| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| 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: vector subtraction, magnitude calculation, vector equation of a line, angle between line and plane using dot product, and finding intersection point by substitution. All parts follow routine procedures with no novel insight required, though the context and multiple steps make it slightly above average difficulty for typical A-level work. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}100-(-200)\\200-100\\100-0\end{pmatrix} = \begin{pmatrix}300\\100\\100\end{pmatrix}\) | E1 | |
| \(AB = \sqrt{300^2 + 100^2 + 100^2} = 332\) m | M1, A1 [3] | accept surds |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = \begin{pmatrix}-200\\100\\0\end{pmatrix} + \lambda\begin{pmatrix}300\\100\\100\end{pmatrix}\) | B1, B1 | oe |
| Angle between \(\begin{pmatrix}3\\1\\1\end{pmatrix}\) and \(\begin{pmatrix}0\\0\\1\end{pmatrix}\) | M1 | and \(\begin{pmatrix}0\\0\\1\end{pmatrix}\) |
| \(\cos\theta = \frac{3\times0+1\times0+1\times1}{\sqrt{11}\sqrt{1}} = \frac{1}{\sqrt{11}}\) | M1, A1 | complete scalar product method (including cosine) for correct vectors |
| \(\theta = 72.45°\) | A1 [6] | \(72.5°\) or better; accept 1.26 radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Meets plane of layer when \((-200+300\lambda)+2(100+100\lambda)+3\times100\lambda = 320\) | M1 | |
| \(800\lambda = 320\), \(\lambda = 2/5\) | A1 | |
| \(\mathbf{r} = \begin{pmatrix}-200\\100\\0\end{pmatrix} + \frac{2}{5}\begin{pmatrix}300\\100\\100\end{pmatrix} = \begin{pmatrix}-80\\140\\40\end{pmatrix}\) | M1 | |
| meets layer at \((-80, 140, 40)\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Normal to plane is \(\begin{pmatrix}1\\2\\3\end{pmatrix}\) | B1 | |
| Angle between \(\begin{pmatrix}3\\1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\2\\3\end{pmatrix}\) | ||
| \(\cos\theta = \frac{3\times1+1\times2+1\times3}{\sqrt{11}\sqrt{14}} = \frac{8}{\sqrt{11}\sqrt{14}} = 0.6446...\) | M1, A1 | complete method |
| \(\theta = 49.86°\) | A1 | |
| angle with layer \(= 40.1°\) | A1 [5] | ft \(90°-\theta\); accept radians |
# Question 7:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}100-(-200)\\200-100\\100-0\end{pmatrix} = \begin{pmatrix}300\\100\\100\end{pmatrix}$ | E1 | |
| $AB = \sqrt{300^2 + 100^2 + 100^2} = 332$ m | M1, A1 [3] | accept surds |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = \begin{pmatrix}-200\\100\\0\end{pmatrix} + \lambda\begin{pmatrix}300\\100\\100\end{pmatrix}$ | B1, B1 | oe |
| Angle between $\begin{pmatrix}3\\1\\1\end{pmatrix}$ and $\begin{pmatrix}0\\0\\1\end{pmatrix}$ | M1 | and $\begin{pmatrix}0\\0\\1\end{pmatrix}$ |
| $\cos\theta = \frac{3\times0+1\times0+1\times1}{\sqrt{11}\sqrt{1}} = \frac{1}{\sqrt{11}}$ | M1, A1 | complete scalar product method (including cosine) for correct vectors |
| $\theta = 72.45°$ | A1 [6] | $72.5°$ or better; accept 1.26 radians |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Meets plane of layer when $(-200+300\lambda)+2(100+100\lambda)+3\times100\lambda = 320$ | M1 | |
| $800\lambda = 320$, $\lambda = 2/5$ | A1 | |
| $\mathbf{r} = \begin{pmatrix}-200\\100\\0\end{pmatrix} + \frac{2}{5}\begin{pmatrix}300\\100\\100\end{pmatrix} = \begin{pmatrix}-80\\140\\40\end{pmatrix}$ | M1 | |
| meets layer at $(-80, 140, 40)$ | A1 [4] | |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Normal to plane is $\begin{pmatrix}1\\2\\3\end{pmatrix}$ | B1 | |
| Angle between $\begin{pmatrix}3\\1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\2\\3\end{pmatrix}$ | | |
| $\cos\theta = \frac{3\times1+1\times2+1\times3}{\sqrt{11}\sqrt{14}} = \frac{8}{\sqrt{11}\sqrt{14}} = 0.6446...$ | M1, A1 | complete method |
| $\theta = 49.86°$ | A1 | |
| angle with layer $= 40.1°$ | A1 [5] | ft $90°-\theta$; accept radians |
---7 A straight pipeline AB passes through a mountain. With respect to axes $\mathrm { O } x y z$, with $\mathrm { O } x$ due East, $\mathrm { O } y$ due North and $\mathrm { O } z$ vertically upwards, A has coordinates $( - 200,100,0 )$ and B has coordinates $( 100,200,100 )$, where units are metres.\\
(i) Verify that $\overrightarrow { \mathrm { AB } } = \left( \begin{array} { l } 300 \\ 100 \\ 100 \end{array} \right)$ and find the length of the pipeline.\\
(ii) Write down a vector equation of the line AB , and calculate the angle it makes with the vertical.
A thin flat layer of hard rock runs through the mountain. The equation of the plane containing this layer is $x + 2 y + 3 z = 320$.\\
(iii) Find the coordinates of the point where the pipeline meets the layer of rock.\\
(iv) By calculating the angle between the line AB and the normal to the plane of the layer, find the angle at which the pipeline cuts through the layer.
\hfill \mbox{\textit{OCR MEI C4 2010 Q7 [18]}}