| 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 structured multi-part 3D vectors question with clear scaffolding. Part (i) involves routine vector subtraction and angle calculation using dot product. Part (ii) is a verification exercise (equation given) followed by standard plane-to-plane angle calculation. Part (iii) uses coplanarity to find k (substitution into plane equation), then requires showing trapezium properties via parallel vectors. While it involves multiple techniques, each step is standard and well-signposted, making it slightly easier than average for A-level. |
| 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 forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}-4\\0\\-2\end{pmatrix},\ \overrightarrow{AC} = \begin{pmatrix}-2\\4\\1\end{pmatrix}\) | B1B1 | condone rows |
| \(\cos BAC = \dfrac{\begin{pmatrix}-4\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}-2\\4\\1\end{pmatrix}}{AB \cdot AC} = \dfrac{(-4)(-2)+0\cdot4+(-2)\cdot1}{\sqrt{20}\sqrt{21}}\) | M1 | dot product evaluated |
| M1 | \(\cos BAC =\) dot product \(/\ | AB\ |
| \(= 0.293\) | A1 | \(0.293\) or \(\cos ABC =\) correct numerical expression as RHS above, or better |
| \(\Rightarrow BAC = 73.0°\) | A1 | or rounds to \(73.0°\) (accept \(73°\) www); or \(1.27\) radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| A: \(x+y-2z+d = 2-6+d=0 \Rightarrow d=4\) | M1 | substituting one point |
| B: \(-2+0-2\times1+4=0\) | DM1 | evaluating for other two points |
| C: \(0+4-2\times4+4=0\) | A1 | \(d=4\) www |
| Normal \(\mathbf{n} = \begin{pmatrix}1\\1\\-2\end{pmatrix}\) | B1 | stated or used as normal anywhere in part (ii) |
| \(\mathbf{n}\cdot\begin{pmatrix}0\\0\\1\end{pmatrix} = \dfrac{-2}{\sqrt{6}} = \cos\theta\) | M1 | finding angle between normal vector and \(\mathbf{k}\) |
| A1 | allow \(\pm2/\sqrt{6}\) or \(144.7°\) for A1; may have deliberately made \(+\)ve to find acute angle | |
| \(\Rightarrow \theta = 144.7°\); acute angle \(= 35.3°\) | A1 | or rounds to \(35.3°\); do not need to find \(144.7°\) explicitly; or \(0.615\) radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At D: \(-2+4-2k+4=0\) | M1 | substituting into plane equation |
| \(\Rightarrow 2k=6,\ k=3\) * | A1 | AG |
| \(\overrightarrow{CD} = \begin{pmatrix}-2\\0\\-1\end{pmatrix} = \dfrac{1}{2}\overrightarrow{AB}\) | M1 | finding vector CD (or vector DC); \(\overrightarrow{CD} = \begin{pmatrix}-2\\0\\-1\end{pmatrix}\) |
| \(\Rightarrow\) CD is parallel to AB | A1 | or DC parallel to AB or BA oe (or hence two parallel sides, if clear which); but A0 if their vector CD is vector DC |
| \(CD:AB = 1:2\) | B1 | mark final answer www; allow \(CD:AB=1/2\), \(\sqrt{5}:\sqrt{20}\) oe; AB is twice CD oe; for B1 allow vector CD used as vector DC |
# Question 2:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}-4\\0\\-2\end{pmatrix},\ \overrightarrow{AC} = \begin{pmatrix}-2\\4\\1\end{pmatrix}$ | B1B1 | condone rows |
| $\cos BAC = \dfrac{\begin{pmatrix}-4\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}-2\\4\\1\end{pmatrix}}{AB \cdot AC} = \dfrac{(-4)(-2)+0\cdot4+(-2)\cdot1}{\sqrt{20}\sqrt{21}}$ | M1 | dot product evaluated |
| | M1 | $\cos BAC =$ dot product $/\|AB\|\cdot\|AC\|$; need to see method for modulae as far as $\sqrt{\ldots}$; use of vectors BA and CA could obtain B0 B0 M1 M1 A1 A1 |
| $= 0.293$ | A1 | $0.293$ or $\cos ABC =$ correct numerical expression as RHS above, or better |
| $\Rightarrow BAC = 73.0°$ | A1 | or rounds to $73.0°$ (accept $73°$ www); or $1.27$ radians |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| A: $x+y-2z+d = 2-6+d=0 \Rightarrow d=4$ | M1 | substituting one point |
| B: $-2+0-2\times1+4=0$ | DM1 | evaluating for other two points |
| C: $0+4-2\times4+4=0$ | A1 | $d=4$ www |
| Normal $\mathbf{n} = \begin{pmatrix}1\\1\\-2\end{pmatrix}$ | B1 | stated or used as normal anywhere in part (ii) |
| $\mathbf{n}\cdot\begin{pmatrix}0\\0\\1\end{pmatrix} = \dfrac{-2}{\sqrt{6}} = \cos\theta$ | M1 | finding angle between normal vector and $\mathbf{k}$ |
| | A1 | allow $\pm2/\sqrt{6}$ or $144.7°$ for A1; may have deliberately made $+$ve to find acute angle |
| $\Rightarrow \theta = 144.7°$; acute angle $= 35.3°$ | A1 | or rounds to $35.3°$; do not need to find $144.7°$ explicitly; or $0.615$ radians |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| At D: $-2+4-2k+4=0$ | M1 | substituting into plane equation |
| $\Rightarrow 2k=6,\ k=3$ * | A1 | AG |
| $\overrightarrow{CD} = \begin{pmatrix}-2\\0\\-1\end{pmatrix} = \dfrac{1}{2}\overrightarrow{AB}$ | M1 | finding vector CD (or vector DC); $\overrightarrow{CD} = \begin{pmatrix}-2\\0\\-1\end{pmatrix}$ |
| $\Rightarrow$ CD is parallel to AB | A1 | or DC parallel to AB or BA oe (or hence two parallel sides, if clear which); but A0 if their vector CD is vector DC |
| $CD:AB = 1:2$ | B1 | mark final answer www; allow $CD:AB=1/2$, $\sqrt{5}:\sqrt{20}$ oe; AB is twice CD oe; for B1 allow vector CD used as vector DC |
---2 A piece of cloth ABDC is attached to the tops of vertical poles $\mathrm { AE } , \mathrm { BF } , \mathrm { DG }$ and CH , where $\mathrm { E } , \mathrm { F } , \mathrm { G }$ and H are at ground level (see Fig. 7). Coordinates are as shown, with lengths in metres. The length of pole DG is $k$ metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b46db958-aa88-47fb-8db3-786472791577-2_916_1255_464_397}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Write down the vectors $\overrightarrow { \mathrm { AB } }$ and $\overrightarrow { \mathrm { AC } }$. Hence calculate the angle BAC .\\
(ii) Verify that the equation of the plane ABC is $x + y - 2 z + d = 0$, where $d$ is a constant to be determined.
Calculate the acute angle the plane makes with the horizontal plane.\\
(iii) Given that $\mathrm { A } , \mathrm { B } , \mathrm { D }$ and C are coplanar, show that $k = 3$.
Hence show that ABDC is a trapezium, and find the ratio of CD to AB .
\hfill \mbox{\textit{OCR MEI C4 Q2 [18]}}