## Study23 Group Title 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. 10 months ago 10 months ago

1. ganeshie8 Group Title

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

2. Study23 Group Title

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

3. ganeshie8 Group Title

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

4. ganeshie8 Group Title

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5. ganeshie8 Group Title

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

6. Study23 Group Title

Is the angle for the second problem correct?

7. ganeshie8 Group Title

nope

8. ganeshie8 Group Title

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

9. Study23 Group Title

Hmmm, that's where I'm getting confused

10. ganeshie8 Group Title

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

11. ganeshie8 Group Title

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12. Study23 Group Title

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 Group Title

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

14. ganeshie8 Group Title

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

15. ganeshie8 Group Title

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16. Study23 Group Title

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

17. Study23 Group Title

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

18. Study23 Group Title

theta

19. ganeshie8 Group Title

look at the figure, make out whats happening

20. ganeshie8 Group Title

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21. ganeshie8 Group Title

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

22. ganeshie8 Group Title

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

23. Study23 Group Title

Hmmm okay I guess that makes sense. Thank you!

24. ganeshie8 Group Title

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

25. Study23 Group Title

oh okay!!!!

26. ganeshie8 Group Title

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

27. Study23 Group Title

?

28. ganeshie8 Group Title

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29. Study23 Group Title

Yea, its in the 2nd quadrant

30. ganeshie8 Group Title

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

31. ganeshie8 Group Title

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

32. Study23 Group Title

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

33. ganeshie8 Group Title

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

34. ganeshie8 Group Title

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35. Study23 Group Title

And 180?

36. ganeshie8 Group Title

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

37. Study23 Group Title

But when do you need to add 180?

38. ganeshie8 Group Title

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. Study23 Group Title

yes

40. ganeshie8 Group Title

41. ganeshie8 Group Title

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42. ganeshie8 Group Title

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43. Study23 Group Title

That makes sense!

44. Study23 Group Title

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

45. ganeshie8 Group Title

np :)

46. Study23 Group Title

:)