The parametric straight line paths of two objects are given. a) do the objects c

Question

The parametric straight line paths of two objects are given. a) do the objects crash (at the same location at the same time)? If so, at what time? b) do the paths of the objects intersect (at the same point at different times)? If so, how close do they get to each other? c) do the objects and paths miss each other? If so, how close do the objects get to each other and how close do their paths get to each other?
Object A is at x=5-5t y=t z=5t Object B is at x=6-3t y=5-2t z=-3+4t The parametric straight line paths of two objects are given. a) do the objects crash (at the same location at the same time)? If so, at what time? b) do the paths of the objects intersect (at the same point at different times)? If so, how close do they get to each other? c) do the objects and paths miss each other? If so, how close do the objects get to each other and how close do their paths get to each other?
Object A is at x=5-5t y=t z=5t Object B is at x=6-3t y=5-2t z=-3+4t a) do the objects crash (at the same location at the same time)? If so, at what time? b) do the paths of the objects intersect (at the same point at different times)? If so, how close do they get to each other? c) do the objects and paths miss each other? If so, how close do the objects get to each other and how close do their paths get to each other?
Object A is at x=5-5t y=t z=5t Object B is at x=6-3t y=5-2t z=-3+4t

Explanation / Answer

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.

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