| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | 3D geometry applications |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question testing standard techniques: verifying a plane equation by substitution, finding a normal vector via cross product, calculating angles between planes, and finding line-plane intersections. All parts follow routine procedures with no novel problem-solving required. The coal seam context adds no mathematical complexity. Slightly easier than average due to the highly structured guidance through each step. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At A: \(3\times0+2\times0+20\times(-15)+300=0\) | M1 | substituting co-ords into equation of plane |
| At B: \(3\times100+2\times0+20\times(-30)+300=0\) | A2,1,0 | for ABC OR using two vectors in the plane form vector product M1A1 then \(3x+2y+20z = c = -300\) A1 OR using vector equation of plane M1, elim both parameters M1, A1 |
| At C: \(3\times0+2\times100+20\times(-25)+300=0\) | ||
| So ABC has equation \(3x+2y+20z+300=0\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{DE} = \begin{pmatrix}100\\0\\-10\end{pmatrix}\), \(\overrightarrow{DF} = \begin{pmatrix}0\\100\\5\end{pmatrix}\) | B1B1 | |
| \(\begin{pmatrix}100\\0\\-10\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\20\end{pmatrix} = 100\times2+0\times-1+-10\times20=200-200=0\) | B1 | need evaluation |
| \(\begin{pmatrix}0\\100\\5\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\20\end{pmatrix} = 0\times2+100\times-1+5\times20=-100+100=0\) | B1 | need evaluation |
| Equation of plane is \(2x-y+20z=c\) | M1 | |
| At D (say) \(c=20\times-40=-800\); so equation is \(2x-y+20z+800=0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Angle is \(\theta\) where \(\cos\theta = \frac{\begin{pmatrix}2\\-1\\20\end{pmatrix}\cdot\begin{pmatrix}3\\2\\20\end{pmatrix}}{\sqrt{2^2+(-1)^2+20^2}\sqrt{3^2+2^2+20^2}} = \frac{404}{\sqrt{405}\sqrt{413}}\) | M1 | formula with correct vectors |
| A1 | top | |
| A1 | bottom | |
| \(\theta = 8.95°\) | A1cao | (or \(0.156\) radians) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| RS: \(\mathbf{r} = \begin{pmatrix}15\\34\\0\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\20\end{pmatrix}\) | B1 | \(\begin{pmatrix}15\\34\\0\end{pmatrix}+\ldots\) |
| \(= \begin{pmatrix}15+3\lambda\\34+2\lambda\\20\lambda\end{pmatrix}\) | B1 | \(\ldots+\lambda\begin{pmatrix}3\\2\\20\end{pmatrix}\) |
| \(3(15+3\lambda)+2(34+2\lambda)+20\cdot20\lambda+300=0\) | M1 | solving with plane |
| \(45+9\lambda+68+4\lambda+400\lambda+300=0\) | ||
| \(413+413\lambda=0\) | A1 | \(\lambda=-1\) |
| \(\lambda=-1\); so S is \((12, 32, -20)\) | A1 | cao |
## Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At A: $3\times0+2\times0+20\times(-15)+300=0$ | M1 | substituting co-ords into equation of plane |
| At B: $3\times100+2\times0+20\times(-30)+300=0$ | A2,1,0 | for ABC **OR** using two vectors in the plane form vector product M1A1 then $3x+2y+20z = c = -300$ A1 **OR** using vector equation of plane M1, elim both parameters M1, A1 |
| At C: $3\times0+2\times100+20\times(-25)+300=0$ | | |
| So ABC has equation $3x+2y+20z+300=0$ | [3] | |
---
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{DE} = \begin{pmatrix}100\\0\\-10\end{pmatrix}$, $\overrightarrow{DF} = \begin{pmatrix}0\\100\\5\end{pmatrix}$ | B1B1 | |
| $\begin{pmatrix}100\\0\\-10\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\20\end{pmatrix} = 100\times2+0\times-1+-10\times20=200-200=0$ | B1 | need evaluation |
| $\begin{pmatrix}0\\100\\5\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\20\end{pmatrix} = 0\times2+100\times-1+5\times20=-100+100=0$ | B1 | need evaluation |
| Equation of plane is $2x-y+20z=c$ | M1 | |
| At D (say) $c=20\times-40=-800$; so equation is $2x-y+20z+800=0$ | A1 | |
---
## Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Angle is $\theta$ where $\cos\theta = \frac{\begin{pmatrix}2\\-1\\20\end{pmatrix}\cdot\begin{pmatrix}3\\2\\20\end{pmatrix}}{\sqrt{2^2+(-1)^2+20^2}\sqrt{3^2+2^2+20^2}} = \frac{404}{\sqrt{405}\sqrt{413}}$ | M1 | formula with correct vectors |
| | A1 | top |
| | A1 | bottom |
| $\theta = 8.95°$ | A1cao | (or $0.156$ radians) |
---
## Question 8(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| RS: $\mathbf{r} = \begin{pmatrix}15\\34\\0\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\20\end{pmatrix}$ | B1 | $\begin{pmatrix}15\\34\\0\end{pmatrix}+\ldots$ |
| $= \begin{pmatrix}15+3\lambda\\34+2\lambda\\20\lambda\end{pmatrix}$ | B1 | $\ldots+\lambda\begin{pmatrix}3\\2\\20\end{pmatrix}$ |
| $3(15+3\lambda)+2(34+2\lambda)+20\cdot20\lambda+300=0$ | M1 | solving with plane |
| $45+9\lambda+68+4\lambda+400\lambda+300=0$ | | |
| $413+413\lambda=0$ | A1 | $\lambda=-1$ |
| $\lambda=-1$; so S is $(12, 32, -20)$ | A1 | cao |
---8 The upper and lower surfaces of a coal seam are modelled as planes ABC and DEF, as shown in Fig. 8. All dimensions are metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8ad99e2a-4cef-40b3-af8d-673b97536227-03_1004_1397_493_374}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
Relative to axes $\mathrm { O } x$ (due east), $\mathrm { O } y$ (due north) and $\mathrm { O } z$ (vertically upwards), the coordinates of the points are as follows.\\
A: (0, 0, -15)\\
B: (100, 0, -30)\\
C: (0, 100, -25)\\
D: (0, 0, -40)\\
E: (100, 0, -50)\\
F: (0, 100, -35)\\
(i) Verify that the cartesian equation of the plane ABC is $3 x + 2 y + 20 z + 300 = 0$.\\
(ii) Find the vectors $\overrightarrow { \mathrm { DE } }$ and $\overrightarrow { \mathrm { DF } }$. Show that the vector $2 \mathbf { i } - \mathbf { j } + 20 \mathbf { k }$ is perpendicular to each of these vectors. Hence find the cartesian equation of the plane DEF .\\
(iii) By calculating the angle between their normal vectors, find the angle between the planes ABC and DEF.
It is decided to drill down to the seam from a point $\mathrm { R } ( 15,34,0 )$ in a line perpendicular to the upper surface of the seam. This line meets the plane ABC at the point S .\\
(iv) Write down a vector equation of the line RS.
Calculate the coordinates of S.
\hfill \mbox{\textit{OCR MEI C4 2008 Q8 [18]}}