\lambda-11\mu=-1$; $14\lambda-2\mu=9$ | | |
| $\lambda=\frac{101}{150},\ \mu=\frac{16}{75}$ | A1 | Correct values |
| Finds ends of chord or substitutes into chord vector: $\begin{pmatrix}2+2\lambda-\mu\\-4+3\lambda-\mu\\1-\lambda-3\mu\end{pmatrix}=\begin{pmatrix}\frac{47}{15}\\-\frac{329}{150}\\-\frac{47}{150}\end{pmatrix}$ | M1 | Uses values to find ends of chord or substitutes into chord vector |
| $d=\sqrt{\left(\frac{47}{15}\right)^2+\left(\frac{329}{150}\right)^2+\left(\frac{47}{150}\right)^2}=\frac{47\sqrt{6}}{30}$ | M1 | Correct completion by finding distance between the two points |
| Any equivalent e.g. $\frac{47\sqrt{6}}{30}$ or awrt $3.84$ | A1 | |
**(5 marks)**
---
## Part (b) Way 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}1\\1\\3\end{pmatrix}=\pm\begin{pmatrix}10\\-7\\-1\end{pmatrix}$ | M1 | Attempt cross product of direction vectors |
| Correct vector | A1 | |
| Sets chord $=k\begin{pmatrix}10\\-7\\-1\end{pmatrix}$, giving 3 equations in 3 unknowns, solves to find $k=\frac{47}{150}$ | M1 | Finds common chord parallel to normal vector |
| $d=\sqrt{\left(\frac{47}{15}\right)^2+\left(\frac{329}{150}\right)^2+\left(\frac{47}{150}\right)^2}=\frac{47\sqrt{6}}{30}$ | M1 | Correct completion by finding length of vector |
| Any equivalent e.g. $\frac{47\sqrt{6}}{30}$ or awrt $3.84$ | A1 | |
**(5 marks)**
---
## Part (c) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3\\8\\13\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}2\\8\\11\end{pmatrix}$ | M1 | Attempt another non-parallel vector in $\Pi$ |
| $\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}2\\8\\11\end{pmatrix}=\begin{pmatrix}41\\-24\\10\end{pmatrix}$ | dM1 | Attempt cross product of two non-parallel vectors in the plane. If method not shown, at least 2 components correct. **Dependent on first M mark.** |
| $\begin{pmatrix}41\\-24\\10\end{pmatrix}\bullet\begin{pmatrix}1\\0\\2\end{pmatrix}=\ldots$ or $\begin{pmatrix}41\\-24\\10\end{pmatrix}\bullet\begin{pmatrix}3\\8\\13\end{pmatrix}=\ldots$ | ddM1 | Attempt scalar product with a point in the plane. **Dependent on both previous method marks.** |
| $41x-24y+10z=61$ | A1 | Any multiple but must be a Cartesian equation. |
**(4 marks)**
---
## Part (c) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3\\8\\13\end{pmatrix}-\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}2\\8\\11\end{pmatrix}$ | M1 | Attempt another vector in $\Pi$ |
| $r=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\-1\end{pmatrix}+\mu\begin{pmatrix}2\\8\\11\end{pmatrix}$ i.e. $x=1+2\lambda+2\mu$ (1), $y=3\lambda+8\mu$ (2), $z=2-\lambda+11\mu$ (3) | dM1 | Forms vector equation of plane. **Dependent on first M mark.** |
| $(1)+2(3): x+2z=5+24\mu$; $(2)+3(3): 3z+y=6+41\mu$ | ddM1 | Eliminates $\lambda$ or $\mu$. **Dependent on both previous method marks.** |
| $\frac{3z+y-6}{41}=\frac{x+2z-5}{24}$ | A1 | Any correct equation but must be a correct Cartesian equation. |
**(4 marks)**
**Total: 13 marks**6. The line $l _ { 1 }$ has equation
$$\mathbf { r } = \mathbf { i } + 2 \mathbf { k } + \lambda ( 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } )$$
where $\lambda$ is a scalar parameter.
The line $l _ { 2 }$ has equation
$$\frac { x + 1 } { 1 } = \frac { y - 4 } { 1 } = \frac { z - 1 } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Prove that the lines $l _ { 1 }$ and $l _ { 2 }$ are skew.
\item Find the shortest distance between the lines $l _ { 1 }$ and $l _ { 2 }$
The plane $\Pi$ contains $l _ { 1 }$ and intersects $l _ { 2 }$ at the point $( 3,8,13 )$.
\item Find a cartesian equation for the plane $\Pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2018 Q6 [13]}}