SOLUTION: From a car traveling east at 40 mi.per hr., an airplane traveling horizontally north at 100 mi.per hr, is visible 1 mi. east, 2 mi. south and 2 mi. up. Find when the two will be

We place the initial position of the car at the origin of the coordinate system (x,y,z) = (0,0,0).


Then the airplane's initial position is (x,y,z) = (1,-2,2).


The trajectory of the car   in time is  (40t,0,0).

The trajectory of the plane in time is  (0,-2+100t,2).


The vector from the car to the airplane in the coordinate form is  (-40t,-2+100t,2).


The square of the length of this vector (= the distance between the objects) is

    d^2(t) =  =  = 

                 = .


The distance d(t) is minimum when d^2(t) is minimum.


d^2(t) is minimum at  t = "  " = - = 0.017241 of an hour = 1.0345 minutes.


At this time moment, the square of the distance is  d^2(t) =  = 8.0069 mi^2,

hence,  the distance itself is  d(t) =  = 2.830 miles.


Compare it with the initial distance  d(0) =  =  = 3 miles.