OCR MEI FP3 2008 June — Question 1 24 marks

Exam BoardOCR MEI
ModuleFP3 (Further Pure Mathematics 3)
Year2008
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeVolume of tetrahedron using scalar triple product
DifficultyChallenging +1.8 This is a comprehensive Further Maths question requiring multiple vector techniques (cross product, scalar triple product, distance formulas, volume ratios) across six parts. While each individual technique is standard for FP3, the extended multi-part nature, the final volume ratio requiring geometric insight about how a plane divides a tetrahedron, and the need to synthesize multiple concepts elevates this significantly above average A-level difficulty.
Spec4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line
1 A tetrahedron ABCD has vertices \(\mathrm { A } ( - 3,5,2 ) , \mathrm { B } ( 3,13,7 ) , \mathrm { C } ( 7,0,3 )\) and \(\mathrm { D } ( 5,4,8 )\).
  1. Find the vector product \(\overrightarrow { \mathrm { AB } } \times \overrightarrow { \mathrm { AC } }\), and hence find the equation of the plane ABC .
  2. Find the shortest distance from \(D\) to the plane \(A B C\).
  3. Find the shortest distance between the lines AB and CD .
  4. Find the volume of the tetrahedron ABCD . The plane \(P\) with equation \(3 x - 2 z + 5 = 0\) contains the point B , and meets the lines AC and AD at E and F respectively.
  5. Find \(\lambda\) and \(\mu\) such that \(\overrightarrow { \mathrm { AE } } = \lambda \overrightarrow { \mathrm { AC } }\) and \(\overrightarrow { \mathrm { AF } } = \mu \overrightarrow { \mathrm { AD } }\). Deduce that E is between A and C , and that F is between A and D.
  6. Hence, or otherwise, show that \(P\) divides the tetrahedron ABCD into two parts having volumes in the ratio 4 to 17.
1(i)
\[\mathbf{AB} \times \mathbf{AC} = \begin{pmatrix} 6 \\ 8 \\ 1 \end{pmatrix} \times \begin{pmatrix} 10 \\ -5 \\ 1 \end{pmatrix} = \begin{pmatrix} 33 \\ 44 \\ -110 \end{pmatrix}\]
B2 - Ignore subsequent working
ABC is: \(3x + 4y - 10z + 9 = 0\)
or \(3x + 4y - 10z = -9\) or \(-3x - 4y + 10z = 9\)
M1 - For \(3x + 4y - 10z\)
A1 - Accept \(33x + 44y - 110z = -99\) etc
Mark: 4
1(ii)
Distance is \(\frac{3x5 + 4x4 - 10x8 + 9}{\sqrt{3^2 + 4^2 + 10^2}} = \frac{40}{\sqrt{125}} = \frac{8}{\sqrt{5}}\)
M1 - Using distance formula (or other complete method)
A1 ft
A1 - Condone negative answer. Accept a.r.t. 3.58
Mark: 3
1(iii)
\[\mathbf{AB} \times \mathbf{CD} = \begin{pmatrix} 6 \\ 8 \\ 1 \end{pmatrix} \times \begin{pmatrix} -2 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 20 \\ -40 \\ 40 \end{pmatrix} \text{ or } \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\]
M1 - Evaluating \(\mathbf{AB} \times \mathbf{CD}\) or method for finding end-points of common perp PQ
A1 - or \(P(\frac{1}{2}, 11, \frac{25}{2})\) & \(Q(\frac{11}{8}, \frac{33}{8}, \frac{31}{8})\) or \(PQ = (\frac{23}{8}, -\frac{55}{8}, \frac{31}{8})\)
Distance is \(\frac{\mathbf{AC} \cdot \mathbf{n}}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{22}{3}\)
M1
A1
Mark: 4
1(iv)
Volume is \(\frac{1}{6}(\mathbf{AB} \times \mathbf{AC}) \cdot \mathbf{AD}\)
M1 - Scalar triple product
A1
\[= \frac{1}{6} \begin{pmatrix} 33 \\ 44 \\ -110 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ -1 \\ 6 \end{pmatrix} = (-) \frac{220}{3}\]
M1
A1 - Accept a.r.t. 73.3
Mark: 4
1(v)
E is \((-3 + 10\lambda, 5 - 5\lambda, 2 + \lambda)\)
\(3(-3 + 10\lambda) - 2(2 + \lambda) + 5 = 0\)
\(\lambda = \frac{2}{7}\)
M1
A1
F is \((-3 + 8\mu, 5 - \mu, 2 + 6\mu)\)
\(3(-3 + 8\mu) - 2(2 + 6\mu) + 5 = 0\)
\(\mu = \frac{2}{3}\)
M1
A1
Since \(0 < \lambda < 1\), E is between A and C
Since \(0 < \mu < 1\), F is between A and D
B1
Mark: 5
1(vi)
\[V_{\text{ABEF}} = \frac{1}{6}(\mathbf{AB} \times \mathbf{AE}) \cdot \mathbf{AF}\]
\[= \frac{1}{6}\mu(\mathbf{AB} \times \mathbf{AC}) \cdot \mathbf{AD}\]
\[= \lambda \mu V_{\text{ABCD}}\]
\[= \frac{3}{1}V_{\text{ABCD}}\]
M1
A1 - \((13\frac{43}{63})\) if numerical
Ratio of volumes is \(\frac{4}{21} : \frac{17}{21} = 4:17\)
M1 - Finding ratio of volumes of two parts
A1 ag
SC1 for \(4 : 17\) deduced from \(\frac{2}{7}\) without working
Mark: 4
## 1(i)
$$\mathbf{AB} \times \mathbf{AC} = \begin{pmatrix} 6 \\ 8 \\ 1 \end{pmatrix} \times \begin{pmatrix} 10 \\ -5 \\ 1 \end{pmatrix} = \begin{pmatrix} 33 \\ 44 \\ -110 \end{pmatrix}$$

**B2** - Ignore subsequent working

**ABC is:** $3x + 4y - 10z + 9 = 0$
or $3x + 4y - 10z = -9$ or $-3x - 4y + 10z = 9$

**M1** - For $3x + 4y - 10z$

**A1** - Accept $33x + 44y - 110z = -99$ etc

**Mark: 4**

## 1(ii)
Distance is $\frac{3x5 + 4x4 - 10x8 + 9}{\sqrt{3^2 + 4^2 + 10^2}} = \frac{40}{\sqrt{125}} = \frac{8}{\sqrt{5}}$

**M1** - Using distance formula (or other complete method)

**A1 ft** 

**A1** - Condone negative answer. Accept a.r.t. 3.58

**Mark: 3**

## 1(iii)
$$\mathbf{AB} \times \mathbf{CD} = \begin{pmatrix} 6 \\ 8 \\ 1 \end{pmatrix} \times \begin{pmatrix} -2 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 20 \\ -40 \\ 40 \end{pmatrix} \text{ or } \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$$

**M1** - Evaluating $\mathbf{AB} \times \mathbf{CD}$ or method for finding end-points of common perp PQ

**A1** - or $P(\frac{1}{2}, 11, \frac{25}{2})$ & $Q(\frac{11}{8}, \frac{33}{8}, \frac{31}{8})$ or $PQ = (\frac{23}{8}, -\frac{55}{8}, \frac{31}{8})$

Distance is $\frac{\mathbf{AC} \cdot \mathbf{n}}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{22}{3}$

**M1**

**A1**

**Mark: 4**

## 1(iv)
Volume is $\frac{1}{6}(\mathbf{AB} \times \mathbf{AC}) \cdot \mathbf{AD}$

**M1** - Scalar triple product

**A1**

$$= \frac{1}{6} \begin{pmatrix} 33 \\ 44 \\ -110 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ -1 \\ 6 \end{pmatrix} = (-) \frac{220}{3}$$

**M1**

**A1** - Accept a.r.t. 73.3

**Mark: 4**

## 1(v)
E is $(-3 + 10\lambda, 5 - 5\lambda, 2 + \lambda)$

$3(-3 + 10\lambda) - 2(2 + \lambda) + 5 = 0$

$\lambda = \frac{2}{7}$

**M1**

**A1**

F is $(-3 + 8\mu, 5 - \mu, 2 + 6\mu)$

$3(-3 + 8\mu) - 2(2 + 6\mu) + 5 = 0$

$\mu = \frac{2}{3}$

**M1**

**A1**

Since $0 < \lambda < 1$, E is between A and C

Since $0 < \mu < 1$, F is between A and D

**B1**

**Mark: 5**

## 1(vi)
$$V_{\text{ABEF}} = \frac{1}{6}(\mathbf{AB} \times \mathbf{AE}) \cdot \mathbf{AF}$$

$$= \frac{1}{6}\mu(\mathbf{AB} \times \mathbf{AC}) \cdot \mathbf{AD}$$

$$= \lambda \mu V_{\text{ABCD}}$$

$$= \frac{3}{1}V_{\text{ABCD}}$$

**M1**

**A1** - $(13\frac{43}{63})$ if numerical

Ratio of volumes is $\frac{4}{21} : \frac{17}{21} = 4:17$

**M1** - Finding ratio of volumes of two parts

**A1 ag**

**SC1** for $4 : 17$ deduced from $\frac{2}{7}$ without working

**Mark: 4**

---
1 A tetrahedron ABCD has vertices $\mathrm { A } ( - 3,5,2 ) , \mathrm { B } ( 3,13,7 ) , \mathrm { C } ( 7,0,3 )$ and $\mathrm { D } ( 5,4,8 )$.\\
(i) Find the vector product $\overrightarrow { \mathrm { AB } } \times \overrightarrow { \mathrm { AC } }$, and hence find the equation of the plane ABC .\\
(ii) Find the shortest distance from $D$ to the plane $A B C$.\\
(iii) Find the shortest distance between the lines AB and CD .\\
(iv) Find the volume of the tetrahedron ABCD .

The plane $P$ with equation $3 x - 2 z + 5 = 0$ contains the point B , and meets the lines AC and AD at E and F respectively.\\
(v) Find $\lambda$ and $\mu$ such that $\overrightarrow { \mathrm { AE } } = \lambda \overrightarrow { \mathrm { AC } }$ and $\overrightarrow { \mathrm { AF } } = \mu \overrightarrow { \mathrm { AD } }$. Deduce that E is between A and C , and that F is between A and D.\\
(vi) Hence, or otherwise, show that $P$ divides the tetrahedron ABCD into two parts having volumes in the ratio 4 to 17.

\hfill \mbox{\textit{OCR MEI FP3 2008 Q1 [24]}}
This paper (5 questions)
View full paper
Q1 24 Q2 24 Q3 24 Q4 24 Q5 24