| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| 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 challenging Further Maths question requiring multiple sophisticated techniques: finding direction vectors, computing cross products, applying the skew lines distance formula to derive a quadratic equation, then finding the angle between two planes using normal vectors. The multi-step nature, algebraic manipulation to reach the given equation, and the need to work with two different values of λ make this significantly harder than standard A-level questions, though the individual techniques are well-defined once the approach is identified. |
| Spec | 4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
The position vectors of the points $A , B , C , D$ are\\
$7 \mathbf { i } + 4 \mathbf { j } - \mathbf { k }$,\\
$3 \mathbf { i } + 5 \mathbf { j } - 2 \mathbf { k }$,\\
$2 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k }$,\\
$2 \mathbf { i } + 7 \mathbf { j } + \lambda \mathbf { k }$\\
respectively. It is given that the shortest distance between the line $A B$ and the line $C D$ is 3 .\\
(i) Show that $\lambda ^ { 2 } - 5 \lambda + 4 = 0$.\\
(ii) Find the acute angle between the planes through $A , B , D$ corresponding to the values of $\lambda$ satisfying the equation in part (i).
\hfill \mbox{\textit{CAIE FP1 2008 Q12 EITHER}}