Question

rOA and rOB are the position vectors of the two particles Aand B at time t.If both particles start moving when t=0,determine whethert the particles collide and,if they do,give the value of t when this occurs and the position vector of the point of collision.If they do not ,find the time and their distance apart when they are closest together .

(a) rOA =3i-7j+(3i+2j)t, rOB=8i-6j+(2i+j)t.

(b) rOA=-5i+2j+9k+(5i-j+2k)t, rOB=i+2j+3k+(3i-j+4k)t.

Answer 1

(a)

$$\vec r_A=\vec r_B$$

$$3\vec i-7\vec j+(3\vec i+2\vec j)t=8\vec i-6\vec j+(2\vec i+\vec j)t$$

$$\begin{matrix} 3+3t=8+2t \\ -7+2t=-6+t \\ \end{matrix}$$

$$\begin{matrix} t=5 \\ t=1 \\ \end{matrix}$$

No solution. The particles will never collide.

$$distance^2=d^2$$

$$=(8+2t-3-3t)^2+(-6+t+7-2t)^2$$

$$=(5-t)^2+(1-t)^2=2t^2-12t+26$$

$$t_{vertex}=\dfrac{12}{2(2)}=3$$

$$d_{min}=\sqrt{2(3)^2-12(3)+26}=2\sqrt{2}$$

$$\vec r_A(3)=12\vec i-\vec j$$

$$\vec r_B(3)=14\vec i-3\vec j$$

(b)

$$\vec r_A=\vec r_B$$

$$-5\vec i+2\vec j+9\vec k+(5\vec i-\vec j+2\vec k)t$$

$$=\vec i+2\vec j+3\vec k+(3\vec i-\vec j+4\vec k)t$$

$$\begin{matrix} -5+5t=1+3t \\ 2-t=2-t \\ 9+2t=3+4t \\ \end{matrix}$$

The solution $t=3$

The particles will collide at time $t=3$ after the start, and

$$\vec r_A=\vec r_B=10\vec i-\vec j+15\vec k$$