Hello, woollybull!

An airplane ascends to that its position relative to the airport control tower

$\displaystyle t$ minutes after take-off is given by the vector: .$\displaystyle \vec r(t)\:=\:-{\bf i}-2{\bf j}+t(4{\bf i}+5{\bf j}+0.6{\bf k}),$

the units being kilometres.

The x and y axis point twards the east and the north respectively.

(a) Calculate the closest distance of the airplane from the control tower during the flight.

(b) To the nearest second, how many seconds after leaving the ground is the airplane

at its closest to the control tower?

I drew a triangle ANQ with A(-1,-2,0), The tower Q(0,0,0), and N unknown.

NQ, the required distance, *being perpendicular to AN.* . . . . no Code:

A* |
*|
*
|*
| *
| *
| *N
| /
| /
|/
- - * - - - - -
Q

We have: .$\displaystyle \vec r(t) \;=\;(4t-1){\bf i} + (5t-1){\bf j} + (0.6t){\bf k} $

The distance is: .$\displaystyle d \:=\:QN \;=\;\sqrt{(4t-1)^2 + (5t-2)^2 + (0.6t)^2}$

. . which we wish to minimize.

This can be accomplished by minimizing the __square__ of $\displaystyle d\!:\;\;D \:=\:d^2$

So we have: .$\displaystyle D \;=\;(4t-1)^2 + (5t-2)^2 + (0.6t)^2 \;=\;41.36t^2 - 28t + 5$

Then: .$\displaystyle D' \:=\:82.72t - 28 \:=\:0$

Therefore: .$\displaystyle t \:=\:\frac{28}{87.72} \:=\:0.33836858\text{ minutes} \:\approx\:20\text{ seconds}$