Two particals located at (8i+7j+2k) m and (11i+j+2k) m respectivly, start moving with constant velocities of (

Question

(-3i+ 5j-6k) m/s and (-7i+8j-11k) m/s respectivly. find the minium sepration between them, and the time after which this occurs.

Answer 1

I hope I did this correctly. Would you let me know?

I set up the equation of the relative distance the particles will have from eachother at time t:

distance equals location of particle one minus location of particle two:

x=
(4t-3)i+
(6-3t)j+
(5t)k

then I distributed the i, j and k to get
4it-3j+
6j-3tj+
5tk

Then I took the derivative dx/dt treating i, j, and k as constants to get
4i-3j+5k
this is the minimum separation
4i-3j+5k

This I then used to set up the following three simultaneous equations to find the time that occurs:

4t-3=4
t=1

6-3t=-3
t=1

5t=5
t=1

j

Answer 2

We are interested in finding the minimum separation between the particles. It makes sense to work every thing relative to particle 1.

First, move origin to particle 1 at time 0- which means subtract its position from all positions. The initial position of particle 2 will be:

(11-8, 1-7, 2-2) =(3, -6, 0).

Now, let us change the frame one where particle 1 is at rest - subtract its velocity from all velocities: The velocity of particle 2 becomes

(-7+3, 8-5, -11+6)=(-4, 3, -5)

The position of particle 2 at any time t :

(3, -6, 0) + (-4, 3, -5)t
=
(3 - 4t, -6 +3t, -5t)

The distance from particle 1 is
sqrt((3-4t)^2 + (-6+3t)^2 + (-5t)^2)
= sqrt(50t^2 -60t +45)

Set the derivative to 0 to find the minimum:

100t - 60 = 0
>>
t = 0.6 sec.