| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 17 |
| 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 requiring standard techniques: finding direction vectors, vector equations of lines, angles between lines and planes using dot products, and line-plane intersection. All parts follow routine procedures with clear scaffolding, though it involves multiple steps across 5 parts. The Snell's Law application is just substitution. Slightly easier than average due to the heavy scaffolding and standard methods throughout. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}-1\\-2\\0\end{pmatrix}\) | B1 | |
| \(\mathbf{r} = \begin{pmatrix}0\\0\\2\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\0\end{pmatrix}\) | B1 | or equivalent alternative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{n} = \begin{pmatrix}1\\0\\1\end{pmatrix}\) | B1 | |
| \(\cos\theta = \dfrac{\begin{pmatrix}1\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\0\end{pmatrix}}{\sqrt{2}\sqrt{5}} = \dfrac{1}{\sqrt{10}}\) | B1, M1, M1 | correct vectors (any multiples); scalar product used; finding inverse of scalar product divided by two modulae |
| \(\Rightarrow \theta = 71.57°\) | A1 | \(72°\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\phi = \dfrac{\begin{pmatrix}-1\\0\\-1\end{pmatrix}\cdot\begin{pmatrix}-2\\-2\\-1\end{pmatrix}}{\sqrt{2}\sqrt{9}} = \dfrac{2+1}{3\sqrt{2}} = \dfrac{1}{\sqrt{2}}\) | M1 | ft their \(\mathbf{n}\) for method |
| \(\Rightarrow \phi = 45°\) * | A1 | \(\pm1/\sqrt{2}\) oe exact |
| E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin 71.57° = k\sin 45°\) | M1 | ft on their \(71.57°\) |
| \(\Rightarrow k = \sin 71.57°/\sin 45° = 1.34\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = \begin{pmatrix}0\\0\\2\end{pmatrix} + \mu\begin{pmatrix}-2\\-2\\-1\end{pmatrix}\) | ||
| \(x=-2\mu,\ z=2-\mu\) | M1 | soi |
| \(x+z=-1\): \(-2\mu+2-\mu=-1\) | M1 | subst in \(x+z=-1\) |
| \(\Rightarrow 3\mu=3,\ \mu=1\) | A1 | |
| \(\Rightarrow\) point of intersection is \((-2,-2,1)\) | A1 | |
| distance travelled through glass \(=\) distance between \((0,0,2)\) and \((-2,-2,1)\) \(= \sqrt{2^2+2^2+1^2} = 3\) cm | B1 | www dep on \(\mu=1\) |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}-1\\-2\\0\end{pmatrix}$ | B1 | |
| $\mathbf{r} = \begin{pmatrix}0\\0\\2\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\0\end{pmatrix}$ | B1 | or equivalent alternative |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{pmatrix}1\\0\\1\end{pmatrix}$ | B1 | |
| $\cos\theta = \dfrac{\begin{pmatrix}1\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\0\end{pmatrix}}{\sqrt{2}\sqrt{5}} = \dfrac{1}{\sqrt{10}}$ | B1, M1, M1 | correct vectors (any multiples); scalar product used; finding inverse of scalar product divided by two modulae |
| $\Rightarrow \theta = 71.57°$ | A1 | $72°$ or better |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\phi = \dfrac{\begin{pmatrix}-1\\0\\-1\end{pmatrix}\cdot\begin{pmatrix}-2\\-2\\-1\end{pmatrix}}{\sqrt{2}\sqrt{9}} = \dfrac{2+1}{3\sqrt{2}} = \dfrac{1}{\sqrt{2}}$ | M1 | ft their $\mathbf{n}$ for method |
| $\Rightarrow \phi = 45°$ * | A1 | $\pm1/\sqrt{2}$ oe exact |
| | E1 | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin 71.57° = k\sin 45°$ | M1 | ft on their $71.57°$ |
| $\Rightarrow k = \sin 71.57°/\sin 45° = 1.34$ | A1 | oe |
## Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = \begin{pmatrix}0\\0\\2\end{pmatrix} + \mu\begin{pmatrix}-2\\-2\\-1\end{pmatrix}$ | | |
| $x=-2\mu,\ z=2-\mu$ | M1 | soi |
| $x+z=-1$: $-2\mu+2-\mu=-1$ | M1 | subst in $x+z=-1$ |
| $\Rightarrow 3\mu=3,\ \mu=1$ | A1 | |
| $\Rightarrow$ point of intersection is $(-2,-2,1)$ | A1 | |
| distance travelled through glass $=$ distance between $(0,0,2)$ and $(-2,-2,1)$ $= \sqrt{2^2+2^2+1^2} = 3$ cm | B1 | www dep on $\mu=1$ |4 When a light ray passes from air to glass, it is deflected through an angle. The light ray ABC starts at point $\mathrm { A } ( 1,2,2 )$, and enters a glass object at point $\mathrm { B } ( 0,0,2 )$. The surface of the glass object is a plane with normal vector $\mathbf { n }$. Fig. 7 shows a cross-section of the glass object in the plane of the light ray and $\mathbf { n }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b46db958-aa88-47fb-8db3-786472791577-4_689_812_341_662}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Find the vector $\overrightarrow { \mathrm { AB } }$ and a vector equation of the line AB .
The surface of the glass object is a plane with equation $x + z = 2$. AB makes an acute angle $\theta$ with the normal to this plane.\\
(ii) Write down the normal vector $\mathbf { n }$, and hence calculate $\theta$, giving your answer in degrees.
The line BC has vector equation $\mathbf { r } = \left( \begin{array} { l } 0 \\ 0 \\ 2 \end{array} \right) + \mu \left( \begin{array} { l } - 2 \\ - 2 \\ - 1 \end{array} \right)$. This line makes an acute angle $\phi$ with the\\
normal to the plane. normal to the plane.\\
(iii) Show that $\phi = 45 ^ { \circ }$.\\
(iv) Snell's Law states that $\sin \theta = k \sin \phi$, where $k$ is a constant called the refractive index. Find $k$.
The light ray leaves the glass object through a plane with equation $x + z = - 1$. Units are centimetres.\\
(v) Find the point of intersection of the line BC with the plane $x + z = - 1$. Hence find the distance the light ray travels through the glass object.
\hfill \mbox{\textit{OCR MEI C4 Q4 [17]}}