[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 position vector (p) of ship P at time (t) hours after 2 pm, we can use its initial position vector and add the product of its velocity and time:

$p = (i + j) + (2i - 3j)t = (1 + 2t)i + (1 - 3t)j$

For ship Q, the initial position is given as (-2j), and its velocity is (3i + 4j). Thus, the expression for (q) is:

$q = (-2j) + (3i + 4j)t = 3ti + (-2 + 4t)j$

The vector (PQ) can be found by taking the difference between the position of ship Q and ship P:

$PQ = q - p = [3ti - (1 + 2t)i] + [(-2 + 4t)j - (1 - 3t)j]$ This simplifies to:

$PQ = (3t - (1 + 2t))i + ((-2 + 4t) - (1 - 3t))j = (t - 1)i + (7t - 3)j$

For Q to be due north of P, the x-components of both position vectors must be equal:

$3t - (1 + 2t) = 0\implies t = 1\text{ hour or }3\text{ pm}$

For Q to be north-west of P, we need the condition that the j-component of PQ is greater than the i-component:

$7t - 3 \geq t - 1\implies 6t \geq 2\implies t \geq \frac{1}{3}\text{ or }2:30 \text{ pm.}$