| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | 3D geometry applications |
| Difficulty | Standard +0.3 This is a structured 3D geometry question with clear scaffolding through multiple parts. Part (i) involves straightforward vector length and angle calculations. Part (ii) requires verifying a given normal vector (routine dot product checks) and finding a plane equation using standard methods. Part (iii) involves verifying a given plane equation and finding the angle between planes using normal vectors. While it requires multiple techniques, each step follows standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| B lies on the axes (i.e. either the x-axis or y-axis) | B1 | Accept: B has coordinates \((x, 0)\) or \((0, y)\) for any non-zero \(x\) or \(y\); or "B lies on one of the coordinate axes". The condition \(d(A,B) = t(A,B)\) means the Pythagorean and taxicab distances are equal, which occurs when movement is purely horizontal or vertical. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(t(A,B) = \sqrt{2} \times d(A,B)\) requires \(\ | x\ | = \ |
| B lies on the lines \(y = x\) or \(y = -x\) (the diagonals) | A1 | Accept equivalent descriptions; B lies on either diagonal through the origin. Exclude the origin itself. |
## Question 7:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| B lies on the axes (i.e. either the x-axis or y-axis) | B1 | Accept: B has coordinates $(x, 0)$ or $(0, y)$ for any non-zero $x$ or $y$; or "B lies on one of the coordinate axes". The condition $d(A,B) = t(A,B)$ means the Pythagorean and taxicab distances are equal, which occurs when movement is purely horizontal or vertical. |
**Part (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $t(A,B) = \sqrt{2} \times d(A,B)$ requires $\|x\| = \|y\|$, i.e. B lies on the lines $y = x$ or $y = -x$ | M1 | For recognising that equality of upper bound requires $\|x\| = \|y\|$ or equivalent working showing $|x| + |y| = \sqrt{2}\sqrt{x^2+y^2}$ leads to $|x|=|y|$ |
| B lies on the lines $y = x$ or $y = -x$ (the diagonals) | A1 | Accept equivalent descriptions; B lies on either diagonal through the origin. Exclude the origin itself. |7 A tent has vertices ABCDEF with coordinates as shown in Fig. 7. Lengths are in metres. The $\mathrm { O } x y$ plane is horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-03_547_987_1580_539}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Find the length of the ridge of the tent DE , and the angle this makes with the horizontal.\\
(ii) Show that the vector $\mathbf { i } - 4 \mathbf { j } + 5 \mathbf { k }$ is normal to the plane through $\mathrm { A } , \mathrm { D }$ and E .
Hence find the equation of this plane. Given that B lies in this plane, find $a$.\\
(iii) Verify that the equation of the plane BCD is $x + z = 8$.
Hence find the acute angle between the planes ABDE and BCD .
\hfill \mbox{\textit{OCR MEI C4 2013 Q7 [17]}}