| 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 +0.8 This is a substantial multi-part 3D coordinate geometry question requiring verification of plane equations, finding unknown coordinates using plane membership, computing normal vectors, finding angles between lines, and working with position vectors along line segments. While the individual techniques are standard A-level content (plane equations, dot products, vector arithmetic), the question requires sustained reasoning across multiple parts with several unknowns to determine, making it moderately challenging but within typical Further Maths scope. |
| 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 | Marks | Guidance |
| \(\overrightarrow{FE} = -2\mathbf{i} + \frac{1}{3}\mathbf{j} + \mathbf{k}\), \(\overrightarrow{FB} = \mathbf{i} - \frac{1}{6}\mathbf{j} - 2\mathbf{k}\) | B1 B1 | or \(\overrightarrow{EF} = 2\mathbf{i} + (-\frac{1}{3})\mathbf{j} - \mathbf{k}\) or \(\overrightarrow{BF} = -\mathbf{i} + \frac{1}{6}\mathbf{j} + 2\mathbf{k}\) |
| \(\cos\theta = \dfrac{-2(1) + (\frac{1}{3})(-\frac{1}{6}) + 1(-2)}{\sqrt{4+\frac{1}{9}+1}\sqrt{1+\frac{1}{36}+4}}\) | M1 | \(\cos\theta = (\overrightarrow{FE}\cdot\overrightarrow{FB})/( |
| \(\theta = \arccos\left(\dfrac{-\frac{73}{18}}{\frac{\sqrt{46}}{3}\times\frac{\sqrt{181}}{6}}\right)\) | A1 | \(\arccos\left(\dfrac{-2-\frac{1}{18}-2}{5.069}\right) = \arccos(\pm-0.800)\) |
| \(\Rightarrow q = 143°\) | A1 | 3sf or better (or 2.5(0) radians or better). Allow candidates who find the acute angle using either \(\overrightarrow{EF}\) with \(\overrightarrow{FB}\) or \(\overrightarrow{FE}\) with \(\overrightarrow{BF}\) and then state the obtuse angle. Do not isw those who find the obtuse angle and then state the acute angle. Note: \(90 + 2\arctan(\frac{1}{2})\) is 0/5 |
| [5] | ||
| OR: \(EF = \frac{\sqrt{46}}{3}\), \(FB = \frac{\sqrt{181}}{6}\), \(EB = \frac{\sqrt{73}}{2}\) | B3,2,1,0 | One mark for each (2.26, 2.24, 4.27) |
| \(\theta = \arccos\left(\dfrac{(\frac{\sqrt{46}}{3})^2 + (\frac{\sqrt{181}}{6})^2 - (\frac{\sqrt{73}}{2})^2}{2(\frac{\sqrt{46}}{3})(\frac{\sqrt{181}}{6})}\right)\) | M1 | cosine rule correct with their EF, FB, EB |
| A1 | \(q = 143°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z\) coordinate of P is \(\frac{5}{2}\) | B1 | stating the correct \(z\)-coordinate of P; ignore incorrect \(x\) and \(y\) coordinates (or stated in a position vector) |
| \(\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ} = \begin{pmatrix}1\\-13/6\\5/2\end{pmatrix} + \frac{1}{3}\left(\begin{pmatrix}0\\1\\3\end{pmatrix} - \begin{pmatrix}1\\-13/6\\5/2\end{pmatrix}\right)\) | M1 | Complete method for finding the \(z\)-coordinate of Q or \(\overrightarrow{OQ} = \overrightarrow{OH} + \frac{2}{3}(\overrightarrow{HP})\) or \(\overrightarrow{OQ} = \frac{2}{3}(\overrightarrow{OP}) + \frac{1}{3}(\overrightarrow{OH})\) |
| so height of Q is \(\frac{8}{3}\) (metres above ground) | A1 | 2.67 or better |
| [3] |
# Question 1:
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{FE} = -2\mathbf{i} + \frac{1}{3}\mathbf{j} + \mathbf{k}$, $\overrightarrow{FB} = \mathbf{i} - \frac{1}{6}\mathbf{j} - 2\mathbf{k}$ | B1 B1 | or $\overrightarrow{EF} = 2\mathbf{i} + (-\frac{1}{3})\mathbf{j} - \mathbf{k}$ or $\overrightarrow{BF} = -\mathbf{i} + \frac{1}{6}\mathbf{j} + 2\mathbf{k}$ |
| $\cos\theta = \dfrac{-2(1) + (\frac{1}{3})(-\frac{1}{6}) + 1(-2)}{\sqrt{4+\frac{1}{9}+1}\sqrt{1+\frac{1}{36}+4}}$ | M1 | $\cos\theta = (\overrightarrow{FE}\cdot\overrightarrow{FB})/(|\overrightarrow{FE}||\overrightarrow{FB}|)$ (oe) follow through their FE and FB. Allow any combination of FE, EF with FB, BF – allow one sign slip only |
| $\theta = \arccos\left(\dfrac{-\frac{73}{18}}{\frac{\sqrt{46}}{3}\times\frac{\sqrt{181}}{6}}\right)$ | A1 | $\arccos\left(\dfrac{-2-\frac{1}{18}-2}{5.069}\right) = \arccos(\pm-0.800)$ |
| $\Rightarrow q = 143°$ | A1 | 3sf or better (or 2.5(0) radians or better). Allow candidates who find the acute angle using either $\overrightarrow{EF}$ with $\overrightarrow{FB}$ or $\overrightarrow{FE}$ with $\overrightarrow{BF}$ and then state the obtuse angle. **Do not isw those who find the obtuse angle and then state the acute angle.** Note: $90 + 2\arctan(\frac{1}{2})$ is 0/5 |
| **[5]** | | |
| **OR:** $EF = \frac{\sqrt{46}}{3}$, $FB = \frac{\sqrt{181}}{6}$, $EB = \frac{\sqrt{73}}{2}$ | B3,2,1,0 | One mark for each (2.26, 2.24, 4.27) |
| $\theta = \arccos\left(\dfrac{(\frac{\sqrt{46}}{3})^2 + (\frac{\sqrt{181}}{6})^2 - (\frac{\sqrt{73}}{2})^2}{2(\frac{\sqrt{46}}{3})(\frac{\sqrt{181}}{6})}\right)$ | M1 | cosine rule correct with their EF, FB, EB |
| | A1 | $q = 143°$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z$ coordinate of P is $\frac{5}{2}$ | B1 | stating the correct $z$-coordinate of P; ignore incorrect $x$ and $y$ coordinates (or stated in a position vector) |
| $\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ} = \begin{pmatrix}1\\-13/6\\5/2\end{pmatrix} + \frac{1}{3}\left(\begin{pmatrix}0\\1\\3\end{pmatrix} - \begin{pmatrix}1\\-13/6\\5/2\end{pmatrix}\right)$ | M1 | Complete method for finding the $z$-coordinate of Q or $\overrightarrow{OQ} = \overrightarrow{OH} + \frac{2}{3}(\overrightarrow{HP})$ or $\overrightarrow{OQ} = \frac{2}{3}(\overrightarrow{OP}) + \frac{1}{3}(\overrightarrow{OH})$ |
| so height of Q is $\frac{8}{3}$ (metres above ground) | A1 | 2.67 or better |
| **[3]** | | |
---1 Fig. 6 shows a lean-to greenhouse ABCDHEFG . With respect to coordinate axes $\mathrm { O } x y z$, the coordinates of the vertices are as shown. All distances are in metres. Ground level is the plane $z = 0$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{27c27c79-9aea-45a4-a000-41aac70ff866-1_798_1296_354_418}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Verify that the equation of the plane through $\mathrm { A } , \mathrm { B }$ and E is $x + 6 y + 12 = 0$.
Hence, given that F lies in this plane, show that $a = - 2 \frac { 1 } { 3 }$.
\item (A) Show that the vector $\left( \begin{array} { r } 1 \\ - 6 \\ 0 \end{array} \right)$ is normal to the plane DHC.\\
(B) Hence find the cartesian equation of this plane.\\
(C) Given that G lies in the plane DHC , find $b$ and the length FG .
\item Find the angle EFB .
A straight wire joins point H to a point P which is half way between E and $\mathrm { F } . \mathrm { Q }$ is a point two-thirds of the way along this wire, so that $\mathrm { HQ } = 2 \mathrm { QP }$.
\item Find the height of Q above the ground.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 Q1 [18]}}