1. ## vectors

in this question, distance is in kilometers, time is in hours.
a balloon is moving at a constant height with a speed of 18kmh^-1, in the direction of the vector (3,4,0)
At time t = 0, the balloon is at point B with coordinates (0,0,5).

(a) Show that the position vector b of the baloon at time t is given by b = (x,y,z) = (0,0,5) + t(10.8, 14.4, 0)

At time t = 0, a helicopter goes to deliver a message to the baloon. The position bector h of the helicopter at time t is given by h = (x,y,z) = (49, 32, 0) + t(-48, -24, 6).

(b) (i) Write down the coordinates of the starting position of the helicopter.
(ii) find the speed of the helicopter.

(c) the helicopter reaches the baloon at point R.
(i) Find the time the helicopter takes to reach the baloon.
(ii) Find the coordinates of R.

Thanks!

2. Originally Posted by samantha_malone
in this question, distance is in kilometers, time is in hours.
a balloon is moving at a constant height with a speed of 18kmh^-1, in the direction of the vector (3,4,0)
At time t = 0, the balloon is at point B with coordinates (0,0,5).

(a) Show that the position vector b of the baloon at time t is given by b = (x,y,z) = (0,0,5) + t(10.8, 14.4, 0)

At time t = 0, a helicopter goes to deliver a message to the baloon. The position bector h of the helicopter at time t is given by h = (x,y,z) = (49, 32, 0) + t(-48, -24, 6).

(b) (i) Write down the coordinates of the starting position of the helicopter.
(ii) find the speed of the helicopter.

(c) the helicopter reaches the baloon at point R.
(i) Find the time the helicopter takes to reach the baloon.
(ii) Find the coordinates of R.

Thanks!
Hello,

first calculate the length of (3, 4, 0):
$|(3, 4, 0)| = \sqrt{3^2 + 4^2 + 0^2} = 5$

You know now that the balloon travels the distance of 18 km in 1 hour:

$18 km = r \cdot |(3,4,0)| = r \cdot 5~ \Longrightarrow ~ r = 3.6$.

That means the speed-vector is: $3.6 \cdot (3,4,0) = (10.8, 14.4, 0)$.

Obviously the balloon travels at a height of 5 km. Therefore the position of the balloon at the time t could be described by:

$b = (x,y,z) = (0,0,5) + t \cdot (10.8, 14.4, 0)$

to (b)(i): Plug in t = 0 and you'll get: (49, 32, 0)
to (b)(ii): This question can't be answered so easily:
1. The speed through the air correspond to the length of the speed vector which is: (-48, -24, 6): $|\overrightarrow{c_{H}}| = \sqrt{(-48)^2+(-24)^2+6^2} = \sqrt{2916}=54~\frac{km}{h}$

2. The speed over ground correspond to the length of the vector (-48, -24, 0): $|(-48, -24, 0)| = \sqrt{48^2+24^2+0^2} = \sqrt{2880}= 24 \cdot \sqrt{5} \approx 53.7~\frac{km}{h}$

to c)(i) If the helicopter meets the balloon it must have exactly the height of 5 km. Because the helicopter needs 1 hour to reach the height of 6 km it will be at 5 km height after $\frac{5}{6} ~ h = 50~ min$

to (c)(ii): Plug in this value for t into the position vector of the helicopter:

$h = (x,y,z) = (49, 32, 0) + \frac{5}{6} \cdot (-48, -24, 6) = (9, 12, 5)$

To check this result plug in the value for t into the equation of the position vector for b:

$b = (x,y,z) = (0,0,5) + \frac{5}{6} \cdot (10.8, 14.4, 0) = (9, 12, 5)$

Congrats to the heli-pilot