Two particles (1 and 2) move with constant speeds v_1 and…
Two particles (1 and 2) move with constant speeds $v_1$ and $v_2$ along two mutually perpendicular straight lines, toward the point where the lines cross. At $t = 0$, particle 1 is a distance $l_1$ from the crossing point and particle 2 is a distance $l_2$ from it. (a) After what time is the distance between the particles smallest? (b) What is that smallest distance?
Explanation of Method
Closest approach of two particles on perpendicular paths.
Step 1: Positions and distance
Measure each particle's position from the crossing point O along its own line. Both move toward O, so
\[ x_1 = l_1 - v_1 t, \]
\[ y_2 = l_2 - v_2 t. \]
(After a particle passes O the value becomes negative, which is fine: it is then on the other side.) By Pythagoras, $d^2(t) = (l_1 - v_1 t)^2 + (l_2 - v_2 t)^2$.
- particle 1
- particle 2
- shortest distance
Step 2: A parabola in $t$
Expand and group by powers of $t$: $d^2(t) = (v_1^2 + v_2^2)\,t^2 - 2\,(v_1 l_1 + v_2 l_2)\,t + (l_1^2 + l_2^2)$.
This is a parabola $At^2 - 2Bt + C$ opening upward, so its lowest point is at the vertex $t^* = B/A$. The distance $d$ is smallest exactly when $d^2$ is smallest.
Step 3: (a) Time of closest approach
\[t^* = \frac{v_1 l_1 + v_2 l_2}{v_1^2 + v_2^2}\]
Step 4: (b) The smallest distance
At the vertex, $d_{\min}^2 = C - \dfrac{B^2}{A}$: $d_{\min}^2 = l_1^2 + l_2^2 - \frac{(v_1 l_1 + v_2 l_2)^2}{v_1^2 + v_2^2}$.
\[ \phantom{d_{\min}^2} = \frac{(l_1^2 + l_2^2)(v_1^2 + v_2^2) - (v_1 l_1 + v_2 l_2)^2}{v_1^2 + v_2^2}. \]
Multiplying out the numerator, most terms cancel and what is left is a perfect square, $(v_2 l_1 - v_1 l_2)^2$. So
\[d_{\min} = \frac{|\,v_2 l_1 - v_1 l_2\,|}{\sqrt{v_1^2 + v_2^2}}\]
Check with the example in the figure
$ l_1 = 4,\ l_2 = 3,\ v_1 = 1,\ v_2 = 2: \quad t^* = \dfrac{4 + 6}{1 + 4} = 2 $
Positions at $t^* = 2$: $ x_1 = 2,\ y_2 = -1,\ d = \sqrt{4 + 1} = \sqrt 5 $
Formula: $ d_{\min} = \dfrac{|2\cdot 4 - 1 \cdot 3|}{\sqrt 5} = \dfrac{5}{\sqrt 5} = \sqrt 5\ \checkmark $
Result
\[\boxed{\,t^* = \frac{v_1 l_1 + v_2 l_2}{v_1^2 + v_2^2}\,}\]
\[\boxed{\,d_{\min} = \frac{|\,v_2 l_1 - v_1 l_2\,|}{\sqrt{v_1^2 + v_2^2}}\,}\]
Remark
Notes. The original solution found the same results by setting the derivative of $d(t)$ to zero; the parabola's vertex gives them without calculus. The simpler form of $d_{\min}$ shows when the particles collide: $d_{\min} = 0$ exactly when $\dfrac{l_1}{v_1} = \dfrac{l_2}{v_2}$, i.e. when both reach O at the same moment.