SOLUTION: Object P moves according to r=(5i,-1j,2k)+t(-1,-3,1), and the object q moves according to r=(-4,-3,4)+t(1,-2,2). The distance units are meters, and t>0 is the time in seconds. (

Algebra ->  Vectors -> SOLUTION: Object P moves according to r=(5i,-1j,2k)+t(-1,-3,1), and the object q moves according to r=(-4,-3,4)+t(1,-2,2). The distance units are meters, and t>0 is the time in seconds. (      Log On


   

Question 1184825: Object P moves according to r=(5i,-1j,2k)+t(-1,-3,1), and the object q moves according to r=(-4,-3,4)+t(1,-2,2). The distance units are meters, and t>0 is the time in seconds.
(a) Find the initial distance between the objects.
(b)Find the shortest distance between the objects, and the time when it occurs

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I understand that the coordinates of P (in meters) are
x=5-t , y=-1-3t , z=2%2Bt ,
and the coordinates of Q (in meters) are
x=-4%2Bt , y=-3-2t , z=-4%2B2t , with t in seconds.
The square of the distance between the two objects is
d%5E2=%285-t%2B4-t%29%5E2%2B%28-1-3t%2B3%2B2t%29%5E2%2B%282%2Bt-4-2t%29%5E2
d%5E2=%289-2t%29%5E2%2B%282-t%29%5E2%2B%28-2-t%29%5E2
d%5E2=81-36t%2B4t%5E2%2B4-4t%2Bt%5E2%2B4%2B4t%2Bt%5E2
d%5E2=89-36t%2B6t%5E2
d%5E2=6t%5E2-36t%2B89
d%5E2=6%28t%5E2-6t%29%2B89
That equation shows that the initial distance between the objects (in meters) is d=highlight%28sqrt%2889%29%29, or approximately highlight%289.4m%29 .
With some more algebraic manipulation, we get
d%5E2=6%28t%5E2-6t%2B9%29%2B89-54
d%5E2=6%28t-3%29%5E2%2B35
That last equation shows that the maximum for d%5E2 , and for d occurs when highlight%28t=3%29seconds .
At that time,
d%5E2=35 and highlight%28d=sqrt%2835%29%29meters or approximately highlight%285.9meters%29