| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.3 This is a standard Core Pure AS vectors question requiring routine application of formulas: (a) finding angle between line and plane using direction vector and normal vector, (b) finding perpendicular distance from point to line using the standard formula. Both parts are textbook exercises with straightforward calculations, making it slightly easier than average for A-level. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04i Shortest distance: between a point and a line |
| V349 SIHI NI IMIMM ION OC | VJYV SIHIL NI LIIIM ION OO | VJYV SIHIL NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts scalar product between direction of \(W\) and normal to road, uses trigonometry to find angle | M1 | Must identify correct strategy |
| \(\left(\begin{pmatrix}1\\2\\-3\end{pmatrix}-\begin{pmatrix}-1\\-1\\-3\end{pmatrix}\right)\cdot\begin{pmatrix}3\\-5\\-18\end{pmatrix}=-9\) | M1 | Allow sign slips as long as intention is clear |
| e.g. \(\begin{pmatrix}2\\3\\0\end{pmatrix}\cdot\begin{pmatrix}3\\-5\\-18\end{pmatrix}=-9\) | A1 | Accept equivalent correct dot products |
| \(\sqrt{(2)^2+(3)^2+(0)^2}\,\sqrt{(3)^2+(-5)^2+(-18)^2}\cos\alpha = -9\) | M1 | Full correct method for acute angle |
| \(\theta = 90 - \arccos\!\left(\dfrac{9}{\sqrt{13}\sqrt{358}}\right)\) or \(\theta = \arcsin\!\left(\dfrac{9}{\sqrt{13}\sqrt{358}}\right)\) | ||
| Angle \(= 7.58°\) (3sf) or \(0.132\) radians (3sf) | A1 | Must state units; allow \(-7.58°\) or \(-0.132\) rad |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W: \begin{pmatrix}-1\\-1\\-3\end{pmatrix}+t\begin{pmatrix}2\\3\\0\end{pmatrix}\) or \(\begin{pmatrix}1\\2\\-3\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\0\end{pmatrix}\) | B1ft | Follow through direction vector for \(W\) from (a) |
| \(C\) to \(W\): \(\left\{\begin{pmatrix}-1\\-1\\-3\end{pmatrix}+t\begin{pmatrix}2\\3\\0\end{pmatrix}-\begin{pmatrix}-1\\-2\\0\end{pmatrix}\right\}\) | M1 | Identifies and forms vector connecting \(C\) to \(W\) |
| \(\begin{pmatrix}2t\\3t+1\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\3\\0\end{pmatrix}=0 \Rightarrow t=\ldots\) | M1 | Scalar product of \(C\) to \(W\) with direction of \(W\); or minimise distance squared |
| \(t=-\dfrac{3}{13}\) or \(\lambda=-\dfrac{16}{13}\) \(\Rightarrow (C\text{ to }W)_{\min}\) is \(-\dfrac{6}{13}\mathbf{i}+\dfrac{4}{13}\mathbf{j}-3\mathbf{k}\) | A1 | Correct vector or correct completion of square |
| \(d=\sqrt{\left(-\dfrac{6}{13}\right)^2+\left(\dfrac{4}{13}\right)^2+(-3)^2}\) or \(d=\sqrt{\dfrac{121}{13}}\) | ddM1 | Correct Pythagoras on vector \(CW\); dependent on both previous M marks |
| Shortest length of pipe \(= 305\) or \(305\) cm or \(3.05\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{AC}=\begin{pmatrix}0\\1\\-3\end{pmatrix}\), \(\mathbf{AB}=\begin{pmatrix}2\\3\\0\end{pmatrix}\) | B1ft | Follow through direction vector for \(W\) |
| \(\mathbf{AC}\cdot\mathbf{AB}=\begin{pmatrix}0\\1\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\3\\0\end{pmatrix}=3\) | M1 | |
| \(\cos CAB = \dfrac{3}{\sqrt{10}\sqrt{13}} \Rightarrow CAB = \ldots\) | M1 | |
| \(CAB = 74.74\ldots°\) | A1 | |
| \(d = \sqrt{10}\sin 74.74\ldots°\) | ddM1 | Dependent on both M marks |
| Shortest length \(= 305\) or \(305\) cm or \(3.05\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{AC}=\begin{pmatrix}0\\1\\-3\end{pmatrix}\), \(\mathbf{AB}=\begin{pmatrix}2\\3\\0\end{pmatrix}\) | B1ft | Follow through direction vector for \(W\) |
| \(\mathbf{AC}\times\mathbf{AB}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&-1&3\\2&3&0\end{vmatrix}=\begin{pmatrix}-9\\6\\2\end{pmatrix}\) | M1 | |
| \( | \mathbf{AC}\times\mathbf{AB} | =\sqrt{9^2+6^2+2^2}=\ldots\) |
| \(=11\) | A1 | |
| \(d=\dfrac{11}{ | \mathbf{AB} | }=\dfrac{11}{\sqrt{2^2+3^2}}=\ldots\) |
| Shortest length \(= 305\) or \(305\) cm or \(3.05\) m | A1 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts scalar product between direction of $W$ and normal to road, uses trigonometry to find angle | M1 | Must identify correct strategy |
| $\left(\begin{pmatrix}1\\2\\-3\end{pmatrix}-\begin{pmatrix}-1\\-1\\-3\end{pmatrix}\right)\cdot\begin{pmatrix}3\\-5\\-18\end{pmatrix}=-9$ | M1 | Allow sign slips as long as intention is clear |
| e.g. $\begin{pmatrix}2\\3\\0\end{pmatrix}\cdot\begin{pmatrix}3\\-5\\-18\end{pmatrix}=-9$ | A1 | Accept equivalent correct dot products |
| $\sqrt{(2)^2+(3)^2+(0)^2}\,\sqrt{(3)^2+(-5)^2+(-18)^2}\cos\alpha = -9$ | M1 | Full correct method for acute angle |
| $\theta = 90 - \arccos\!\left(\dfrac{9}{\sqrt{13}\sqrt{358}}\right)$ or $\theta = \arcsin\!\left(\dfrac{9}{\sqrt{13}\sqrt{358}}\right)$ | | |
| Angle $= 7.58°$ (3sf) or $0.132$ radians (3sf) | A1 | Must state units; allow $-7.58°$ or $-0.132$ rad |
### Part (b) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W: \begin{pmatrix}-1\\-1\\-3\end{pmatrix}+t\begin{pmatrix}2\\3\\0\end{pmatrix}$ or $\begin{pmatrix}1\\2\\-3\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\0\end{pmatrix}$ | B1ft | Follow through direction vector for $W$ from (a) |
| $C$ to $W$: $\left\{\begin{pmatrix}-1\\-1\\-3\end{pmatrix}+t\begin{pmatrix}2\\3\\0\end{pmatrix}-\begin{pmatrix}-1\\-2\\0\end{pmatrix}\right\}$ | M1 | Identifies and forms vector connecting $C$ to $W$ |
| $\begin{pmatrix}2t\\3t+1\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\3\\0\end{pmatrix}=0 \Rightarrow t=\ldots$ | M1 | Scalar product of $C$ to $W$ with direction of $W$; or minimise distance squared |
| $t=-\dfrac{3}{13}$ or $\lambda=-\dfrac{16}{13}$ $\Rightarrow (C\text{ to }W)_{\min}$ is $-\dfrac{6}{13}\mathbf{i}+\dfrac{4}{13}\mathbf{j}-3\mathbf{k}$ | A1 | Correct vector or correct completion of square |
| $d=\sqrt{\left(-\dfrac{6}{13}\right)^2+\left(\dfrac{4}{13}\right)^2+(-3)^2}$ or $d=\sqrt{\dfrac{121}{13}}$ | ddM1 | Correct Pythagoras on vector $CW$; dependent on both previous M marks |
| Shortest length of pipe $= 305$ or $305$ cm or $3.05$ m | A1 | |
### Part (b) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AC}=\begin{pmatrix}0\\1\\-3\end{pmatrix}$, $\mathbf{AB}=\begin{pmatrix}2\\3\\0\end{pmatrix}$ | B1ft | Follow through direction vector for $W$ |
| $\mathbf{AC}\cdot\mathbf{AB}=\begin{pmatrix}0\\1\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\3\\0\end{pmatrix}=3$ | M1 | |
| $\cos CAB = \dfrac{3}{\sqrt{10}\sqrt{13}} \Rightarrow CAB = \ldots$ | M1 | |
| $CAB = 74.74\ldots°$ | A1 | |
| $d = \sqrt{10}\sin 74.74\ldots°$ | ddM1 | Dependent on both M marks |
| Shortest length $= 305$ or $305$ cm or $3.05$ m | A1 | |
### Part (b) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AC}=\begin{pmatrix}0\\1\\-3\end{pmatrix}$, $\mathbf{AB}=\begin{pmatrix}2\\3\\0\end{pmatrix}$ | B1ft | Follow through direction vector for $W$ |
| $\mathbf{AC}\times\mathbf{AB}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&-1&3\\2&3&0\end{vmatrix}=\begin{pmatrix}-9\\6\\2\end{pmatrix}$ | M1 | |
| $|\mathbf{AC}\times\mathbf{AB}|=\sqrt{9^2+6^2+2^2}=\ldots$ | M1 | |
| $=11$ | A1 | |
| $d=\dfrac{11}{|\mathbf{AB}|}=\dfrac{11}{\sqrt{2^2+3^2}}=\ldots$ | ddM1 | Dependent on both M marks |
| Shortest length $= 305$ or $305$ cm or $3.05$ m | A1 | |\begin{enumerate}
\item Part of the mains water system for a housing estate consists of water pipes buried beneath the ground surface. The water pipes are modelled as straight line segments. One water pipe, $W$, is buried beneath a particular road. With respect to a fixed origin $O$, the road surface is modelled as a plane with equation $3 x - 5 y - 18 z = 7$, and $W$ passes through the points $A ( - 1 , - 1 , - 3 )$ and $B ( 1,2 , - 3 )$. The units are in metres.\\
(a) Use the model to calculate the acute angle between $W$ and the road surface.
\end{enumerate}
A point $C ( - 1 , - 2,0 )$ lies on the road. A section of water pipe needs to be connected to $W$ from $C$.\\
(b) Using the model, find, to the nearest cm, the shortest length of pipe needed to connect $C$ to $W$.
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\hfill \mbox{\textit{Edexcel CP AS 2018 Q4 [11]}}