Closest Approach In Two-dimensional Relative Motion

Minimum separation problems are most cleanly solved in the relative frame, where one ship is treated as stationary and the other moves with the relative velocity along a straight line; the closest approach is the perpendicular distance from the stationary ship to that line. Place ship B at the origin and ship A initially at $(100,0)$. Ship A's velocity is $(0,5)$ and ship B's is $(5,0)$, so A's velocity relative to B is $(0-5,5-0)=(-5,5)\text{ m/s}$. In B's frame, A starts at $(100,0)$ and travels along the direction $(-5,5)$, i.e. $(-1,1)$ normalized. The closest approach is the perpendicular distance from the origin to the line through $(100,0)$ with direction $(-1,1)$. That line is $x+y=100$, and the distance from the origin is $\frac{|0+0-100|}{\sqrt{1^2+1^2}}=\frac{100}{\sqrt{2}}=50\sqrt{2}\text{ m}$. The option 100 m is the initial separation, not the minimum. The option 50 m and 100√2 m arise from arithmetic mis-scaling of the perpendicular distance. This is the JEE Advanced relative-frame technique for closest approach. As a check, $50\sqrt{2}\approx 70.7\text{ m}$ is sensibly less than the starting 100 m, confirming the ships first draw nearer before separating.