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Explanation:

Step1: Set up position - vector

Let the position of JBLM be the origin ((0,0)). The destination is at the point ((139, - 170)) (taking east as the positive x - axis and north as the positive y - axis). The position of Mt. Rainier is ((56,-40)). The position of the plane at time (t) (in hours) is given by the vector (\vec{r}(t)=(139t,-170t)) (since speed (v = 800) km/h and the direction vector from JBLM to the destination has components based on the x and y displacements).

Step2: Calculate the distance - squared function

The distance (d) between the plane at position (\vec{r}(t)) and Mt. Rainier ((56,-40)) is given by (d^{2}=(139t - 56)^{2}+(-170t + 40)^{2}). Expand this: [ \begin{align*} d^{2}&=(139t - 56)^{2}+(-170t + 40)^{2}\ &=139^{2}t^{2}-2\times139\times56t + 56^{2}+170^{2}t^{2}-2\times170\times40t + 40^{2}\ &=(139^{2}+170^{2})t^{2}-(2\times139\times56 + 2\times170\times40)t+(56^{2}+40^{2}) \end{align*} ] where (139^{2}=19321), (170^{2}=28900), (2\times139\times56 = 2\times7784=15568), (2\times170\times40 = 13600), (56^{2}=3136), (40^{2}=1600). So (d^{2}=(19321 + 28900)t^{2}-(15568+13600)t+(3136 + 1600)=48221t^{2}-29168t + 4736).

Step3: Minimize the distance - squared function

Since minimizing (d^{2}) is equivalent to minimizing (d), and for a quadratic function (y = ax^{2}+bx + c) ((a = 48221), (b=-29168), (c = 4736)), the minimum occurs at (t=-\frac{b}{2a}). [t=\frac{29168}{2\times48221}=\frac{29168}{96442}\approx0.3024] hours.

Step4: Convert time to minutes

Since (1) hour ( = 60) minutes, (t) in minutes is (t_{min}=0.3024\times60 = 18.144\approx18.14) minutes.

Answer:

(18.14) minutes