| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Common perpendicular to two skew lines |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring students to find the common perpendicular between two skew lines using vector methods. It demands setting up parametric equations, applying perpendicularity conditions (dot products = 0), solving simultaneous equations, and computing the shortest distance. While the techniques are standard for FM students, the multi-step coordination and algebraic manipulation elevate it significantly above routine exercises. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&3&0\\4&0&-1\end{vmatrix} = -3\mathbf{i}+4\mathbf{j}-12\mathbf{j}\) | M1A1 | Uses vector product to find vector perpendicular to both lines |
| \(BA = \mathbf{i}+10\mathbf{j}-11\mathbf{j}\) | ||
| perp.dist. \(= \frac{-3\times1+4\times10+12\times11}{\sqrt{3^2+4^2+12^2}} = \frac{169}{\sqrt{169}} = 13\) (AG) | M1A1 | Finds \(BA\) and scalar product with unit perpendicular vector; Part mark: 4; No penalty for sign errors in \(\mathbf{n}\) which lead to correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{p} = \begin{pmatrix}8+4\lambda\\8+3\lambda\\-7\end{pmatrix}\), \(\mathbf{q} = \begin{pmatrix}7+4\mu\\-2\\4-\mu\end{pmatrix}\) | ||
| \(\mathbf{PQ} = \begin{pmatrix}-1-4\lambda+4\mu\\-10-3\lambda\\11-\mu\end{pmatrix} = t\begin{pmatrix}-3\\4\\-12\end{pmatrix}\) | B1 M1A1 | |
| \(\Rightarrow t=-1,\ \lambda=-2,\ \mu=-1\) | M1 | |
| \(\mathbf{p} = 2\mathbf{j}-7\mathbf{k}\), \(\mathbf{q} = 3\mathbf{i}-2\mathbf{j}+5\mathbf{k}\) | A1 | Part mark: 5; Award B1B1B1 if \(t\) assumed to be \(\pm1\) i.e. 3/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Finds two parameter representation for \(\mathbf{PQ}\) | ||
| \(\mathbf{PQ} = \begin{pmatrix}-1-4\lambda+4\mu\\-10-3\lambda\\11-\mu\end{pmatrix}\) | B1 | |
| \(16\mu - 25\lambda = 34\) | M1A1 | Uses scalar product between \(\mathbf{PQ}\) and direction vector of at least one line and equates to zero |
| \(17\mu - 16\lambda = 15\) | A1 | |
| \(\mu = -1,\ \lambda = -2\) | M1A1 | Solves simultaneously |
| \(\mathbf{p} = 2\mathbf{j}-7\mathbf{k}\), \(\mathbf{q} = 3\mathbf{i}-2\mathbf{j}+5\mathbf{k}\) | A1 | Part mark: 7 |
| \(\sqrt{3^2+(-4)^2+12^2} = 13\) (AG) | M1A1 | Obtains length of \(PQ\); Part mark: 2 |
## Question 6(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&3&0\\4&0&-1\end{vmatrix} = -3\mathbf{i}+4\mathbf{j}-12\mathbf{j}$ | M1A1 | Uses vector product to find vector perpendicular to both lines |
| $BA = \mathbf{i}+10\mathbf{j}-11\mathbf{j}$ | | |
| perp.dist. $= \frac{-3\times1+4\times10+12\times11}{\sqrt{3^2+4^2+12^2}} = \frac{169}{\sqrt{169}} = 13$ (AG) | M1A1 | Finds $BA$ and scalar product with unit perpendicular vector; Part mark: 4; No penalty for sign errors in $\mathbf{n}$ which lead to correct result |
## Question 6(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{p} = \begin{pmatrix}8+4\lambda\\8+3\lambda\\-7\end{pmatrix}$, $\mathbf{q} = \begin{pmatrix}7+4\mu\\-2\\4-\mu\end{pmatrix}$ | | |
| $\mathbf{PQ} = \begin{pmatrix}-1-4\lambda+4\mu\\-10-3\lambda\\11-\mu\end{pmatrix} = t\begin{pmatrix}-3\\4\\-12\end{pmatrix}$ | B1 M1A1 | |
| $\Rightarrow t=-1,\ \lambda=-2,\ \mu=-1$ | M1 | |
| $\mathbf{p} = 2\mathbf{j}-7\mathbf{k}$, $\mathbf{q} = 3\mathbf{i}-2\mathbf{j}+5\mathbf{k}$ | A1 | Part mark: 5; Award B1B1B1 if $t$ assumed to be $\pm1$ i.e. 3/5 |
### Question 6(ii) Alternative Solution:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Finds two parameter representation for $\mathbf{PQ}$ | | |
| $\mathbf{PQ} = \begin{pmatrix}-1-4\lambda+4\mu\\-10-3\lambda\\11-\mu\end{pmatrix}$ | B1 | |
| $16\mu - 25\lambda = 34$ | M1A1 | Uses scalar product between $\mathbf{PQ}$ and direction vector of at least one line and equates to zero |
| $17\mu - 16\lambda = 15$ | A1 | |
| $\mu = -1,\ \lambda = -2$ | M1A1 | Solves simultaneously |
| $\mathbf{p} = 2\mathbf{j}-7\mathbf{k}$, $\mathbf{q} = 3\mathbf{i}-2\mathbf{j}+5\mathbf{k}$ | A1 | Part mark: 7 |
| $\sqrt{3^2+(-4)^2+12^2} = 13$ (AG) | M1A1 | Obtains length of $PQ$; Part mark: 2 |6 The line $l _ { 1 }$ passes through the point with position vector $8 \mathbf { i } + 8 \mathbf { j } - 7 \mathbf { k }$ and is parallel to the vector $4 \mathbf { i } + 3 \mathbf { j }$. The line $l _ { 2 }$ passes through the point with position vector $7 \mathbf { i } - 2 \mathbf { j } + 4 \mathbf { k }$ and is parallel to the vector $4 \mathbf { i } - \mathbf { k }$. The point $P$ on $l _ { 1 }$ and the point $Q$ on $l _ { 2 }$ are such that $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$. In either order,\\
(i) show that $P Q = 13$,\\
(ii) find the position vectors of $P$ and $Q$.
\hfill \mbox{\textit{CAIE FP1 2011 Q6 [9]}}