Another Vector Question

1. Sep 17, 2007

ThomasHW

1. The problem statement, all variables and given/known data
http://www.tunerspec.ca/school/question7.jpg

3. 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.

2. Sep 17, 2007

learningphysics

Get the angle of vector OB from the positive x - axis going counterclockwise...

3. Sep 17, 2007

ThomasHW

How? :uhh:

4. Sep 17, 2007

learningphysics

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

5. Sep 17, 2007

ThomasHW

Using

$$tan x = \frac{711j}{418i}$$

$$x = 59.55^{\circ}$$

6. Sep 17, 2007

learningphysics

good. what's the relationship between that OA and angle... and the OB angle?

7. Sep 17, 2007

ThomasHW

I'm not sure...

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

8. Sep 17, 2007

dontdisturbmycircles

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 $$|r_{oa}+r_{ab}|$$ 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 $$|r_{oa}+r_{ab}|$$ to get the components.

Last edited: Sep 18, 2007
9. Sep 18, 2007

learningphysics

Use the angle AOB, and subtract it away from the angle of OA... that gives the angle of OB...

10. Sep 18, 2007

ThomasHW

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. Sep 18, 2007

learningphysics

yup. looks right to me.

12. Sep 18, 2007

ThomasHW

Thank you for the help. :)