| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane through three points |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring cross products, plane equations, and distance calculations between points/lines. While it involves several steps and FM content, each part follows standard algorithmic procedures (cross product for normal, perpendicular distance formulas, skew line distance) without requiring novel geometric insight or proof techniques. The 'show that' in part (i) provides the answer to check against, reducing difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04g 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-90\\160\\20\end{pmatrix} \times \begin{pmatrix}3\\198\\15.6\end{pmatrix} = \begin{pmatrix}-1464\\1464\\-18300\end{pmatrix} \left[= -732\begin{pmatrix}2\\-2\\25\end{pmatrix}\right]\) | M1 | Evaluation of vector product |
| Give A1 for one component correct | A1 | A1 for any non-zero multiple correctly obtained |
| \(2x - 2y + 25z = 2(15) - 2(-60) + 25(20)\) | M1 | For \(2x - 2y + 25z = d\) |
| Equation of ABC is \(2x - 2y + 25z = 650\) | E1 | Evidence of substitution required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} \times \mathbf{d}_A = \begin{pmatrix}-90\\160\\20\end{pmatrix} \times \begin{pmatrix}15\\14\\-2\end{pmatrix} = \begin{pmatrix}-600\\120\\-3660\end{pmatrix}\) | M1 | Appropriate vector product |
| Give A1 if one error | A1 | |
| \(\left | \overrightarrow{AB} \times \mathbf{d}_A\right | = \sqrt{600^2 + 120^2 + 3660^2}\) |
| \(\left | \mathbf{d}_A\right | = \sqrt{15^2 + 14^2 + 2^2}\) |
| Distance is \(\dfrac{\left | \overrightarrow{AB} \times \mathbf{d}_A\right | }{\left |
| Distance is 180 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\begin{pmatrix}15+15\lambda\\-60+14\lambda\\20-2\lambda\end{pmatrix} - \begin{pmatrix}-75\\100\\40\end{pmatrix}\right] \cdot \begin{pmatrix}15\\14\\-2\end{pmatrix} = 0\) | M1A1 | |
| \(\lambda = 2\) | M1, A1 | Obtaining a value of \(\lambda\) |
| Distance is \(\sqrt{(120)^2 + (-132)^2 + (-24)^2}\) | M1 | |
| Distance is 180 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{d} = \begin{pmatrix}15\\14\\-2\end{pmatrix} \times \begin{pmatrix}8\\7\\-2\end{pmatrix} = \begin{pmatrix}-14\\14\\-7\end{pmatrix} \left[= 7\begin{pmatrix}-2\\2\\-1\end{pmatrix}\right]\) | M1 | Vector product of direction vectors |
| Give A1 if one error | A2 | |
| \(\begin{pmatrix}3\\198\\15.6\end{pmatrix} \cdot \begin{pmatrix}-2\\2\\-1\end{pmatrix}\) | M1 | Appropriate scalar product |
| \(\overrightarrow{AC} \cdot \hat{\mathbf{d}} = \dfrac{(15.6)(-1)}{\sqrt{2^2+2^2+1^2}} = \dfrac{374.4}{3}\) | M1 | Fully correct method for finding distance |
| Distance is 124.8 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}15\\-60\\20\end{pmatrix} + \lambda\begin{pmatrix}15\\14\\-2\end{pmatrix} = \begin{pmatrix}-75\\100\\40\end{pmatrix} + \mu\begin{pmatrix}93\\38\\p-40\end{pmatrix}\) | M1 | Must use different parameters |
| \(15+15\lambda = -75+93\mu\), \(-60+14\lambda = 100+38\mu\) | A1 | Both equations correct |
| \(\lambda = 25,\quad \mu = 5\) | M1 | Obtaining value of \(\lambda\) or \(\mu\) |
| \(20 - 50 = 40 + 5(p-40)\) | M1 | Or other method for finding \(p\) |
| \(p = 26\) | A1 | |
| Q is \((15+375,\ -60+350,\ 20-50)\) | M1 | |
| Q is \((390,\ 290,\ -30)\) | A1 |
# Question 1:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-90\\160\\20\end{pmatrix} \times \begin{pmatrix}3\\198\\15.6\end{pmatrix} = \begin{pmatrix}-1464\\1464\\-18300\end{pmatrix} \left[= -732\begin{pmatrix}2\\-2\\25\end{pmatrix}\right]$ | M1 | Evaluation of vector product |
| Give A1 for one component correct | A1 | A1 for any non-zero multiple correctly obtained |
| $2x - 2y + 25z = 2(15) - 2(-60) + 25(20)$ | M1 | For $2x - 2y + 25z = d$ |
| Equation of ABC is $2x - 2y + 25z = 650$ | E1 | Evidence of substitution required |
**[5 marks]**
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} \times \mathbf{d}_A = \begin{pmatrix}-90\\160\\20\end{pmatrix} \times \begin{pmatrix}15\\14\\-2\end{pmatrix} = \begin{pmatrix}-600\\120\\-3660\end{pmatrix}$ | M1 | Appropriate vector product |
| Give A1 if one error | A1 | |
| $\left|\overrightarrow{AB} \times \mathbf{d}_A\right| = \sqrt{600^2 + 120^2 + 3660^2}$ | M1 | |
| $\left|\mathbf{d}_A\right| = \sqrt{15^2 + 14^2 + 2^2}$ | | |
| Distance is $\dfrac{\left|\overrightarrow{AB} \times \mathbf{d}_A\right|}{\left|\mathbf{d}_A\right|}$ | M1 | |
| Distance is 180 m | A1 | |
**[6 marks]**
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\begin{pmatrix}15+15\lambda\\-60+14\lambda\\20-2\lambda\end{pmatrix} - \begin{pmatrix}-75\\100\\40\end{pmatrix}\right] \cdot \begin{pmatrix}15\\14\\-2\end{pmatrix} = 0$ | M1A1 | |
| $\lambda = 2$ | M1, A1 | Obtaining a value of $\lambda$ |
| Distance is $\sqrt{(120)^2 + (-132)^2 + (-24)^2}$ | M1 | |
| Distance is 180 m | A1 | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{d} = \begin{pmatrix}15\\14\\-2\end{pmatrix} \times \begin{pmatrix}8\\7\\-2\end{pmatrix} = \begin{pmatrix}-14\\14\\-7\end{pmatrix} \left[= 7\begin{pmatrix}-2\\2\\-1\end{pmatrix}\right]$ | M1 | Vector product of direction vectors |
| Give A1 if one error | A2 | |
| $\begin{pmatrix}3\\198\\15.6\end{pmatrix} \cdot \begin{pmatrix}-2\\2\\-1\end{pmatrix}$ | M1 | Appropriate scalar product |
| $\overrightarrow{AC} \cdot \hat{\mathbf{d}} = \dfrac{(15.6)(-1)}{\sqrt{2^2+2^2+1^2}} = \dfrac{374.4}{3}$ | M1 | Fully correct method for finding distance |
| Distance is 124.8 m | A1 | |
**[6 marks]**
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}15\\-60\\20\end{pmatrix} + \lambda\begin{pmatrix}15\\14\\-2\end{pmatrix} = \begin{pmatrix}-75\\100\\40\end{pmatrix} + \mu\begin{pmatrix}93\\38\\p-40\end{pmatrix}$ | M1 | Must use different parameters |
| $15+15\lambda = -75+93\mu$, $-60+14\lambda = 100+38\mu$ | A1 | Both equations correct |
| $\lambda = 25,\quad \mu = 5$ | M1 | Obtaining value of $\lambda$ or $\mu$ |
| $20 - 50 = 40 + 5(p-40)$ | M1 | Or other method for finding $p$ |
| $p = 26$ | A1 | |
| Q is $(15+375,\ -60+350,\ 20-50)$ | M1 | |
| Q is $(390,\ 290,\ -30)$ | A1 | |
**[7 marks]**
---1 A mine contains several underground tunnels beneath a hillside. The hillside is a plane, all the tunnels are straight and the width of the tunnels may be neglected. A coordinate system is chosen with the $z$-axis pointing vertically upwards and the units are metres. Three points on the hillside have coordinates $\mathrm { A } ( 15 , - 60,20 )$, $B ( - 75,100,40 )$ and $C ( 18,138,35.6 )$.\\
(i) Find the vector product $\overrightarrow { \mathrm { AB } } \times \overrightarrow { \mathrm { AC } }$ and hence show that the equation of the hillside is $2 x - 2 y + 25 z = 650$.
The tunnel $T _ { \mathrm { A } }$ begins at A and goes in the direction of the vector $15 \mathbf { i } + 14 \mathbf { j } - 2 \mathbf { k }$; the tunnel $T _ { \mathrm { C } }$ begins at C and goes in the direction of the vector $8 \mathbf { i } + 7 \mathbf { j } - 2 \mathbf { k }$. Both these tunnels extend a long way into the ground.\\
(ii) Find the least possible length of a tunnel which connects B to a point in $T _ { \mathrm { A } }$.\\
(iii) Find the least possible length of a tunnel which connects a point in $T _ { \mathrm { A } }$ to a point in $T _ { \mathrm { C } }$.\\
(iv) A tunnel starts at B , passes through the point ( $18,138 , p$ ) vertically below C , and intersects $T _ { \mathrm { A } }$ at the point Q . Find the value of $p$ and the coordinates of Q .
\hfill \mbox{\textit{OCR MEI FP3 2012 Q1 [24]}}