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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 xaxis:
(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.
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 xaxis: (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.

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ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1mag = \(\sqrt{x^2+y^2}\) angle = \(\tan^{1} \frac{y}{x}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know, but I'm confused with the "counter clockwise from the xaxis" part of the problem.

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1angle is always measured in counter clockwise, so dont wry about it

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381000651274:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1btw, ur magnitudes are correct, for angle just plug it in the formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is the angle for the second problem correct?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm, that's where I'm getting confused

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1yeah :) just add 180 to that angle, as tan period is \(\pi\)

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381001032077:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay! 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
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, add 180 to that negative :) whats the big deal !

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1sorry wait, negative angle is Clockwise, so from positive x direction, it wud be 360\(\theta\)

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381001359771:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But how can it be in the 4th quadrant when .5,1 is in 2nd quadrant?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wouldn't 360thetha give me an angle over 360?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1look at the figure, make out whats happening

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381001694372:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1the whole thing is 360 the vector is \(\theta\) fro positive xaxis clockwise direction

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1so the vector wud be \(360\theta\) from positive xaxis counter clockwise direction

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm okay I guess that makes sense. Thank you!

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1in \(\theta\), minus just tells us the direction is clockwise

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1hey wat about this q of urs : ``` But how can it be in the 4th quadrant when .5,1 is in 2nd quadrant? ```

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381001958479:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yea, its in the 2nd quadrant

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1angle = \(\large \tan^{1}\frac{y}{x}\) \(\large \tan^{1}\frac{1}{.5}\) \(63.43\) \(296.57\)

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1add/subtract 180 to get to 2nd quadrant as tan period is \(\pi\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So when do I use 180 v 360? Does the counter clock wise from +x affect that?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1when the angle is NEGATIVE, that means its measured in CLOCKWISE so subtract it from 360 to get COUNTERCLOCKWISE

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381002350796:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1180 comes from the fact that period of tan is 180 tan(x+180) = tan(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But when do you need to add 180?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1in 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 ?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1since adding/subtracting 180 to tan wont change its value, add/subtract 180 to get to 2nd quadrant

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381002627185:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1381002685399:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you so much for your help! I understand this now!
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