Solving Vector Question: Finding rAB Coordinates | TunerSpec.ca

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Homework Help Overview

The discussion revolves around finding the coordinates of vector rAB in a vector problem context. Participants are attempting to determine the x and y coordinates based on given angles and lengths, using trigonometric relationships and the cosine law.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring how to calculate angles from the positive x-axis and how these relate to the coordinates of vector OB. There are discussions about using the cosine law to find angles and subsequently the coordinates of point B.

Discussion Status

Some participants have provided calculations and angles, while others express uncertainty about their results. There is a mix of attempts to clarify relationships between angles and lengths, with some guidance offered on using trigonometric functions to derive coordinates.

Contextual Notes

Participants are working with a diagram that is referenced but not visible in the discussion. There are concerns about the validity of calculated lengths and angles based on the diagram's appearance.

ThomasHW
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Homework Statement


http://www.tunerspec.ca/school/question7.jpg

The Attempt at a Solution


Not sure where to start. I want to get the x & y coordinates of rAB but I'm not sure how.
 
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Get the angle of vector OB from the positive x - axis going counterclockwise...
 
How? :rolleyes:
 
ThomasHW said:
How? :rolleyes:

Can you get the angle of OA from the positive x-axis?
 
Using

[tex]tan x = \frac{711j}{418i}[/tex]

[tex]x = 59.55^{\circ}[/tex]
 
ThomasHW said:
Using

[tex]tan x = \frac{711j}{418i}[/tex]

[tex]x = 59.55^{\circ}[/tex]

good. what's the relationship between that OA and angle... and the OB angle?
 
I'm not sure...

If this ends up being easy I'm going to feel stupid. :-p
 
Label the angle which is 59.55*. You have a triangle and 3 sides are known, use the cosine law to determine any angle right? From that you can get the angle that [tex]|r_{oa}+r_{ab}|[/tex] makes with the origin.

Then you can use the angle and sine/cosine to find the coords of the point B or you can set up a unit vector and multiply by the magnitude of the [tex]|r_{oa}+r_{ab}|[/tex] to get the components.
 
Last edited:
ThomasHW said:
I'm not sure...

If this ends up being easy I'm going to feel stupid. :-p

Oops... I thought I posted a reply. Sorry about that!

Use the angle AOB, and subtract it away from the angle of OA... that gives the angle of OB...
 
  • #10
I'm getting a length of 824.77m for side OA. It looks too big according to the diagram, is it right?

Assuming it is...

I used the cosine law to determine the angle for AOB (43.99) and subtracted that from angle x (determined previously.) This gave me an angle of 15.56 from the x-axis to OB. Then using the length of 606m and the angle of 15.56 I determined the x and y coordinates to be 583.79m and 162.56m, respectively.

Is that right?
 
  • #11
ThomasHW said:
I'm getting a length of 824.77m for side OA. It looks too big according to the diagram, is it right?

Assuming it is...

I used the cosine law to determine the angle for AOB (43.99) and subtracted that from angle x (determined previously.) This gave me an angle of 15.56 from the x-axis to OB. Then using the length of 606m and the angle of 15.56 I determined the x and y coordinates to be 583.79m and 162.56m, respectively.

Is that right?

yup. looks right to me.
 
  • #12
Thank you for the help. :)
 

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