anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
mag = \(\sqrt{x^2+y^2}\) angle = \(\tan^{-1} \frac{y}{x}\)
anonymous
  • anonymous
I know, but I'm confused with the "counter clockwise from the x-axis" part of the problem.
ganeshie8
  • ganeshie8
angle is always measured in counter clockwise, so dont wry about it

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ganeshie8
  • ganeshie8
|dw:1381000651274:dw|
ganeshie8
  • ganeshie8
btw, ur magnitudes are correct, for angle just plug it in the formula
anonymous
  • anonymous
Is the angle for the second problem correct?
ganeshie8
  • ganeshie8
nope
ganeshie8
  • ganeshie8
clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180
anonymous
  • anonymous
Hmmm, that's where I'm getting confused
ganeshie8
  • ganeshie8
yeah :) just add 180 to that angle, as tan period is \(\pi\)
ganeshie8
  • ganeshie8
|dw:1381001032077:dw|
anonymous
  • 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????
ganeshie8
  • ganeshie8
yeah, add 180 to that negative :) whats the big deal !
ganeshie8
  • ganeshie8
sorry wait, negative angle is Clockwise, so from positive x direction, it wud be 360-\(\theta\)
ganeshie8
  • ganeshie8
|dw:1381001359771:dw|
anonymous
  • anonymous
But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant?
anonymous
  • anonymous
wouldn't 360--thetha give me an angle over 360?
anonymous
  • anonymous
theta
ganeshie8
  • ganeshie8
look at the figure, make out whats happening
ganeshie8
  • ganeshie8
|dw:1381001694372:dw|
ganeshie8
  • ganeshie8
the whole thing is 360 the vector is \(\theta\) fro positive x-axis clockwise direction
ganeshie8
  • ganeshie8
so the vector wud be \(360-\theta\) from positive x-axis counter clockwise direction
anonymous
  • anonymous
Hmmm okay I guess that makes sense. Thank you!
ganeshie8
  • ganeshie8
in \(-\theta\), minus just tells us the direction is clockwise
anonymous
  • anonymous
oh okay!!!!
ganeshie8
  • ganeshie8
hey wat about this q of urs :- ``` But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant? ```
anonymous
  • anonymous
?
ganeshie8
  • ganeshie8
|dw:1381001958479:dw|
anonymous
  • anonymous
Yea, its in the 2nd quadrant
ganeshie8
  • ganeshie8
angle = \(\large \tan^{-1}\frac{y}{x}\) \(\large \tan^{-1}\frac{-1}{.5}\) \(-63.43\) \(296.57\)
ganeshie8
  • ganeshie8
add/subtract 180 to get to 2nd quadrant as tan period is \(\pi\)
anonymous
  • anonymous
So when do I use 180 v 360? Does the counter clock wise from +x affect that?
ganeshie8
  • ganeshie8
when the angle is NEGATIVE, that means its measured in CLOCKWISE so subtract it from 360 to get COUNTER-CLOCKWISE
ganeshie8
  • ganeshie8
|dw:1381002350796:dw|
anonymous
  • anonymous
And 180?
ganeshie8
  • ganeshie8
180 comes from the fact that period of tan is 180 tan(x+180) = tan(x)
anonymous
  • anonymous
But when do you need to add 180?
ganeshie8
  • 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 ?
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
since adding/subtracting 180 to tan wont change its value, add/subtract 180 to get to 2nd quadrant
ganeshie8
  • ganeshie8
|dw:1381002627185:dw|
ganeshie8
  • ganeshie8
|dw:1381002685399:dw|
anonymous
  • anonymous
That makes sense!
anonymous
  • anonymous
Thank you so much for your help! I understand this now!
ganeshie8
  • ganeshie8
np :)
anonymous
  • anonymous
:)

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