two boats p and q are at points whose position vector are 4i+8j and 4i+3j respectively. both of the boats are moving at a constant velocity of p is 4i+j and q is 2i+5j. find the position vectors of p&q and pq after t hours, and hence express the distance between p and q interms of t. show that the minimum distance between the two boats is 15kilometers

Answer 1

p(t) = 4i+8j + (4i+j)t
q(t) = 4i+3j + (2i+5j)t
the distance pq is
d(t) = p(t)-q(t) = 5j + (2i-4j)t = (2t)i + (5-4t)j
|d| = √((2t)^2 + (5-4t)^2)
= √(20t^2-40t+25)

Hmmm. I get √5 as the minimum distance.

Better check my math.