anonymous 3 years ago Need help with these two vector problems! Given the following x and y components, find the magnitude and direction of the vector, counter clockwise from the x-axis: (1) x comp: -.5, y comp: 1. I found a magnitude of 1.12, but I'm having trouble with the angle. (2) x comp: -1.5 and y comp: -1; I got (1.80, 33.7 degrees, not sure if this is correct, though. I was confused with the negatives and to find the angle.

1. ganeshie8

mag = $$\sqrt{x^2+y^2}$$ angle = $$\tan^{-1} \frac{y}{x}$$

2. anonymous

I know, but I'm confused with the "counter clockwise from the x-axis" part of the problem.

3. ganeshie8

angle is always measured in counter clockwise, so dont wry about it

4. ganeshie8

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5. ganeshie8

btw, ur magnitudes are correct, for angle just plug it in the formula

6. anonymous

Is the angle for the second problem correct?

7. ganeshie8

nope

8. ganeshie8

clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180

9. anonymous

Hmmm, that's where I'm getting confused

10. ganeshie8

yeah :) just add 180 to that angle, as tan period is $$\pi$$

11. ganeshie8

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12. anonymous

Okay! I just realized as you said that I incorrectly drew the latter problem! The main problem I'm having is the angle for the first problem. When I set it up, (inverse tan) I got a negative angle????

13. ganeshie8

yeah, add 180 to that negative :) whats the big deal !

14. ganeshie8

sorry wait, negative angle is Clockwise, so from positive x direction, it wud be 360-$$\theta$$

15. ganeshie8

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16. anonymous

But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant?

17. anonymous

wouldn't 360--thetha give me an angle over 360?

18. anonymous

theta

19. ganeshie8

look at the figure, make out whats happening

20. ganeshie8

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21. ganeshie8

the whole thing is 360 the vector is $$\theta$$ fro positive x-axis clockwise direction

22. ganeshie8

so the vector wud be $$360-\theta$$ from positive x-axis counter clockwise direction

23. anonymous

Hmmm okay I guess that makes sense. Thank you!

24. ganeshie8

in $$-\theta$$, minus just tells us the direction is clockwise

25. anonymous

oh okay!!!!

26. ganeshie8

hey wat about this q of urs :-  But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant? 

27. anonymous

?

28. ganeshie8

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29. anonymous

Yea, its in the 2nd quadrant

30. ganeshie8

angle = $$\large \tan^{-1}\frac{y}{x}$$ $$\large \tan^{-1}\frac{-1}{.5}$$ $$-63.43$$ $$296.57$$

31. ganeshie8

add/subtract 180 to get to 2nd quadrant as tan period is $$\pi$$

32. anonymous

So when do I use 180 v 360? Does the counter clock wise from +x affect that?

33. ganeshie8

when the angle is NEGATIVE, that means its measured in CLOCKWISE so subtract it from 360 to get COUNTER-CLOCKWISE

34. ganeshie8

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35. anonymous

And 180?

36. ganeshie8

180 comes from the fact that period of tan is 180 tan(x+180) = tan(x)

37. anonymous

But when do you need to add 180?

38. ganeshie8

in this particular case, we clearly knw that the vector is in 2nd quadrant. but when we did the formula we got the angle in 4th quadrant, right ?

39. anonymous

yes

40. ganeshie8

41. ganeshie8

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42. ganeshie8

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43. anonymous

That makes sense!

44. anonymous

Thank you so much for your help! I understand this now!

45. ganeshie8

np :)

46. anonymous

:)