In: Advanced Math
r1(t) = (5-5t , t , 5t) , r2(t) = (6-3t , 5-2t ,-3+4t)
(a) for crashing , r1(t) = r2(t)
=> (5-5t , t , 5t) = (6-3t , 5-2t ,-3+4t)
Now , Equating x coordinates we get 5-5t = 6-3t => -2t = 1 => t = -0.5 which is not possible , since time cannot be negative , So objects cannot crash
(b)
Let they cross path at time t1 and t2 respectively
r1(t1) = r2(t2)
=> (5-5t1 , t1 , 5t1) = (6-3t2, 5-2t2,-3+4t2)
Now Equating x coordinates : 5-5t1 = 6-3t2 => 3t2 - 5t1 = 1 ----------(1)
Equating y coordinates : t1 = 5-2t2 ------- (2)
Put t1 from (2) in (1) we get
3t2 - 5(5-2t2) = 1 => 3t2 -25+10t2 = 1 => 13t2 = 26 => t2 = 2
t1 = 5 - 2*2 = 1
Verifying for z coordinates : r1_z = 5t1 = 5*1 = 5 and r2_z = -3+4t2 = -3+4*2 = 5 , Clearly r1_z = r2_z ,
Therefore paths of objects intersect.
Now we will found how close do the objects get to each other at any time t
Distance between them at any time t>=0 , d = |r1 - r2| = |(5-5t , t , 5t) - (6-3t , 5-2t ,-3+4t)| = |(-1-2t ,3t-5,t+3 )|
d2 = (-1-2t)2 + (3t-5)2 + (t+3)2 = 1+4t2+4t +9t2+25-30t + t2+6t+9 = 14t2 - 20t +35
d2 ' = 0 => 28 t - 20 => t = 20/28 = 5/7
d2 at 5/7 = 14*(5/7)2 -20 *5/7 +35 = 195/7 => d = (195/7)1/2
Minimum distance between objects at any time t is (195/7)1/2.
(c) The objects never meet but cross paths at some time. (i.e. their paths do not miss each other).
Minimum distance between objects at any time t is (195/7)1/2.
Minimum distance between paths of objects is 0 as they intersect.