Shortest distance between two moving points

Question

So I found this question on the Internet, which turned out more tricky than I thought:

" The position of boat A is given by $x(t)=3-t$ and $y(t)=2t-4$ The position of boat B is $x(t)=4-3t)$ and $y(t)=3-2t)$ respectively. Find the value of $t$ for which the boats are closes to each other. (distances are in kilometers and time in hours)"

I solved the question by finding $|B-A|$ and differentiating so that I can see when the distance is at minimum. I got the right answer but then realised that I'm not allowed to use calculus... and here's my problem. Is there any way to solve this using vector or other geometric operations? And am I right in looking at the algorithm to find the distance between lines?

Answer 1

The key word is perpendicularity. Given a parametrised line $$ \cases{x(t)=at+b\\y(t)=ct+d}, $$ the point on that line where it's closest to the origin is is given by $t_0$, which has the property that the position vector $(x(t_0),y(t_0))$ is perpendicular to the direction vector $(a,c)$. Do that for the line $B-A$, and you get your answer.

Answer 2

Once you write an equation for $d=\sqrt{(x_A(t)-x_B(t))^2+(y_A(t)-y_B(t))^2}$, then consider $d^2$.

$d^2=20t^2-60t+50$

$0=20t^2-60t+50-d^2$

$0=2t^2-6t+5-\frac{d^2}{10}$

The axis of symmetry of a parabola is $t=-\frac{b}{2a}$. The minimum or maximum must be at the vertex.

We know it is a minimum because ...?

(If this isn't enough let me know.)