What is the equation for the line AB and the distance XY?

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Homework Help Overview

The discussion centers around a problem involving the positions of three towns (A, B, and X) and the path of an airplane flying from A to B. The task includes finding the equation of the line AB and the distance from point Y on this line to town X.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations representing the line AB and the reasoning behind the parameters used in these equations. There are attempts to find the point of intersection and the distance XY, with some questioning the calculations and assumptions made.

Discussion Status

Some participants have identified potential errors in calculations and are seeking clarification on the derivation of the equations used. There is recognition of the need to verify the parameters in the equations, but no consensus has been reached regarding the correct distance.

Contextual Notes

Participants note discrepancies between their calculated distance and the value provided in the textbook, indicating a need for further exploration of the setup and calculations involved.

Peter G.
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Hi,

So, there are three towns: A, B and X

Town A is 240 km East and 70 km North of O.
Town B is 480 km East and 250 km North of O.
Town X is 339 km East and 238 km North of O.

At A, the airplane changes direction so it now flies towards B. Point Y in the path AB is the closest the airplane ever is to town X.

They then ask us to show that AB is perpendicular to -3i + 4j, which I can do.

They then ask us to find Distance XY.

To do so, I did the following:

r = (240 + 240t) + (140+280t)

r = (339 - 3s) + (238 + 4s)

I then solved to find the point of intersection, which should, supposedly, yield the position of Y.

I got t as 0.46, meaning Y would be 350.4i + 222.8j.

I then went on to find the vector XY and work out its magnitude, which gave me 19 km.

The book, however, claims it is 75km.

Can anyone help me spot where I went wrong?

Thanks in advance!
 
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Peter G. said:
Hi,

So, there are three towns: A, B and X

Town A is 240 km East and 70 km North of O.
Town B is 480 km East and 250 km North of O.
Town X is 339 km East and 238 km North of O.

At A, the airplane changes direction so it now flies towards B. Point Y in the path AB is the closest the airplane ever is to town X.

They then ask us to show that AB is perpendicular to -3i + 4j, which I can do.

They then ask us to find Distance XY.

To do so, I did the following:

r = (240 + 240t) + (140+280t)
What is your reasoning for the above?
Peter G. said:
r = (339 - 3s) + (238 + 4s)

I then solved to find the point of intersection, which should, supposedly, yield the position of Y.

I got t as 0.46, meaning Y would be 350.4i + 222.8j.

I then went on to find the vector XY and work out its magnitude, which gave me 19 km.

The book, however, claims it is 75km.

Can anyone help me spot where I went wrong?

Thanks in advance!
 
Sorry, I meant:

r= (240+240t) + (140+180t).

The calculations were performed based on 180 not 280t
 
Peter G. said:
Sorry, I meant:

r= (240+240t) + (140+180t).

The calculations were performed based on 180 not 280t
What I'm asking is, where does this equation come from, especially the 240t and 180t terms?
 
Sorry, I got my mistake now. The method was right, I just had gotten the numbers wrong again.

I'll explain that equation (which should read (240 + 240t)+(70+180t)

It is the equation for the line AB. A point on the line (A) is 240i + 70j whereas the direction, AB, is 240i + 180j