A boat B is moving with constant velocity - Edexcel - A-Level Maths Mechanics - Question 7 - 2007 - Paper 1

The position vector of B at time t hours after noon can be expressed as:

$b = (3i - 4j) + t \cdot v$, where v is the velocity found previously.

Substituting the value of v:

$b = (3i - 4j) + t(2i + 6j) = (3 + 2t)i + (-4 + 6t)j$ km.

Since the position vector of C is given by: $c = (9i + 20j) + ((6i + 6j)\lambda),$ we need to express the position vector at the same time t when both B and C intercept.

Equating the i-component for interception: $3 + 2t = 9 + 6λ$

And for the j-component: $-4 + 6t = 20 + 6λ$.

Solving both equations will yield the required value of λ.

The speed of B is given as: $v_B = \sqrt{(2)^2 + (6)^2} = \sqrt{40} = 2\sqrt{10} \text{ km/h}.$ For boat C, the velocity is derived from the coefficients in the position vector: $v_C = \sqrt{(6)^2 + (2)^2} = \sqrt{40} = 2\sqrt{10} \text{ km/h}.$

Therefore, both boats B and C have the same speed of $2\sqrt{10}$ km/h before intercepting.