| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.2 This is a comprehensive Further Maths vectors question covering multiple standard techniques (cross product, parallel lines, shortest distance between skew lines, intersection conditions). While it requires several steps and different methods, each part follows well-established procedures taught in FP3. The cross product calculation is routine, finding k for parallel lines is straightforward, and the shortest distance formulas are standard. Part (iv) requires solving simultaneous equations but is methodical rather than requiring novel insight. |
| Spec | 4.04b 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 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix}-1\\4\\3\end{pmatrix}\), \(\overrightarrow{CD} = \begin{pmatrix}-k\\4\\k+2\end{pmatrix}\) | M1 | Finding both vectors |
| \(\overrightarrow{AB} \times \overrightarrow{CD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&4&3\\-k&4&k+2\end{vmatrix}\) | M1 | Attempt at cross product |
| \(= \begin{pmatrix}4(k+2)-12\\-3k+(-k-2+3k)\\-4+4k\end{pmatrix} = \begin{pmatrix}4k-4\\-k-2\\4k-4\end{pmatrix}\) | A1 A1 | Each component correct |
| Answer | Marks |
|---|---|
| \(k=1\) (since \(\overrightarrow{AB} \times \overrightarrow{CD} = \mathbf{0}\) requires \(4k-4=0\)) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| With \(k=1\): \(\overrightarrow{CD} = \begin{pmatrix}-1\\4\\3\end{pmatrix}\), unit vector \(= \frac{1}{\sqrt{26}}\begin{pmatrix}-1\\4\\3\end{pmatrix}\) | M1 | |
| \(\overrightarrow{AC} = \begin{pmatrix}3\\8\\-4\end{pmatrix}\) | M1 | Finding vector between lines |
| \(\overrightarrow{AC} \times \hat{n}\): component of \(\overrightarrow{AC}\) perpendicular to \(\overrightarrow{AB}\) | M1 | Correct method for distance |
| \(\overrightarrow{AC} \times \overrightarrow{AB} = \begin{pmatrix}8\cdot3-(-4)\cdot4\\(-4)(-1)-3\cdot3\\3\cdot4-8\cdot(-1)\end{pmatrix} = \begin{pmatrix}40\\-5\\20\end{pmatrix}\) | A1 | |
| Distance \(= \frac{\sqrt{40^2+5^2+20^2}}{\sqrt{26}} = \frac{\sqrt{2025}}{\sqrt{26}} = \frac{45}{\sqrt{26}}\) | A1 A1 | \(\approx 8.83\) |
| Answer | Marks | Guidance |
|---|---|---|
| Normal to plane is \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}40\\-5\\20\end{pmatrix}\), simplify to \(\begin{pmatrix}8\\-1\\4\end{pmatrix}\) | M1 | |
| Using point A(-2,-3,2): \(8(-2)-1(-3)+4(2)+d=0\) | M1 | Substituting point |
| \(-16+3+8+d=0 \Rightarrow d=5\) | A1 | |
| \(8x - y + 4z + 5 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Lines: \(\mathbf{r} = \begin{pmatrix}-2\\-3\\2\end{pmatrix}+s\begin{pmatrix}-1\\4\\3\end{pmatrix}\), \(\mathbf{r} = \begin{pmatrix}k\\5\\-2\end{pmatrix}+t\begin{pmatrix}-k\\4\\k+2\end{pmatrix}\) | B1 | |
| \(\overrightarrow{AC} = \begin{pmatrix}k+2\\8\\-4\end{pmatrix}\) | M1 | |
| \(\overrightarrow{AB} \times \overrightarrow{CD} = \begin{pmatrix}4k-4\\-k-2\\4k-4\end{pmatrix}\), magnitude \(= \sqrt{2(4k-4)^2+(k+2)^2}\) | M1 | |
| Distance \(= \frac{ | \overrightarrow{AC} \cdot (\overrightarrow{AB}\times\overrightarrow{CD}) | }{ |
| Numerator: \((k+2)(4k-4)+8(-k-2)+(-4)(4k-4)\) \(= 4k^2-4k+8k-8-8k-16-16k+16 = 4k^2-20k-8\) | A1 | |
| Distance \(= \frac{ | 4k^2-20k-8 | }{\sqrt{2(4k-4)^2+(k+2)^2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| For intersection, set \(\begin{pmatrix}-2-s\\-3+4s\\2+3s\end{pmatrix} = \begin{pmatrix}k-kt\\5+4t\\-2+(k+2)t\end{pmatrix}\) | M1 | |
| From equations: solve simultaneously | M1 | |
| \(4k^2-20k-8=0 \Rightarrow k^2-5k-2=0\)... | M1 | Setting numerator = 0 |
| From \(-3+4s=5+4t\) and \(2+3s=-2+(k+2)t\) and \(-2-s=k-kt\) | ||
| \(k=6\), with \(s=2, t=1\) | A1 A1 | |
| Point of intersection: \((-4, 5, 8)\) | A1 |
# Question 1: (Option 1: Vectors)
A(-2,-3,2), B(-3,1,5), C(k,5,-2), D(0,9,k)
## Part (i)
| $\overrightarrow{AB} = \begin{pmatrix}-1\\4\\3\end{pmatrix}$, $\overrightarrow{CD} = \begin{pmatrix}-k\\4\\k+2\end{pmatrix}$ | M1 | Finding both vectors |
|---|---|---|
| $\overrightarrow{AB} \times \overrightarrow{CD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&4&3\\-k&4&k+2\end{vmatrix}$ | M1 | Attempt at cross product |
| $= \begin{pmatrix}4(k+2)-12\\-3k+(-k-2+3k)\\-4+4k\end{pmatrix} = \begin{pmatrix}4k-4\\-k-2\\4k-4\end{pmatrix}$ | A1 A1 | Each component correct |
## Part (ii)(A)
| $k=1$ (since $\overrightarrow{AB} \times \overrightarrow{CD} = \mathbf{0}$ requires $4k-4=0$) | B1 | |
## Part (ii)(B)
| With $k=1$: $\overrightarrow{CD} = \begin{pmatrix}-1\\4\\3\end{pmatrix}$, unit vector $= \frac{1}{\sqrt{26}}\begin{pmatrix}-1\\4\\3\end{pmatrix}$ | M1 | |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}3\\8\\-4\end{pmatrix}$ | M1 | Finding vector between lines |
| $\overrightarrow{AC} \times \hat{n}$: component of $\overrightarrow{AC}$ perpendicular to $\overrightarrow{AB}$ | M1 | Correct method for distance |
| $\overrightarrow{AC} \times \overrightarrow{AB} = \begin{pmatrix}8\cdot3-(-4)\cdot4\\(-4)(-1)-3\cdot3\\3\cdot4-8\cdot(-1)\end{pmatrix} = \begin{pmatrix}40\\-5\\20\end{pmatrix}$ | A1 | |
| Distance $= \frac{\sqrt{40^2+5^2+20^2}}{\sqrt{26}} = \frac{\sqrt{2025}}{\sqrt{26}} = \frac{45}{\sqrt{26}}$ | A1 A1 | $\approx 8.83$ |
## Part (ii)(C)
| Normal to plane is $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}40\\-5\\20\end{pmatrix}$, simplify to $\begin{pmatrix}8\\-1\\4\end{pmatrix}$ | M1 | |
|---|---|---|
| Using point A(-2,-3,2): $8(-2)-1(-3)+4(2)+d=0$ | M1 | Substituting point |
| $-16+3+8+d=0 \Rightarrow d=5$ | A1 | |
| $8x - y + 4z + 5 = 0$ | | |
## Part (iii)
| Lines: $\mathbf{r} = \begin{pmatrix}-2\\-3\\2\end{pmatrix}+s\begin{pmatrix}-1\\4\\3\end{pmatrix}$, $\mathbf{r} = \begin{pmatrix}k\\5\\-2\end{pmatrix}+t\begin{pmatrix}-k\\4\\k+2\end{pmatrix}$ | B1 | |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}k+2\\8\\-4\end{pmatrix}$ | M1 | |
| $\overrightarrow{AB} \times \overrightarrow{CD} = \begin{pmatrix}4k-4\\-k-2\\4k-4\end{pmatrix}$, magnitude $= \sqrt{2(4k-4)^2+(k+2)^2}$ | M1 | |
| Distance $= \frac{|\overrightarrow{AC} \cdot (\overrightarrow{AB}\times\overrightarrow{CD})|}{|\overrightarrow{AB}\times\overrightarrow{CD}|}$ | M1 | |
| Numerator: $(k+2)(4k-4)+8(-k-2)+(-4)(4k-4)$ $= 4k^2-4k+8k-8-8k-16-16k+16 = 4k^2-20k-8$ | A1 | |
| Distance $= \frac{|4k^2-20k-8|}{\sqrt{2(4k-4)^2+(k+2)^2}}$ | A1 | Accept equivalent forms |
## Part (iv)
| For intersection, set $\begin{pmatrix}-2-s\\-3+4s\\2+3s\end{pmatrix} = \begin{pmatrix}k-kt\\5+4t\\-2+(k+2)t\end{pmatrix}$ | M1 | |
|---|---|---|
| From equations: solve simultaneously | M1 | |
| $4k^2-20k-8=0 \Rightarrow k^2-5k-2=0$... | M1 | Setting numerator = 0 |
| From $-3+4s=5+4t$ and $2+3s=-2+(k+2)t$ and $-2-s=k-kt$ | | |
| $k=6$, with $s=2, t=1$ | A1 A1 | |
| Point of intersection: $(-4, 5, 8)$ | A1 | |
---1 Four points have coordinates $\mathrm { A } ( - 2 , - 3,2 ) , \mathrm { B } ( - 3,1,5 ) , \mathrm { C } ( k , 5 , - 2 )$ and $\mathrm { D } ( 0,9 , k )$.
\begin{enumerate}[label=(\roman*)]
\item Find the vector product $\overrightarrow { \mathrm { AB } } \times \overrightarrow { \mathrm { CD } }$.
\item For the case when AB is parallel to CD ,\\
(A) state the value of $k$,\\
(B) find the shortest distance between the parallel lines AB and CD ,\\
(C) find, in the form $a x + b y + c z + d = 0$, the equation of the plane containing AB and CD .
\item When AB is not parallel to CD , find the shortest distance between the lines AB and CD , in terms of $k$.
\item Find the value of $k$ for which the line AB intersects the line CD , and find the coordinates of the point of intersection in this case.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2006 Q1 [24]}}