Port A is defined to be the origin of a set of coordinate axes and port B is located at the point (70, 30), where distances are measured in kilometres. A ship S1 sails from port A at 10:00 in a straight line such that its position t hours after 10:00 is given by $\vec r = t{10 \choose 20}$
A speedboat S2 is capable of three times the speed of S1 and is to meet S1 by traveling the shortest possible distance. What is the latest time that S2 can leave port B?
My working:
I need to relate both the vectors, so I know I need to parameterize. As well as it was given that S2 speed is 3 times S1 so I know I must include this into the equation too.
as follows:
$\vec r = {70 \choose 30} + (t-p){30 \choose 60}$
However, this is still incorrect! the answer is:
$\vec r = {70 \choose 30} + (t-k){-60 \choose 30}$
can someone please explain why is that? How and where did the answer get negative from? as well as why is (x, y) interchanged? Is the directional vector the gradient?