Crossing Paths in a Space Station

Best Technique To Solve This Problem is By using Concept of Relative Motion

Let us Consider our Refrence frame at B (Say Bob) , Then in this frame B is at rest , So

Equation of motion of A (Alfred) is a straight Line Motion w.r.t Bob (B) So it's equation is :

$y-0\quad =\quad \cfrac { -3 }{ 5 } (x-\cfrac { 3L }{ 2 } )\\ \\ 3x+5y\quad =\quad \cfrac { 9L }{ 2 }$ .

Closest Distance between them is equal to Perpendicular distance of (bob) B(0,L) from the Line which is :

${ d }_{ min }=\left| \cfrac { 0+\quad 5L\quad -\cfrac { 9L }{ 2 } }{ \sqrt { { 3 }^{ 2 }+{ 5 }^{ 2 } } } \right| \\ \\ \boxed { { d }_{ min }\quad =\quad \cfrac { L }{ 2\sqrt { 34 } } } \quad$ .

Let the lower-left corner of the room be the origin of our x- y- coordinate system. Then, Alfred and Bob's position vectors, as functions of time, can be written as:

$\vec{r}_{A}(t) = <(\frac{3L}{2}),(3vt)>$ , and $\vec{r}_{B}(t) = <(5vt),(L)>$

Then, the displacement vector pointing from Bob to Alfred can be written:

$\Delta\vec{r} = <(\frac{3L}{2}-5vt),(3vt-L)>$

Now we need to minimize the distance. To make the math simpler, we mimimize the square of the distance instead. That is:

$r^{2} = (\frac{3L}{2}-5vt)^{2}+(3vt-L)^{2}$

and:

$\frac{d}{dt}r^{2} = 2r\frac{dr}{dt} = -21vL + 68v^{2}t$

Setting this equal to 0, we can cancel out the $2r$ , as I've already verified that $r=0$ never occurs. Solving for $t$ , we find that the time of occurence where Alfred and Bob are closest together is:

$t = \frac{21L}{68v}$

Now we simply plug this into our distance formula, and simplify to get:

$r_{min} = \frac{L}{2\sqrt{34}}$

So, $\boxed{a+b = 36}$