| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.2 This is a multi-part applied vectors question requiring standard techniques (point-to-line distance, skew lines distance formula, parametric equations) with straightforward calculations. While it has many parts and requires careful bookkeeping with the context, each individual step uses routine Further Maths methods without requiring novel insight or particularly complex reasoning. The applied context adds mild difficulty but the mathematical content is typical FP3 material. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}3\\4\\8\end{pmatrix} \times \begin{pmatrix}2\\1\\0\end{pmatrix} = \begin{pmatrix}-8\\16\\-5\end{pmatrix}\) | M1 | Appropriate vector product |
| Correctly evaluated | A1 | Correctly evaluated |
| Distance is \(\dfrac{\sqrt{8^2+16^2+5^2}}{\sqrt{2^2+1^2+0^2}} = \dfrac{\sqrt{345}}{\sqrt{5}}\) | A1 | Dividing by \(\sqrt{2^2+1^2+0^2}\); Sign error in vector product can earn M1A0A1A1 |
| \(= \sqrt{69} \approx 8.31\) (km) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}3+2\lambda\\4+\lambda\\8\end{pmatrix}\) is perpendicular to \(\begin{pmatrix}2\\1\\0\end{pmatrix}\) | M1 | |
| \(2(3+2\lambda)+(4+\lambda)=0\) | A1 | |
| \(\lambda=-2\); shortest vector is \(\begin{pmatrix}-1\\2\\8\end{pmatrix}\) | A1 | |
| Distance is \(\sqrt{1^2+2^2+8^2} = \sqrt{69}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2\\1\\0\end{pmatrix} \times \begin{pmatrix}-8\\-4\\-1\end{pmatrix} = \begin{pmatrix}-1\\2\\0\end{pmatrix}\) | M1 | Vector product of directions |
| Correctly evaluated | A1 | Correctly evaluated |
| \(\begin{pmatrix}77\\36\\2\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\0\end{pmatrix} = -5\) or \(\begin{pmatrix}3\\4\\8\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\0\end{pmatrix} = -5\) | M1, A1 | Appropriate scalar product; For \((-) 5\); Dependent on previous M1 |
| Distance is \(\dfrac{5}{\sqrt{1^2+2^2+0^2}}\) | M1 | Dividing by \(\sqrt{1^2+2^2+0^2}\); Dependent on M1M1 |
| \(= \dfrac{5}{\sqrt{5}} = \sqrt{5} = 2.236... = 2.24\) (km) (correct to 3 sf) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}80-8\mu\\40-4\mu\\10-\mu\end{pmatrix} - \begin{pmatrix}3+2\lambda\\4+\lambda\\8\end{pmatrix} = \begin{pmatrix}77-2\lambda-8\mu\\36-\lambda-4\mu\\2-\mu\end{pmatrix}\) | B1 | |
| \(\overrightarrow{AB} \cdot \begin{pmatrix}2\\1\\0\end{pmatrix} = 0\) and \(\overrightarrow{AB} \cdot \begin{pmatrix}-8\\-4\\-1\end{pmatrix} = 0\) | M1 | |
| \(2(77-2\lambda-8\mu)+(36-\lambda-4\mu)=0\) | A1 | |
| \(-8(77-2\lambda-8\mu)-4(36-\lambda-4\mu)-(2-\mu)=0\); \(5\lambda+20\mu=190\); \(20\lambda+81\mu=762\) and hence \(\lambda=30,\ \mu=2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB}\) is parallel to \(\begin{pmatrix}-1\\2\\0\end{pmatrix}\) | M1 | |
| \(36-\lambda-4\mu = -2(77-2\lambda-8\mu)\); \(2-\mu=0\); \(5\lambda+20\mu=190\); \(2-\mu=0\) and hence \(\lambda=30,\ \mu=2\) | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = (\pm)\begin{pmatrix}1\\-2\\0\end{pmatrix}\) | M1 | |
| \(77-2\lambda-8\mu=1\); \(36-\lambda-4\mu=-2\); \(2-\mu=0\) | A1 | |
| \(\lambda=30,\ \mu=2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Closest points are \(A(63,34,8)\) and \(B(64,32,8)\); P is at A at time | A1 | |
| \(t_1 = \dfrac{\sqrt{5}\lambda}{900}\) or \(\dfrac{\sqrt{60^2+30^2+0^2}}{900} = \dfrac{\sqrt{4500}}{900} = \dfrac{\sqrt{5}}{30}\) | M1 | Method for finding time |
| Q is at B at time \(t_2 = \dfrac{9\mu}{270}\) or \(\dfrac{\sqrt{16^2+8^2+2^2}}{270} = \dfrac{18}{270} = \dfrac{2}{30}\); These times are different, so the planes are never this close | E1 | Both times correct, and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{q} = \begin{pmatrix}80\\40\\10\end{pmatrix} + \dfrac{270t}{9}\begin{pmatrix}-8\\-4\\-1\end{pmatrix} = \begin{pmatrix}80-240t\\40-120t\\10-30t\end{pmatrix}\) | M1, A1 | Speed and unit direction vectors |
| \(\mathbf{r} = \begin{pmatrix}29\\19\\5.5\end{pmatrix} + \dfrac{285t}{19}\begin{pmatrix}18\\6\\1\end{pmatrix} = \begin{pmatrix}29+270t\\19+90t\\5.5+15t\end{pmatrix}\) | A1 | |
| Q, R will collide if \(80-240t=29+270t\); \(40-120t=19+90t\); \(10-30t=5.5+15t\) | M1 | One equation sufficient for M1 |
| All three equations have solution \(t=0.1\) | A1, A1 | For \(t=0.1\) obtained; Shown to satisfy all three correctly shown; Point of collision is \((56,28,7)\) |
| Planes would collide; so R must alter course or speed | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Paths intersect if \(80-8\mu=29+18\nu\); \(40-4\mu=19+6\nu\); \(10-\mu=5.5+\nu\) | M1, A1 | Two correct equations |
| \(\mu=3,\ \nu=1.5\) | A1 | Must check third equation |
| All equations satisfied, paths intersect at \(X(56,28,7)\) | A1 | |
| Q is at X at time \(t=\dfrac{9\mu}{270}\) or \(\dfrac{\sqrt{24^2+12^2+3^2}}{270}=\dfrac{27}{270}=0.1\) | M1, A1 | Method for finding time; For \(t=0.1\) |
| R is at X at time \(t=\dfrac{19\nu}{285}\) or \(\dfrac{\sqrt{27^2+9^2+1.5^2}}{285}=\dfrac{28.5}{285}=0.1\) | ||
| Planes would collide; so R must alter course or speed | E1 | Correctly shown; Dependent on all previous marks |
# Question 1:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}3\\4\\8\end{pmatrix} \times \begin{pmatrix}2\\1\\0\end{pmatrix} = \begin{pmatrix}-8\\16\\-5\end{pmatrix}$ | M1 | Appropriate vector product |
| Correctly evaluated | A1 | Correctly evaluated |
| Distance is $\dfrac{\sqrt{8^2+16^2+5^2}}{\sqrt{2^2+1^2+0^2}} = \dfrac{\sqrt{345}}{\sqrt{5}}$ | A1 | Dividing by $\sqrt{2^2+1^2+0^2}$; Sign error in vector product can earn M1A0A1A1 |
| $= \sqrt{69} \approx 8.31$ (km) | A1 | |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}3+2\lambda\\4+\lambda\\8\end{pmatrix}$ is perpendicular to $\begin{pmatrix}2\\1\\0\end{pmatrix}$ | M1 | |
| $2(3+2\lambda)+(4+\lambda)=0$ | A1 | |
| $\lambda=-2$; shortest vector is $\begin{pmatrix}-1\\2\\8\end{pmatrix}$ | A1 | |
| Distance is $\sqrt{1^2+2^2+8^2} = \sqrt{69}$ | A1 | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\1\\0\end{pmatrix} \times \begin{pmatrix}-8\\-4\\-1\end{pmatrix} = \begin{pmatrix}-1\\2\\0\end{pmatrix}$ | M1 | Vector product of directions |
| Correctly evaluated | A1 | Correctly evaluated |
| $\begin{pmatrix}77\\36\\2\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\0\end{pmatrix} = -5$ or $\begin{pmatrix}3\\4\\8\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\0\end{pmatrix} = -5$ | M1, A1 | Appropriate scalar product; For $(-) 5$; Dependent on previous M1 |
| Distance is $\dfrac{5}{\sqrt{1^2+2^2+0^2}}$ | M1 | Dividing by $\sqrt{1^2+2^2+0^2}$; Dependent on M1M1 |
| $= \dfrac{5}{\sqrt{5}} = \sqrt{5} = 2.236... = 2.24$ (km) (correct to 3 sf) | E1 | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}80-8\mu\\40-4\mu\\10-\mu\end{pmatrix} - \begin{pmatrix}3+2\lambda\\4+\lambda\\8\end{pmatrix} = \begin{pmatrix}77-2\lambda-8\mu\\36-\lambda-4\mu\\2-\mu\end{pmatrix}$ | B1 | |
| $\overrightarrow{AB} \cdot \begin{pmatrix}2\\1\\0\end{pmatrix} = 0$ and $\overrightarrow{AB} \cdot \begin{pmatrix}-8\\-4\\-1\end{pmatrix} = 0$ | M1 | |
| $2(77-2\lambda-8\mu)+(36-\lambda-4\mu)=0$ | A1 | |
| $-8(77-2\lambda-8\mu)-4(36-\lambda-4\mu)-(2-\mu)=0$; $5\lambda+20\mu=190$; $20\lambda+81\mu=762$ and hence $\lambda=30,\ \mu=2$ | A1 | |
**OR method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB}$ is parallel to $\begin{pmatrix}-1\\2\\0\end{pmatrix}$ | M1 | |
| $36-\lambda-4\mu = -2(77-2\lambda-8\mu)$; $2-\mu=0$; $5\lambda+20\mu=190$; $2-\mu=0$ and hence $\lambda=30,\ \mu=2$ | A1, A1 | |
**OR method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = (\pm)\begin{pmatrix}1\\-2\\0\end{pmatrix}$ | M1 | |
| $77-2\lambda-8\mu=1$; $36-\lambda-4\mu=-2$; $2-\mu=0$ | A1 | |
| $\lambda=30,\ \mu=2$ | A1 | |
Closest points are $A(63, 34, 8)$ and $B(64, 32, 8)$
| Answer | Marks | Guidance |
|--------|-------|----------|
| Closest points are $A(63,34,8)$ and $B(64,32,8)$; P is at A at time | A1 | |
| $t_1 = \dfrac{\sqrt{5}\lambda}{900}$ or $\dfrac{\sqrt{60^2+30^2+0^2}}{900} = \dfrac{\sqrt{4500}}{900} = \dfrac{\sqrt{5}}{30}$ | M1 | Method for finding time |
| Q is at B at time $t_2 = \dfrac{9\mu}{270}$ or $\dfrac{\sqrt{16^2+8^2+2^2}}{270} = \dfrac{18}{270} = \dfrac{2}{30}$; These times are different, so the planes are never this close | E1 | Both times correct, and conclusion |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{q} = \begin{pmatrix}80\\40\\10\end{pmatrix} + \dfrac{270t}{9}\begin{pmatrix}-8\\-4\\-1\end{pmatrix} = \begin{pmatrix}80-240t\\40-120t\\10-30t\end{pmatrix}$ | M1, A1 | Speed and unit direction vectors |
| $\mathbf{r} = \begin{pmatrix}29\\19\\5.5\end{pmatrix} + \dfrac{285t}{19}\begin{pmatrix}18\\6\\1\end{pmatrix} = \begin{pmatrix}29+270t\\19+90t\\5.5+15t\end{pmatrix}$ | A1 | |
| Q, R will collide if $80-240t=29+270t$; $40-120t=19+90t$; $10-30t=5.5+15t$ | M1 | One equation sufficient for M1 |
| All three equations have solution $t=0.1$ | A1, A1 | For $t=0.1$ obtained; Shown to satisfy all three correctly shown; Point of collision is $(56,28,7)$ |
| Planes would collide; so R must alter course or speed | E1 | |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Paths intersect if $80-8\mu=29+18\nu$; $40-4\mu=19+6\nu$; $10-\mu=5.5+\nu$ | M1, A1 | Two correct equations |
| $\mu=3,\ \nu=1.5$ | A1 | Must check third equation |
| All equations satisfied, paths intersect at $X(56,28,7)$ | A1 | |
| Q is at X at time $t=\dfrac{9\mu}{270}$ or $\dfrac{\sqrt{24^2+12^2+3^2}}{270}=\dfrac{27}{270}=0.1$ | M1, A1 | Method for finding time; For $t=0.1$ |
| R is at X at time $t=\dfrac{19\nu}{285}$ or $\dfrac{\sqrt{27^2+9^2+1.5^2}}{285}=\dfrac{28.5}{285}=0.1$ | | |
| Planes would collide; so R must alter course or speed | E1 | Correctly shown; Dependent on all previous marks |
---1 Positions in space around an aerodrome are modelled by a coordinate system with a point on the runway as the origin, O . The $x$-axis is east, the $y$-axis is north and the $z$-axis is vertically upwards. Units of distance are kilometres. Units of time are hours.\\
At time $t = 0$, an aeroplane, P , is at $( 3,4,8 )$ and is travelling in a direction $\left( \begin{array} { l } 2 \\ 1 \\ 0 \end{array} \right)$ at a constant speed of\\
$900 \mathrm { kmh } ^ { - 1 }$.\\
(i) Find the least distance of the path of P from the point O .
At time $t = 0$, a second aeroplane, Q , is at $( 80,40,10 )$. It is travelling in a straight line towards the point O . Its speed is constant at $270 \mathrm { kmh } ^ { - 1 }$.\\
(ii) Show that the shortest distance between the paths of the two aeroplanes is 2.24 km correct to three significant figures.\\
(iii) By finding the points on the paths where the shortest distance occurs and the times at which the aeroplanes are at these points, show that in fact the aeroplanes are never this close.\\
(iv) A third aeroplane, R , is at position $( 29,19,5.5 )$ at time $t = 0$ and is travelling at $285 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ in a direction $\left( \begin{array} { c } 18 \\ 6 \\ 1 \end{array} \right)$. Given that Q is in the process of landing and cannot change course, show that R needs to be instructed to alter course or change speed.
\hfill \mbox{\textit{OCR MEI FP3 2016 Q1 [24]}}