| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| 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 +1.8 This is a substantial Further Maths question requiring multiple vector techniques: finding line of intersection of two planes, shortest distance between skew lines using the scalar triple product formula, distance between parallel lines, and finding intersection of two lines in 3D. While each individual technique is standard for FP3, the multi-part nature (4 parts building on each other) and the need to correctly apply several different formulas makes this significantly harder than average A-level content, though not exceptionally difficult for Further Maths students who have practiced these methods. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. \(23x - 23y = 46\) | M1A1 | Eliminating one of \(x\), \(y\), \(z\) |
| \(x = t\), \(y = t - 2\) | M1A1 ft | |
| \(3t - 4(t-2) - 2z = 8\) | ||
| \(x = t\), \(y = t-2\), \(z = -\frac{1}{2}t\) | A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{PQ} = \begin{pmatrix}-1+7\mu \\ 12-14\mu \\ 5+4\mu\end{pmatrix} - \begin{pmatrix}2\lambda \\ -2+2\lambda \\ -\lambda\end{pmatrix}\), \(\overrightarrow{PQ} \cdot \mathbf{d} = \overrightarrow{PQ} \cdot \overrightarrow{AB} = 0\) | M1 | Two equations for \(\lambda\) and \(\mu\) |
| \(2(-1+7\mu-2\lambda)+2(14-14\mu-2\lambda)-(5+4\mu+\lambda)=0\) | A1 ft | |
| \(7(-1+7\mu-2\lambda)-14(14-14\mu-2\lambda)+4(5+4\mu+\lambda)=0\), \(\lambda=27/25\), \(\mu=47/75\) | A1 ft | |
| \( | \overrightarrow{PQ} | = \sqrt{(92/75)^2+(230/75)^2+(644/75)^2} = 9.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AX} \cdot \mathbf{d} = \begin{pmatrix}6+2\lambda+1 \\ -2+2\lambda-12 \\ 9-\lambda-5\end{pmatrix} \cdot \begin{pmatrix}2\\2\\-1\end{pmatrix} = 0\) | M1 | |
| \(2(7+2\lambda)+2(2\lambda-14)-(4-\lambda)=0\), \(\lambda=2\) | A1 ft | |
| \(\overrightarrow{AX} = \begin{pmatrix}11\\-10\\2\end{pmatrix}\) | M1 | |
| \(AX = \sqrt{11^2+10^2+2^2} = 15\) | M1, A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}7\\-14\\4\end{pmatrix} \cdot \left[\begin{pmatrix}k+1\\-12\\-3\end{pmatrix} \times \begin{pmatrix}2\\2\\-1\end{pmatrix}\right] = 0\) | M1 | Appropriate scalar triple product equated to zero |
| \(\begin{pmatrix}7\\-14\\4\end{pmatrix} \cdot \begin{pmatrix}18\\k-5\\2k+26\end{pmatrix} = 0\) | M1, A1 | |
| \(126-14k+70+8k+104=0\), \(k=50\) | A1 | Equation for \(k\) |
| At D: \(\begin{pmatrix}-1\\12\\5\end{pmatrix}+\lambda\begin{pmatrix}51\\-12\\-3\end{pmatrix} = \begin{pmatrix}6\\-2\\9\end{pmatrix}+\mu\begin{pmatrix}2\\2\\-1\end{pmatrix}\) | M1 | Condone use of same parameter on both sides |
| \(-1+51\lambda = 6+2\mu\); \(12-12\lambda=-2+2\mu\); \(5-3\lambda=9-\mu\) | A1 ft | |
| \(\lambda=\frac{1}{3}\), \(\mu=5\) | M1, A1 | Two equations for \(\lambda\) and \(\mu\) |
| D is \((6+2\mu,\ -2+2\mu,\ 9-\mu)\) i.e. (16, 8, 4) | M1, A1 | 8 Obtaining coordinates of D |
# Question 1:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. $23x - 23y = 46$ | M1A1 | Eliminating one of $x$, $y$, $z$ |
| $x = t$, $y = t - 2$ | M1A1 ft | |
| $3t - 4(t-2) - 2z = 8$ | | |
| $x = t$, $y = t-2$, $z = -\frac{1}{2}t$ | A1 | **5 marks** |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PQ} = \begin{pmatrix}-1+7\mu \\ 12-14\mu \\ 5+4\mu\end{pmatrix} - \begin{pmatrix}2\lambda \\ -2+2\lambda \\ -\lambda\end{pmatrix}$, $\overrightarrow{PQ} \cdot \mathbf{d} = \overrightarrow{PQ} \cdot \overrightarrow{AB} = 0$ | M1 | Two equations for $\lambda$ and $\mu$ |
| $2(-1+7\mu-2\lambda)+2(14-14\mu-2\lambda)-(5+4\mu+\lambda)=0$ | A1 ft | |
| $7(-1+7\mu-2\lambda)-14(14-14\mu-2\lambda)+4(5+4\mu+\lambda)=0$, $\lambda=27/25$, $\mu=47/75$ | A1 ft | |
| $|\overrightarrow{PQ}| = \sqrt{(92/75)^2+(230/75)^2+(644/75)^2} = 9.2$ | M1A1 ft, A1 | **6** Expression for shortest distance |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AX} \cdot \mathbf{d} = \begin{pmatrix}6+2\lambda+1 \\ -2+2\lambda-12 \\ 9-\lambda-5\end{pmatrix} \cdot \begin{pmatrix}2\\2\\-1\end{pmatrix} = 0$ | M1 | |
| $2(7+2\lambda)+2(2\lambda-14)-(4-\lambda)=0$, $\lambda=2$ | A1 ft | |
| $\overrightarrow{AX} = \begin{pmatrix}11\\-10\\2\end{pmatrix}$ | M1 | |
| $AX = \sqrt{11^2+10^2+2^2} = 15$ | M1, A1 | **5 marks** |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}7\\-14\\4\end{pmatrix} \cdot \left[\begin{pmatrix}k+1\\-12\\-3\end{pmatrix} \times \begin{pmatrix}2\\2\\-1\end{pmatrix}\right] = 0$ | M1 | Appropriate scalar triple product equated to zero |
| $\begin{pmatrix}7\\-14\\4\end{pmatrix} \cdot \begin{pmatrix}18\\k-5\\2k+26\end{pmatrix} = 0$ | M1, A1 | |
| $126-14k+70+8k+104=0$, $k=50$ | A1 | Equation for $k$ |
| At D: $\begin{pmatrix}-1\\12\\5\end{pmatrix}+\lambda\begin{pmatrix}51\\-12\\-3\end{pmatrix} = \begin{pmatrix}6\\-2\\9\end{pmatrix}+\mu\begin{pmatrix}2\\2\\-1\end{pmatrix}$ | M1 | Condone use of same parameter on both sides |
| $-1+51\lambda = 6+2\mu$; $12-12\lambda=-2+2\mu$; $5-3\lambda=9-\mu$ | A1 ft | |
| $\lambda=\frac{1}{3}$, $\mu=5$ | M1, A1 | Two equations for $\lambda$ and $\mu$ |
| D is $(6+2\mu,\ -2+2\mu,\ 9-\mu)$ i.e. **(16, 8, 4)** | M1, A1 | **8** Obtaining coordinates of D |
---1 The point $\mathrm { A } ( - 1,12,5 )$ lies on the plane $P$ with equation $8 x - 3 y + 10 z = 6$. The point $\mathrm { B } ( 6 , - 2,9 )$ lies on the plane $Q$ with equation $3 x - 4 y - 2 z = 8$. The planes $P$ and $Q$ intersect in the line $L$.\\
(i) Find an equation for the line $L$.\\
(ii) Find the shortest distance between $L$ and the line AB .
The lines $M$ and $N$ are both parallel to $L$, with $M$ passing through A and $N$ passing through B .\\
(iii) Find the distance between the parallel lines $M$ and $N$.
The point C has coordinates $( k , 0,2 )$, and the line AC intersects the line $N$ at the point D .\\
(iv) Find the value of $k$, and the coordinates of D .
\hfill \mbox{\textit{OCR MEI FP3 2009 Q1 [24]}}