| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| 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 standard Further Maths vectors question requiring the cross product formula for distance between skew lines (part i) and finding a plane equation (part ii). While it involves multiple steps and Further Maths content, the techniques are routine applications of formulas taught in FP1 with no novel geometric insight required. |
| 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 lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}1\\2\\7\end{pmatrix}\) | B1 | |
| \(\overrightarrow{OC}\times\overrightarrow{AB} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\1&2&7\end{vmatrix} = \begin{pmatrix}-9\\-6\\3\end{pmatrix} = t\begin{pmatrix}3\\2\\-1\end{pmatrix}\) | M1 A1 | Finds direction of common perpendicular |
| \(\frac{\begin{pmatrix}1\\-1\\0\end{pmatrix}\cdot\begin{pmatrix}3\\2\\-1\end{pmatrix}}{\sqrt{3^2+2^2+1^2}} = \frac{1}{\sqrt{14}} = 0.267\) | M1 A1 | Uses formula for shortest distance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\3&2&-1\end{vmatrix} = t\begin{pmatrix}-1\\4\\5\end{pmatrix}\) | M1 A1 | Finds normal to plane |
| \(-(0)+4(0)+5(0) = 0\) | M1 | Uses point on plane |
| \(-x + 4y + 5z = 0\) | A1 | AEF |
## Question 6:
### Part 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}1\\2\\7\end{pmatrix}$ | B1 | |
| $\overrightarrow{OC}\times\overrightarrow{AB} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\1&2&7\end{vmatrix} = \begin{pmatrix}-9\\-6\\3\end{pmatrix} = t\begin{pmatrix}3\\2\\-1\end{pmatrix}$ | M1 A1 | Finds direction of common perpendicular |
| $\frac{\begin{pmatrix}1\\-1\\0\end{pmatrix}\cdot\begin{pmatrix}3\\2\\-1\end{pmatrix}}{\sqrt{3^2+2^2+1^2}} = \frac{1}{\sqrt{14}} = 0.267$ | M1 A1 | Uses formula for shortest distance |
**Total: 5 marks**
### Part 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{n} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\3&2&-1\end{vmatrix} = t\begin{pmatrix}-1\\4\\5\end{pmatrix}$ | M1 A1 | Finds normal to plane |
| $-(0)+4(0)+5(0) = 0$ | M1 | Uses point on plane |
| $-x + 4y + 5z = 0$ | A1 | AEF |
**Total: 4 marks**6 With $O$ as the origin, the points $A , B , C$ have position vectors
$$\mathbf { i } - \mathbf { j } , \quad 2 \mathbf { i } + \mathbf { j } + 7 \mathbf { k } , \quad \mathbf { i } - \mathbf { j } + \mathbf { k }$$
respectively.\\
(i) Find the shortest distance between the lines $O C$ and $A B$.\\
(ii) Find the cartesian equation of the plane containing the line $O C$ and the common perpendicular of the lines $O C$ and $A B$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q6 [9]}}