The Closest Approach of Two Objects

The gap between them, as a vector

Object A is at rA = (1 + 3t, 2 + 2t) and object B at rB = (10 − t, 6 − t), with distances in kilometers and t in hours. They never collide: the x-coordinates agree only at t = 2.25, and there the y-coordinates are 6.5 and 3.75. So how close do they come, and when?

Subtract one position vector from the other, coordinate by coordinate. The vector from A to B is AB = rB − rA = (10 − t − 1 − 3t, 6 − t − 2 − 2t) = (9 − 4t, 4 − 3t). At t = 0 it is (9, 4), and every hour it changes by (−4, −3), which is B's velocity minus A's: the velocity of B relative to A.

Square the length

The distance between the objects is the length of AB. Its square is easier to work with, because it has no square root: $|AB{|}^2={\left(9-4t\right)}^2+{\left(4-3t\right)}^2$|AB|² = (9 − 4t)² + (4 − 3t)².

Expand each bracket: ${\left(9-4t\right)}^2=81-72t+16t^2$(9 − 4t)² = 81 − 72t + 16t² and ${\left(4-3t\right)}^2=16-24t+9t^2$(4 − 3t)² = 16 − 24t + 9t². Add them: $|AB{|}^2=25t^2-96t+97$|AB|² = 25t² − 96t + 97, a quadratic in t.

A distance is never negative, and a bigger distance always has a bigger square. So the time at which $|AB{|}^2$|AB|² is smallest is the time at which |AB| is smallest.

Find the lowest point

The coefficient of $t^2$t² is 25, which is positive, so the graph of the quadratic opens upward and turns once, at its lowest point. Differentiate with respect to t: $\frac{d}{dt}\left(25t^2-96t+97\right)=50t-96$d/dt (25t² − 96t + 97) = 50t − 96. At the turning point the derivative is 0, so 50t = 96 and t = 1.92.

At t = 1.92, $|AB{|}^2=25\times 1.92^2-96\times 1.92+97=4.84$|AB|² = 25 × 1.92² − 96 × 1.92 + 97 = 4.84. Take the square root at the end: the closest approach is $\sqrt{4.84}=2.2$√4.84 = 2.2 km, 1.92 hours after the start.

Check it against nearby times: $|AB{|}^2$|AB|² is 26 at t = 1, 5 at t = 2 and 34 at t = 3, all bigger than 4.84. At t = 2 the distance is $\sqrt{5}$√5, about 2.236 km, only a little more than 2.2.

The gold curve is $|AB{|}^2=25t^2-96t+97$|AB|² = 25t² − 96t + 97, which starts at 97, above the top of the window, and the plain curve below it is the distance |AB| itself. Both are lowest at t = 1.92, where $|AB{|}^2=4.84$|AB|² = 4.84 and |AB| = 2.2.

The positions at the closest moment

At t = 1.92, A is at (1 + 5.76, 2 + 3.84) = (6.76, 5.84) and B is at (10 − 1.92, 6 − 1.92) = (8.08, 4.08). The gap is (1.32, −1.76), and $\sqrt{1.32^2+1.76^2}=\sqrt{4.84}=2.2$√(1.32² + 1.76²) = √4.84 = 2.2.

A moves along the gold path from (1, 2) and B along the plain one from (10, 6). The arrow is the gap from A to B at t = 1.92, when it is shortest: 2.2 long, and at right angles to the direction in which B moves relative to A.

Two more routes to the same moment

Completing the square gives the same answer without calculus. $25t^2-96t+97=25{\left(t-1.92\right)}^2+97-25\times 1.92^2=25{\left(t-1.92\right)}^2+4.84$25t² − 96t + 97 = 25(t − 1.92)² + 97 − 25 × 1.92² = 25(t − 1.92)² + 4.84. A square is never negative, so the least value is 4.84, when t = 1.92.

The gap is also shortest when it is perpendicular to the relative velocity (−4, −3). Before that moment B is still closing on A; after it, B is drawing away. Perpendicular means the dot product is 0: (−4)(9 − 4t) + (−3)(4 − 3t) = 25t − 48 = 0, so t = 1.92 again. Check: (1.32, −1.76) · (−4, −3) = −5.28 + 5.28 = 0.

When the turning point comes too early

If the quadratic turns at a negative time, the objects were nearest before the clock started. Suppose the gap were (9 + 4t, 4 + 3t). Then $|AB{|}^2=25t^2+96t+97$|AB|² = 25t² + 96t + 97, which turns at t = −1.92. From t = 0 on, the objects only draw apart, so the closest they come from then on is at t = 0, $\sqrt{97}$√97, about 9.849 km.

The usual mistakes

Leaving off the square root. The quadratic is the squared distance; its least value 4.84 still needs a square root, giving 2.2.

Dropping the 2 when differentiating. The derivative of $25t^2$25t² is 50t, so $25t^2-96t+97$25t² − 96t + 97 turns at $t=\frac{96}{50}=1.92$t = 96/50 = 1.92, not at $\frac{96}{25}$96/25.

Minimizing each coordinate on its own. 9 − 4t is 0 at t = 2.25 and 4 − 3t is 0 at $t=\frac{4}{3}$t = 4/3, and neither is the moment the distance is least. The coordinates must be combined into $|AB{|}^2$|AB|² first.

Trying whole numbers of hours. t = 2 gives $\sqrt{5}$√5, close but not the least; the least value falls at t = 1.92.

Ships and aircraft

In the first application below, two ships on steady courses do not collide, and the perpendicular condition finds their closest approach. In the second, two aircraft on a radar screen must stay 25 km apart, and completing the square finds the least separation.