Equations of Lines — Question 10

Question 10

Two particles move for $t\ge 0$ according to $$\begin{aligned} \vec r_A(t)&=\left\langle 0,1,2\right\rangle+t\left\langle 2,-1,1\right\rangle,\\ \vec r_B(t)&=\left\langle 6,-2,5\right\rangle+t\left\langle -1,1,0\right\rangle. \end{aligned}$$

Tasks

1. Determine whether their geometric paths intersect.

2. Determine whether the particles collide.

3. Find the time and value of their minimum separation for $t\ge 0$.

Question 10 – Solution

Strategy Path intersection allows different parameters, whereas collision uses the same time. Minimize the squared norm of the relative-position vector.

Paths versus collision Solving $\vec r_A(s)=\vec r_B(u)$ gives $s=3$, $u=0$, so the paths intersect at $\boxed{(6,-2,5)}$. The particles reach that point at different times and therefore do not collide. Equating positions at the same $t$ would require all components of the relative vector below to vanish, which never occurs.

Separation At a common time $t$, $$\vec r_A(t)-\vec r_B(t)=\left\langle -6+3t,3-2t,-3+t\right\rangle.$$ Hence $$D^2(t)=(-6+3t)^2+(3-2t)^2+(-3+t)^2=14t^2-54t+54.$$ The vertex occurs at $$t=\frac{54}{28}=\frac{27}{14}>0.$$ Substitution gives $D^2_{\min}=27/14$, so $$\boxed{D_{\min}=\sqrt{\frac{27}{14}}=\frac{3\sqrt{42}}{14}}.$$

Verification The quadratic opens upward, and the minimizing time lies in the allowed domain. The nonzero minimum confirms independently that no collision occurs.