| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line-plane intersection and related angle/perpendicularity |
| Difficulty | Standard +0.3 This is a standard A-level vectors question requiring routine techniques: finding a plane equation from three points using cross product, calculating angle between planes using normal vectors, and finding intersection distance. Part (d) requires minimal critical thinking. All methods are textbook procedures with no novel insight needed, making it slightly easier than average. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Any two of: \(\pm k\overrightarrow{AB} = \pm k(5\mathbf{i}+25\mathbf{j}+5\mathbf{k})\), \(\pm k\overrightarrow{AC} = \pm k(-15\mathbf{i}+15\mathbf{j}-10\mathbf{k})\), \(\pm k\overrightarrow{BC} = \pm k(-20\mathbf{i}-10\mathbf{j}-15\mathbf{k})\) | M1 | 3.3 |
| Let normal \(= a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\); \((a\mathbf{i}+b\mathbf{j}+c\mathbf{k})\cdot(\mathbf{i}+5\mathbf{j}+\mathbf{k})=0\), \((a\mathbf{i}+b\mathbf{j}+c\mathbf{k})\cdot(-3\mathbf{i}+3\mathbf{j}-2\mathbf{k})=0 \Rightarrow a+5b+c=0,\ -3a+3b-2c=0 \Rightarrow a=\ldots,\ b=\ldots,\ c=\ldots\) | M1 | 1.1b |
| Alternative (cross product): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&5&1\\-3&3&-2\end{vmatrix} = (-10-3)\mathbf{i}-(-2+3)\mathbf{j}+(3+15)\mathbf{k}\) | ||
| \(\mathbf{n} = k(-13\mathbf{i} - \mathbf{j} + 18\mathbf{k})\) | A1 | 1.1b |
| \((-13\mathbf{i}-\mathbf{j}+18\mathbf{k})\cdot(10\mathbf{i}+5\mathbf{j}-50\mathbf{k}) = \ldots\) | M1 | 1.1b |
| \(\mathbf{r}\cdot(13\mathbf{i}+\mathbf{j}-18\mathbf{k}) = 1035\) or \(\mathbf{r}\cdot(-13\mathbf{i}-\mathbf{j}+18\mathbf{k}) = -1035\); \(\mathbf{r}\cdot(325\mathbf{i}+25\mathbf{j}-450\mathbf{k}) = 25875\) | A1 | 2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Attempts scalar product between normal vector and \(\mathbf{k}\), uses trigonometry | M1 | 3.1b |
| \((-13\mathbf{i}-\mathbf{j}+18\mathbf{k})\cdot\mathbf{k} = -18 = \sqrt{13^2+1^2+18^2}\cos\alpha\) | M1 | 1.1b |
| \(\cos\alpha = \frac{-18}{\sqrt{494}} \Rightarrow \alpha = 144.08\ldots \Rightarrow \theta = 36°\) | A1 | 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Distance required is \( | \lambda | \) where \(\begin{pmatrix}13\\1\\-18\end{pmatrix}\cdot\begin{pmatrix}5\\12\\\lambda\end{pmatrix} = 1035\) |
| \( | \lambda | = 53.2\ \text{m}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| E.g. The mineral layer will not be perfectly flat/smooth and will not form a plane; The mineral layer will have a depth and this should be taken into account | B1 | 3.5b |
# Question 6:
## Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| Any two of: $\pm k\overrightarrow{AB} = \pm k(5\mathbf{i}+25\mathbf{j}+5\mathbf{k})$, $\pm k\overrightarrow{AC} = \pm k(-15\mathbf{i}+15\mathbf{j}-10\mathbf{k})$, $\pm k\overrightarrow{BC} = \pm k(-20\mathbf{i}-10\mathbf{j}-15\mathbf{k})$ | M1 | 3.3 |
| Let normal $= a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$; $(a\mathbf{i}+b\mathbf{j}+c\mathbf{k})\cdot(\mathbf{i}+5\mathbf{j}+\mathbf{k})=0$, $(a\mathbf{i}+b\mathbf{j}+c\mathbf{k})\cdot(-3\mathbf{i}+3\mathbf{j}-2\mathbf{k})=0 \Rightarrow a+5b+c=0,\ -3a+3b-2c=0 \Rightarrow a=\ldots,\ b=\ldots,\ c=\ldots$ | M1 | 1.1b |
| Alternative (cross product): $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&5&1\\-3&3&-2\end{vmatrix} = (-10-3)\mathbf{i}-(-2+3)\mathbf{j}+(3+15)\mathbf{k}$ | | |
| $\mathbf{n} = k(-13\mathbf{i} - \mathbf{j} + 18\mathbf{k})$ | A1 | 1.1b |
| $(-13\mathbf{i}-\mathbf{j}+18\mathbf{k})\cdot(10\mathbf{i}+5\mathbf{j}-50\mathbf{k}) = \ldots$ | M1 | 1.1b |
| $\mathbf{r}\cdot(13\mathbf{i}+\mathbf{j}-18\mathbf{k}) = 1035$ or $\mathbf{r}\cdot(-13\mathbf{i}-\mathbf{j}+18\mathbf{k}) = -1035$; $\mathbf{r}\cdot(325\mathbf{i}+25\mathbf{j}-450\mathbf{k}) = 25875$ | A1 | 2.5 |
## Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| Attempts scalar product between normal vector and $\mathbf{k}$, uses trigonometry | M1 | 3.1b |
| $(-13\mathbf{i}-\mathbf{j}+18\mathbf{k})\cdot\mathbf{k} = -18 = \sqrt{13^2+1^2+18^2}\cos\alpha$ | M1 | 1.1b |
| $\cos\alpha = \frac{-18}{\sqrt{494}} \Rightarrow \alpha = 144.08\ldots \Rightarrow \theta = 36°$ | A1 | 3.2a |
## Part (c):
| Working | Mark | Notes |
|---------|------|-------|
| Distance required is $|\lambda|$ where $\begin{pmatrix}13\\1\\-18\end{pmatrix}\cdot\begin{pmatrix}5\\12\\\lambda\end{pmatrix} = 1035$ | M1 | 3.4 |
| $|\lambda| = 53.2\ \text{m}$ | A1 | 1.1b |
## Part (d):
| Working | Mark | Notes |
|---------|------|-------|
| E.g. The mineral layer will **not** be perfectly flat/smooth and will not form a plane; The mineral layer will have a **depth** and this should be taken into account | B1 | 3.5b |
**Total: (5+3+2+1) = 11 marks**\begin{enumerate}
\item A mining company has identified a mineral layer below ground.
\end{enumerate}
The mining company wishes to drill down to reach the mineral layer and models the situation as follows.
With respect to a fixed origin $O$,
\begin{itemize}
\item the ground is modelled as a horizontal plane with equation $z = 0$
\item the mineral layer is modelled as part of the plane containing the points $A ( 10,5 , - 50 ) , B ( 15,30 , - 45 )$ and $C ( - 5,20 , - 60 )$, where the units are in metres\\
(a) Determine an equation for the plane containing $A , B$ and $C$, giving your answer in the form r.n $= d$\\
(b) Determine, according to the model, the acute angle between the ground and the plane containing the mineral layer. Give your answer to the nearest degree.
\end{itemize}
The mining company plans to drill vertically downwards from the point $( 5,12,0 )$ on the ground to reach the mineral layer.\\
(c) Using the model, determine, in metres to 1 decimal place, the distance the mining company will need to drill in order to reach the mineral layer.\\
(d) State a limitation of the assumption that the mineral layer can be modelled as a plane.
\hfill \mbox{\textit{Edexcel CP AS 2021 Q6 [11]}}