How Do You Find When and Where Two Moving Objects Are Closest to Each Other?

Find the relative position vector from one object to the other, minimize its squared magnitude to obtain the time of closest approach, then substitute that time into both original position vectors to find both locations.

The Method

For objects moving with constant velocities, suppose

$$\mathbf{r}_A(t)=\mathbf{a}+t\mathbf{u}, \qquad \mathbf{r}_B(t)=\mathbf{b}+t\mathbf{v}.$$

Their relative position is

$$\Delta\mathbf{r}(t)=\mathbf{r}_A(t)-\mathbf{r}_B(t).$$

This vector points from object B to object A at the same time, so its magnitude is their separation. Instead of minimizing the distance $d(t)=\lvert\Delta\mathbf{r}(t)\rvert$, minimize the squared distance

$$D(t)=d(t)^2=\Delta\mathbf{r}(t)\cdot\Delta\mathbf{r}(t).$$

This removes the square root without changing where the minimum occurs. Expand $D(t)$, solve $D'(t)=0$, and verify a minimum using $D''(t)>0$ or the upward-opening quadratic shape. Check that the time lies in the permitted domain. For a closed interval, compare the stationary value with both endpoints.

For constant velocities, write $\mathbf{p}=\mathbf{a}-\mathbf{b}$ and $\mathbf{w}=\mathbf{u}-\mathbf{v}$, so $\Delta\mathbf{r}=\mathbf{p}+t\mathbf{w}$. Differentiating gives . Hence , provided . This compact formula is equivalent to expanding the quadratic. If , the objects have equal velocities and their separation is constant, so every allowed time gives the same distance.

Worked Example

Let

$$\mathbf{r}_A(t)=\begin{pmatrix}0\\0\end{pmatrix}+t\begin{pmatrix}2\\1\end{pmatrix}, \qquad \mathbf{r}_B(t)=\begin{pmatrix}6\\0\end{pmatrix}+t\begin{pmatrix}-1\\2\end{pmatrix}.$$

Then

$$\Delta\mathbf{r}(t)=\begin{pmatrix}3t-6\\-t\end{pmatrix}, \qquad D(t)=(3t-6)^2+t^2=10t^2-36t+36.$$

Therefore,

$$D'(t)=20t-36=0 \Rightarrow t=1.8.$$

Since $D''(t)=20>0$, this gives the minimum. Substitution gives

$$\mathbf{r}_A(1.8)=\begin{pmatrix}3.6\\1.8\end{pmatrix}, \qquad \mathbf{r}_B(1.8)=\begin{pmatrix}4.2\\3.6\end{pmatrix}.$$

The minimum distance is $\sqrt{D(1.8)}=\sqrt{3.6}\approx1.90$ units. At closest approach, the relative-position vector is perpendicular to the relative-velocity vector, providing a useful check.

Exam Technique

This is AHL 3.12 Vector applications to kinematics, so it is HL-only and all relevant papers are GDC-active. Show the relative vector and squared-distance function before giving a GDC minimum. State the time, both positions, minimum distance, and appropriate units.

Keep full calculator precision until the final answers. If a negative stationary time is obtained but $t\geq0$, the closest permitted approach occurs at $t=0$.

The common misconception is that intersecting paths or closest approach implies a collision. A collision requires both objects to occupy the same position at the same time, so the minimum distance must be zero.