OCR MEI C4 — Question 9 17 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
Type3D geometry applications
DifficultyStandard +0.3 This is a structured multi-part question on 3D coordinate geometry with clear scaffolding. Parts (i)-(ii) involve routine midpoint and ratio division calculations. Part (iii) requires dot product verification (standard technique). Parts (iv)-(v) use the perpendicular vector found earlier to write a plane equation and find an angle—both standard applications. While it has multiple parts (5 marks likely), each step follows logically from the previous with no novel insight required, making it slightly easier than average.
Spec1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles
9 Beside a major route into a county town the authorities decide to build a large pyramid. Fig. 9.1 shows this pyramid, ABCDE O is the centre point of the horizontal base BCDE . A coordinate system is defined with O as the origin. The \(x\)-axis and \(y\)-axis are horizontal and the \(z\)-axis is vertical, as shown in Fig. 9.1 The vertices of the pyramid are $$A ( 0,0,6 ) , B ( - 4 , - 4,0 ) , C ( 4 , - 4,0 ) , D ( 4,4,0 ) \text { and } E ( - 4,4,0 ) .$$ \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{78993065-a6cd-4b77-b21f-c9ccc82fb37a-4_668_878_493_623} \captionsetup{labelformat=empty} \caption{Fig.9.1}
\end{figure} The pyramid is supported by a vertical pole OA and there are also support rods from O to points on the triangular faces \(\mathrm { ABC } , \mathrm { ACD } , \mathrm { ADE }\) and AEB . One of the rods, ON , is shown in fig.9.2 which shows one quarter of the pyramid. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{78993065-a6cd-4b77-b21f-c9ccc82fb37a-4_428_675_1521_831} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
\end{figure} M is the mid-point of the line BC .
  1. Write down the coordinates of M.
  2. Write down the vector \(\overrightarrow { \mathrm { AM } }\) and hence the coordinates of the point N which divides \(\overrightarrow { \mathrm { AM } }\) so that the ratio \(\mathrm { AN } : \mathrm { NM } = 2 : 1\).
  3. Show that ON is perpendicular to both AM and BC .
  4. Hence write down the equation of the plane ABC in its simplest form.
  5. Find the angle between the face ABC and the ground.
Question 9(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(M(0, -4, 0)\)B1
Total: 1 mark
Question 9(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\overrightarrow{AM} = \begin{pmatrix}0\\-4\\-6\end{pmatrix}\)B1
\(\overrightarrow{AN} = \frac{9}{13}\begin{pmatrix}0\\-4\\-6\end{pmatrix}\)M1, A1
\(\Rightarrow \overrightarrow{ON} = \overrightarrow{OA} + \overrightarrow{AN} = \begin{pmatrix}0\\0\\6\end{pmatrix} + \frac{9}{13}\begin{pmatrix}0\\-4\\-6\end{pmatrix}\)M1
\(= \begin{pmatrix}0\\-\frac{36}{13}\\\frac{24}{13}\end{pmatrix} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix}\)A1
Total: 5 marks
Question 9(iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\overrightarrow{ON} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix} \Rightarrow \overrightarrow{ON}\cdot\overrightarrow{AM} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix}\cdot\begin{pmatrix}0\\-4\\-6\end{pmatrix}\)M1 Dot product
\(= \frac{12}{13}(0 + 12 - 12) = 0\)
\(\overrightarrow{ON}\cdot\overrightarrow{BC} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix}\cdot\begin{pmatrix}8\\0\\0\end{pmatrix} = \frac{12}{13}(0+0+0) = 0\)B1, A1 For BC; Both \(= 0\)
Total: 3 marks
Question 9(iv):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Normal has direction \([0, -3, 2]\)B1
So equation is \(-3y + 2z = d\)M1, A1
Substitute any point: \(d = 12\), i.e. \(-3y + 2z = 12\)A1
Total: 4 marks
Question 9(vi):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Angle is AMOB1
\(AMO = \tan^{-1}\frac{6}{4} = 56.3°\)M1, A1
OR: Angle between \(\mathbf{n}\) and \(\overrightarrow{OA} = \arccos\left(\frac{0+0+2}{\sqrt{13}\sqrt{1}}\right) = \arccos\frac{2}{\sqrt{13}}\)A1
Total: 4 marks 
## Question 9(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(0, -4, 0)$ | B1 | |
| **Total: 1 mark** | | |

---

## Question 9(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AM} = \begin{pmatrix}0\\-4\\-6\end{pmatrix}$ | B1 | |
| $\overrightarrow{AN} = \frac{9}{13}\begin{pmatrix}0\\-4\\-6\end{pmatrix}$ | M1, A1 | |
| $\Rightarrow \overrightarrow{ON} = \overrightarrow{OA} + \overrightarrow{AN} = \begin{pmatrix}0\\0\\6\end{pmatrix} + \frac{9}{13}\begin{pmatrix}0\\-4\\-6\end{pmatrix}$ | M1 | |
| $= \begin{pmatrix}0\\-\frac{36}{13}\\\frac{24}{13}\end{pmatrix} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix}$ | A1 | |
| **Total: 5 marks** | | |

---

## Question 9(iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{ON} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix} \Rightarrow \overrightarrow{ON}\cdot\overrightarrow{AM} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix}\cdot\begin{pmatrix}0\\-4\\-6\end{pmatrix}$ | M1 | Dot product |
| $= \frac{12}{13}(0 + 12 - 12) = 0$ | | |
| $\overrightarrow{ON}\cdot\overrightarrow{BC} = \frac{12}{13}\begin{pmatrix}0\\-3\\2\end{pmatrix}\cdot\begin{pmatrix}8\\0\\0\end{pmatrix} = \frac{12}{13}(0+0+0) = 0$ | B1, A1 | For BC; Both $= 0$ |
| **Total: 3 marks** | | |

---

## Question 9(iv):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Normal has direction $[0, -3, 2]$ | B1 | |
| So equation is $-3y + 2z = d$ | M1, A1 | |
| Substitute any point: $d = 12$, i.e. $-3y + 2z = 12$ | A1 | |
| **Total: 4 marks** | | |

---

## Question 9(vi):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Angle is AMO | B1 | |
| $AMO = \tan^{-1}\frac{6}{4} = 56.3°$ | M1, A1 | |
| OR: Angle between $\mathbf{n}$ and $\overrightarrow{OA} = \arccos\left(\frac{0+0+2}{\sqrt{13}\sqrt{1}}\right) = \arccos\frac{2}{\sqrt{13}}$ | A1 | |
| **Total: 4 marks** | | |
9 Beside a major route into a county town the authorities decide to build a large pyramid. Fig. 9.1 shows this pyramid, ABCDE O is the centre point of the horizontal base BCDE . A coordinate system is defined with O as the origin. The $x$-axis and $y$-axis are horizontal and the $z$-axis is vertical, as shown in Fig. 9.1 The vertices of the pyramid are

$$A ( 0,0,6 ) , B ( - 4 , - 4,0 ) , C ( 4 , - 4,0 ) , D ( 4,4,0 ) \text { and } E ( - 4,4,0 ) .$$

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{78993065-a6cd-4b77-b21f-c9ccc82fb37a-4_668_878_493_623}
\captionsetup{labelformat=empty}
\caption{Fig.9.1}
\end{center}
\end{figure}

The pyramid is supported by a vertical pole OA and there are also support rods from O to points on the triangular faces $\mathrm { ABC } , \mathrm { ACD } , \mathrm { ADE }$ and AEB . One of the rods, ON , is shown in fig.9.2 which shows one quarter of the pyramid.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{78993065-a6cd-4b77-b21f-c9ccc82fb37a-4_428_675_1521_831}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{figure}

M is the mid-point of the line BC .\\
(i) Write down the coordinates of M.\\
(ii) Write down the vector $\overrightarrow { \mathrm { AM } }$ and hence the coordinates of the point N which divides $\overrightarrow { \mathrm { AM } }$ so that the ratio $\mathrm { AN } : \mathrm { NM } = 2 : 1$.\\
(iii) Show that ON is perpendicular to both AM and BC .\\
(iv) Hence write down the equation of the plane ABC in its simplest form.\\
(v) Find the angle between the face ABC and the ground.

\hfill \mbox{\textit{OCR MEI C4  Q9 [17]}}
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