[In this question i and j are unit vectors due east and due north respectively: Position vectors are given relative to a fixed origin O.] Two ships P and Q are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

To find the bearing on which ship Q is moving, we first establish its velocity vector, which is given as $(3i + 4j)$ km h$^{-1}$. The bearing is determined by the angle $heta$ formed with the positive direction of the x-axis (east). We use the tangent function:

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}$

Calculating $heta$, we have:

$\theta = \tan^{-1}(\frac{4}{3}) \approx 53.13^{\circ}$

Since bearings are measured clockwise from north, we convert this angle:
Bearing = $90^{\circ} - 53.13^{\circ} = 36.87^{\circ}$.
Thus, rounded to the nearest degree, the bearing is $37^{\circ}$.