1. ## Vector question

An airplane ascends to that its position relative to the airport control tower t minutes after take-off is given by the vector r=-1i-2j+t(4i+5j+0.6k), the units being kilometres. The x and y axis point twards the east and the north respectively. Calculate the closest distance of the airplane from the airport control tower during the flight.
To the nearest second,how many seconds after leaving the ground is the airplane at its closest to the airport 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.

Q-A = 1i+2j
The vector in the direction from A to N is p=4i+5j+0.6k
|p|^2=4^2+5^2+(0.6)^2=41.36
|p|=6.43
The unit vector in the line AN is u=4/6.43i +5/6.43j + 0.6/6.43k=0.622i +0.778j +0.093k
AN is the dot product of Q-A and u,
AN=(i,2j,0k).(0.622i,0.778j,0.093k)=0.622+1.556=2. 178.

QA^2=1+4=5
QN^2=AQ^2-AN^2=5-(2.178)^2=0.256
QN=sqroot(0.256)=0.51km.

The book gives 1.68km.
To calculate the time after departure that this is the distance I used the cosine formula a^2=b^2+c^2-2bccosA and got the angle between QA and NQ. I found the displacement in x to be 0.1167 and substituted this into r=-1i-2j+t(4i+5j+0.6k).

I solved t to be 16.7 or 17 seconds.

The book gives 20 seconds.

Thanks for any help.

2. Hello, woollybull!

An airplane ascends to that its position relative to the airport control tower
$t$ minutes after take-off is given by the vector: . $\vec r(t)\:=\:-{\bf i}-2{\bf j}+t(4{\bf i}+5{\bf j}+0.6{\bf k}),$
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: . $\vec r(t) \;=\;(4t-1){\bf i} + (5t-1){\bf j} + (0.6t){\bf k}$

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

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

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

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

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

3. ## Aha!

Thank you very much for that!

4. ## In need of further help.

While I follow what you have done, the question asks to find the distance first and then the time it takes. Is it possible to find the distance before finding time?
The book only gives one similar type of example, given below.

Find the distance of the point Q with coordinates (1,2,3) from the straight line with equation r=3i+4j-2k+ t(i-2j+2k)

let p=i-2j+2k find u in the direction of p

The solution lets A be the point (3,4,-2) and N be the foot of the perpendicular from Q to the line. It then follows on the triangle ANQ.

AQ=(i+2j+3k)-(3i+4j-2k)
=-2i-2j+5k
AQ^2= (-2)^2+(-2)^2+5^2=33
u=1/3i-2/3j+2/3k

AN=AQcosx=AQ.u=4
so QN^2=AQ^2-AN^2=33-16=17
QN=sqroot17

I am unable to see the differences from the example above and the question from the first post. Even though minimising the distance makes sense, why cant I apply the ideas of the example to the original question.

Thanks.