| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | 3D geometry applications |
| Difficulty | Standard +0.3 This is a standard multi-part 3D vectors question covering routine techniques: writing a line equation, finding a normal vector via cross product, deriving a plane equation, finding intersection points, and calculating angles. While it requires multiple steps and careful arithmetic, each part follows textbook methods without requiring novel insight or particularly challenging problem-solving, making it slightly easier than average. |
| 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 plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\underline{r} = \begin{pmatrix}12\\10\\10\end{pmatrix} + \lambda\begin{pmatrix}-2\\-2\\-3\end{pmatrix}\) | M1 A1 | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}0\\3\\1\end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix}5\\0\\-1\end{pmatrix}\) | M1 A1 | Find vectors on surface |
| \(\overrightarrow{AB}\cdot\begin{pmatrix}3\\-5\\15\end{pmatrix} = 0\), \(\overrightarrow{AC}\cdot\begin{pmatrix}3\\-5\\15\end{pmatrix} = 0\) | M1 A1 | Scalar product |
| Since \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are non-parallel vectors in the plane and perpendicular to the vector given, it must be perpendicular to the plane | E1 | For explanation of result \(= 0\) |
| \((\underline{r} - \underline{a})\cdot\begin{pmatrix}3\\-5\\15\end{pmatrix} = 0\) | M1 A1 | Or any other valid procedure |
| \(\Rightarrow 3x - 5y + 15z = 13\) | A1 | |
| Total: 8 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute parametric equation of line into plane | M1 | |
| \(3(12-2\lambda) - 5(10-2\lambda) + 15(10-3\lambda) = 13\) | A1 | Substitution correct |
| \(\Rightarrow \lambda = 3\) | A1 | \(\lambda\) correct |
| Hence the point is \((6, 4, 1)\) | A1 | Point |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta = \frac{\left | \begin{pmatrix}3\\-5\\15\end{pmatrix}\cdot\begin{pmatrix}-2\\-2\\-3\end{pmatrix}\right | }{\left |
| \(= \left | \frac{-6+10-45}{\sqrt{9+25+225}\cdot\sqrt{4+4+9}}\right | \) |
| \(90 - \theta = \arcsin\left(\frac{41}{\sqrt{259}\sqrt{17}}\right)\) | A1 | |
| \(= 38.16\ldots°\); Required angle is \(38.2°\) to 1 d.p. | A1 | |
| Total: 4 marks |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\underline{r} = \begin{pmatrix}12\\10\\10\end{pmatrix} + \lambda\begin{pmatrix}-2\\-2\\-3\end{pmatrix}$ | M1 A1 | |
| **Total: 2 marks** | | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}0\\3\\1\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}5\\0\\-1\end{pmatrix}$ | M1 A1 | Find vectors on surface |
| $\overrightarrow{AB}\cdot\begin{pmatrix}3\\-5\\15\end{pmatrix} = 0$, $\overrightarrow{AC}\cdot\begin{pmatrix}3\\-5\\15\end{pmatrix} = 0$ | M1 A1 | Scalar product |
| Since $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are non-parallel vectors in the plane and perpendicular to the vector given, it must be perpendicular to the plane | E1 | For explanation of result $= 0$ |
| $(\underline{r} - \underline{a})\cdot\begin{pmatrix}3\\-5\\15\end{pmatrix} = 0$ | M1 A1 | Or any other valid procedure |
| $\Rightarrow 3x - 5y + 15z = 13$ | A1 | |
| **Total: 8 marks** | | |
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute parametric equation of line into plane | M1 | |
| $3(12-2\lambda) - 5(10-2\lambda) + 15(10-3\lambda) = 13$ | A1 | Substitution correct |
| $\Rightarrow \lambda = 3$ | A1 | $\lambda$ correct |
| Hence the point is $(6, 4, 1)$ | A1 | Point |
| **Total: 5 marks** | | |
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = \frac{\left|\begin{pmatrix}3\\-5\\15\end{pmatrix}\cdot\begin{pmatrix}-2\\-2\\-3\end{pmatrix}\right|}{\left|\begin{pmatrix}3\\-5\\15\end{pmatrix}\right|\left|\begin{pmatrix}-2\\-2\\-3\end{pmatrix}\right|}$ | M1 | Scalar product |
| $= \left|\frac{-6+10-45}{\sqrt{9+25+225}\cdot\sqrt{4+4+9}}\right|$ | M1 | Substitute in formula |
| $90 - \theta = \arcsin\left(\frac{41}{\sqrt{259}\sqrt{17}}\right)$ | A1 | |
| $= 38.16\ldots°$; Required angle is $38.2°$ to 1 d.p. | A1 | |
| **Total: 4 marks** | | |9 A laser beam is aimed from a point ( $12,10,10$ ) in the direction $- 2 \mathbf { i } - 2 \mathbf { j } - 3 \mathbf { k }$ towards a plane surface.\\
(i) Give the equation of the path of the laser beam in vector form.
The points $\mathrm { A } ( 1,1,1 ) , \mathrm { B } ( 1,4,2 )$ and $\mathrm { C } ( 6,1,0 )$ lie on the plane.\\
(ii) Show that the vector $3 \mathbf { i } - 5 \mathbf { j } + 15 \mathbf { k }$ is perpendicular to the plane and hence find the cartesian equation of the plane.\\
(iii) Find the coordinate of the point where the laser beam hits the surface of the plane.\\
(iv) Find the angle between the laser beam and the plane.
\section*{Insert for question 6.}
The graph of $y = \tan x$ is given below.\\
On this graph sketch the graph of $y = \cot x$.\\
Show clearly where your graph crosses the graph of $y = \tan x$ and indicate the asymptotes. [4]\\
\includegraphics[max width=\textwidth, alt={}, center]{23771896-942c-4a1d-ab95-6b6d3cc5643c-5_853_1555_703_262}
\hfill \mbox{\textit{OCR MEI C4 Q9 [18]}}