Find Location of Aeroplane After 60km Travel

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The discussion focuses on calculating the location of an airplane after it travels 60 km from its starting point at (1,2,0) in the direction of the vector (-11,10,2). Part A successfully determines how close the airplane gets to an aerial mast at (0,2,1). In Part B, participants explore methods to find the airplane's new coordinates after the specified distance, with some confusion about vector calculations and direction. Ultimately, the correct approach involves using the direction vector's length to scale it to 60 km and then adding it to the starting point. The final coordinates of the airplane after traveling 60 km are confirmed as (-43,42,8).
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An aeroplane takes off from (1,2,0) and climbs in the direction (-11,10,2)

part a) how close does the areo plane get to the top of an aerial mast at (0,2,1)

AB=b-a = (0,2,1)-(1,2,0) = (-1,0,1)

D(hat)= (-11,10,2)/15

AB.D(hat) = (11/15 + 2/15 = 13/15

root(AB - AB.D(hat)^2 )= root(2 - (13/15)^2)

=root(281)/15=1.1175
this is correct

-----
This is the part I'm having trouble with
PARTB)

Find the location of the areoplane after it has traveled 60km

60^2 = AB-AB.D(hat)
A = (1,2,0) B =(x,y,z)

what to do now?
 
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vorcil said:
An aeroplane takes off from (1,2,0) and climbs in the direction (-11,10,2)

Find the location of the areoplane after it has traveled 60km

60^2 = AB-AB.D(hat)
A = (1,2,0) B =(x,y,z)

what to do now?

I don't understand what you're trying to do here. You can find a vector in the direction
of movement by subtracting (1,2,0) from (-11,10,2).
Then find the length of this vector and find a number to multiply the vector with, to
get the length equal to 60.
Finally add the vector to the starting point
 
willem2 said:
I don't understand what you're trying to do here. You can find a vector in the direction
of movement by subtracting (1,2,0) from (-11,10,2).
Then find the length of this vector and find a number to multiply the vector with, to
get the length equal to 60.
Finally add the vector to the starting point

Sorry I don't understand how you're way works any better!

subtracting 1,2,0 from -11,10,2?
1,2,0 is a point
-11,10,2 is the direction of the line from that point

What do i do!

-

1,2,0 - making up another point with 10(1,2,0)
a=1,2,0 b=10,20,0
AB=b-a (9,18,0)
|AB| = 20.124

60/20.124 = 2.981

2.981(9,18,0) = (26.83,53.66,0)

This is me trying to use your method above.

-

The awnser is -43,32,8 but how do i get to it!

-
 
Last edited:
parametric equation(1,2,0)+t(-11,10,2)
components squared (1-11t)^2,(2+10t)^2+(2t^2) = 60^2

(1-11t)(1-11t) = (1-22t+121t^2)
(2+10t)(2+10t) =(4+40t+100t^2)
(4t^2)

sum the parts

t+18t+225t^2 = 60^2
using the quadratic formula

I get
t=3.955
i'm rounding to 4

substituting
1-11t = x
2+10t = y
2t=z

1-44=-43
2+40=42
=8
(-43,42,8)

which is right according to the answers XD
 
Sorry, I misread the question I tought the airplane went from (1,2,0) in the direction of the point (-11,10,2), but (-11,10,2) is already the direction of the line of travel.

You can find the length of (-11,10,2) and multiply this vector by (60/length)
to make its length equal to 60. Finally add it to the starting point.
If you do this you get the exact solution without any unjustified rounding of 3.955 to 4.
What you calculate, is the distance from (0,0,0) but what is asked is the distance travelled, so the distance from the starting point (1,2,0)
 
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