Equations of Lines — Question 10

Question 10

Two particles move according to $$\vec r_A(t)=\left\langle 1,0,2\right\rangle+t\left\langle 2,1,-1\right\rangle,\qquad \vec r_B(t)=\left\langle 5,2,0\right\rangle+t\left\langle-1,0,1\right\rangle,\qquad t\ge 0.$$ Do their paths intersect? Do the particles collide? Find their minimum separation for $t\ge 0$.

Question 10 – Solution

Strategy Path intersection permits different parameters; collision requires the same time. Minimum separation comes from minimizing the squared relative distance.

Paths Solving $\vec r_A(s)=\vec r_B(u)$ gives $s=2,u=0$, so the paths intersect at $(5,2,0)$. The particles do not collide there because they reach it at different times. Directly equating positions at the same $t$ is inconsistent.

Minimum separation The relative position is $$\vec r_A(t)-\vec r_B(t)=\left\langle-4+3t,-2+t,2-2t\right\rangle.$$ Thus $D^2(t)=14t^2-36t+24$, minimized at $t=36/28=9/7$. The minimum squared distance is $6/7$, so $$\boxed{D_{\min}=\sqrt{6/7}=\frac{\sqrt{42}}7}.$$

Verification The minimizing time is nonnegative, and the quadratic has positive leading coefficient.