| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Kinematics with position vectors |
| Difficulty | Standard +0.8 This is a standard M4 relative velocity problem requiring vector calculus to minimize distance between particles. While it involves multiple steps (finding position vectors as functions of time, computing distance squared, differentiating to find minimum), the technique is well-established and follows a predictable method. The 3D vectors and algebraic manipulation add moderate complexity, placing it slightly above average difficulty for Further Maths mechanics. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form4.04a Line equations: 2D and 3D, cartesian and vector forms |
A particle $A$ has constant velocity $(3\mathbf{i} + \mathbf{j})$ m s$^{-1}$ and a particle $B$ has constant velocity $(\mathbf{i} - \mathbf{k})$ m s$^{-1}$. At time $t = 0$ seconds, the position vectors of the particles $A$ and $B$ with respect to a fixed origin $O$ are $(-6\mathbf{i} + 4\mathbf{j} - 3\mathbf{k})$ m and $(-2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$ m respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that, in the subsequent motion, the minimum distance between $A$ and $B$ is $4\sqrt{2}$ m. [6]
\item Find the position vector of $A$ at the instant when the distance between $A$ and $B$ is a minimum. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2014 Q1 [8]}}