| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a Further Maths question requiring the cross product formula for shortest distance between skew lines, then finding a plane equation using a normal vector. While the techniques are standard for FP3, the multi-step vector manipulation and need to work with Cartesian line equations in 3D places it above average difficulty. |
| Spec | 4.04b 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 |
|---|---|---|
| (i) \((\mathbf{i} - \mathbf{j} - 2\mathbf{k}) \times (-4\mathbf{i} - 14\mathbf{j} + 2\mathbf{k}) = -30\mathbf{i} + 6\mathbf{j} - 18\mathbf{k}\) | M1 | For vector product of direction vectors |
| A1 | For correct vector for common perp | |
| \((5\mathbf{i} + \mathbf{j} + 5\mathbf{k}) - (\mathbf{i} + 11\mathbf{j} + 2\mathbf{k}) = 4\mathbf{i} - 10\mathbf{j} + 3\mathbf{k}\) | B1 | For calculating the difference of positions |
| \(d = \frac{ | (4\mathbf{i} - 10\mathbf{j} + 3\mathbf{k}) \cdot (5\mathbf{i} - \mathbf{j} + 3\mathbf{k}) | }{ |
| A1 | For correct exact answer | |
| 5 | ||
| (ii) Normal vector for plane is \(5\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) | B1√ | For stating or using the normal vector |
| Point on plane is \(5\mathbf{i} + \mathbf{j} + 5\mathbf{k}\) | B1 | For using any point of \(l_1\) |
| Equation is \(5x - y + 3z = 25 - 1 + 15\) i.e. \(5x - y + 3z = 39\) | M1 | For using relevant direction and point |
| A1 | For a correct equation | |
| 4 | ||
| 9 |
**(i)** $(\mathbf{i} - \mathbf{j} - 2\mathbf{k}) \times (-4\mathbf{i} - 14\mathbf{j} + 2\mathbf{k}) = -30\mathbf{i} + 6\mathbf{j} - 18\mathbf{k}$ | M1 | For vector product of direction vectors
| A1 | For correct vector for common perp
$(5\mathbf{i} + \mathbf{j} + 5\mathbf{k}) - (\mathbf{i} + 11\mathbf{j} + 2\mathbf{k}) = 4\mathbf{i} - 10\mathbf{j} + 3\mathbf{k}$ | B1 | For calculating the difference of positions
$d = \frac{|(4\mathbf{i} - 10\mathbf{j} + 3\mathbf{k}) \cdot (5\mathbf{i} - \mathbf{j} + 3\mathbf{k})|}{|5\mathbf{i} - \mathbf{j} + 3\mathbf{k}|} = \frac{39}{\sqrt{35}}$ | M1 | For calculation of the projection
| A1 | For correct exact answer
| **5** |
**(ii)** Normal vector for plane is $5\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ | B1√ | For stating or using the normal vector
Point on plane is $5\mathbf{i} + \mathbf{j} + 5\mathbf{k}$ | B1 | For using any point of $l_1$
Equation is $5x - y + 3z = 25 - 1 + 15$ i.e. $5x - y + 3z = 39$ | M1 | For using relevant direction and point
| A1 | For a correct equation
| **4** |
| **9** |5 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations
$$\frac { x - 5 } { 1 } = \frac { y - 1 } { - 1 } = \frac { z - 5 } { - 2 } \quad \text { and } \quad \frac { x - 1 } { - 4 } = \frac { y - 11 } { - 14 } = \frac { z - 2 } { 2 } .$$
(i) Find the exact value of the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.\\
(ii) Find an equation for the plane containing $l _ { 1 }$ and parallel to $l _ { 2 }$ in the form $a x + b y + c z = d$.
\hfill \mbox{\textit{OCR FP3 Q5 [9]}}