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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.

Mathematics
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mag = \(\sqrt{x^2+y^2}\) angle = \(\tan^{-1} \frac{y}{x}\)
I know, but I'm confused with the "counter clockwise from the x-axis" part of the problem.
angle is always measured in counter clockwise, so dont wry about it

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Other answers:

|dw:1381000651274:dw|
btw, ur magnitudes are correct, for angle just plug it in the formula
Is the angle for the second problem correct?
nope
clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180
Hmmm, that's where I'm getting confused
yeah :) just add 180 to that angle, as tan period is \(\pi\)
|dw:1381001032077:dw|
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????
yeah, add 180 to that negative :) whats the big deal !
sorry wait, negative angle is Clockwise, so from positive x direction, it wud be 360-\(\theta\)
|dw:1381001359771:dw|
But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant?
wouldn't 360--thetha give me an angle over 360?
theta
look at the figure, make out whats happening
|dw:1381001694372:dw|
the whole thing is 360 the vector is \(\theta\) fro positive x-axis clockwise direction
so the vector wud be \(360-\theta\) from positive x-axis counter clockwise direction
Hmmm okay I guess that makes sense. Thank you!
in \(-\theta\), minus just tells us the direction is clockwise
oh okay!!!!
hey wat about this q of urs :- ``` But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant? ```
?
|dw:1381001958479:dw|
Yea, its in the 2nd quadrant
angle = \(\large \tan^{-1}\frac{y}{x}\) \(\large \tan^{-1}\frac{-1}{.5}\) \(-63.43\) \(296.57\)
add/subtract 180 to get to 2nd quadrant as tan period is \(\pi\)
So when do I use 180 v 360? Does the counter clock wise from +x affect that?
when the angle is NEGATIVE, that means its measured in CLOCKWISE so subtract it from 360 to get COUNTER-CLOCKWISE
|dw:1381002350796:dw|
And 180?
180 comes from the fact that period of tan is 180 tan(x+180) = tan(x)
But when do you need to add 180?
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 ?
yes
since adding/subtracting 180 to tan wont change its value, add/subtract 180 to get to 2nd quadrant
|dw:1381002627185:dw|
|dw:1381002685399:dw|
That makes sense!
Thank you so much for your help! I understand this now!
np :)
:)

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