Question 1194697
.
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 nearest together.
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<pre>
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) = {{{(-40t)^2 + (-2+100t)^2 + 2^2}}} = {{{11600t^2 - 2*2*100t + 4 + 4}}} = 

                 = {{{11600t^2 - 400t + 8}}}.


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


d^2(t) is minimum at  t = " {{{-b/(2a)}}} " = -{{{(-400/(2*11600))}}} = 0.017241 of an hour = 1.0345 minutes.


At this time moment, the square of the distance is  d^2(t) = {{{11600*0.0345^2-400*0.0345 + 8}}} = 8.0069 mi^2,

hence,  the distance itself is  d(t) = {{{sqrt(8.0069)}}} = 2.830 miles.


Compare it with the initial distance  d(0) = {{{sqrt(1^2 + 2^2 + 2^2)}}} = {{{sqrt(9)}}} = 3 miles.
</pre>

Solved.


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On finding minimum of a quadratic function, &nbsp;see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-complete-the-square-of-a-quadratic-function-to-find-its-minimum-maximum.lesson>HOW TO complete the square to find the minimum/maximum of a quadratic function</A>

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