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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 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.
 6 months ago
 6 months 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.
 6 months ago
 6 months ago

This Question is Closed

ganeshie8Best ResponseYou've already chosen the best response.1
mag = \(\sqrt{x^2+y^2}\) angle = \(\tan^{1} \frac{y}{x}\)
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
I know, but I'm confused with the "counter clockwise from the xaxis" part of the problem.
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
angle is always measured in counter clockwise, so dont wry about it
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381000651274:dw
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
btw, ur magnitudes are correct, for angle just plug it in the formula
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
Is the angle for the second problem correct?
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
Hmmm, that's where I'm getting confused
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
yeah :) just add 180 to that angle, as tan period is \(\pi\)
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381001032077:dw
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
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????
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
yeah, add 180 to that negative :) whats the big deal !
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
sorry wait, negative angle is Clockwise, so from positive x direction, it wud be 360\(\theta\)
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381001359771:dw
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
But how can it be in the 4th quadrant when .5,1 is in 2nd quadrant?
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
wouldn't 360thetha give me an angle over 360?
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
look at the figure, make out whats happening
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381001694372:dw
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
the whole thing is 360 the vector is \(\theta\) fro positive xaxis clockwise direction
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
so the vector wud be \(360\theta\) from positive xaxis counter clockwise direction
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
Hmmm okay I guess that makes sense. Thank you!
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
in \(\theta\), minus just tells us the direction is clockwise
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
hey wat about this q of urs : ``` But how can it be in the 4th quadrant when .5,1 is in 2nd quadrant? ```
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381001958479:dw
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
Yea, its in the 2nd quadrant
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
angle = \(\large \tan^{1}\frac{y}{x}\) \(\large \tan^{1}\frac{1}{.5}\) \(63.43\) \(296.57\)
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
add/subtract 180 to get to 2nd quadrant as tan period is \(\pi\)
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
So when do I use 180 v 360? Does the counter clock wise from +x affect that?
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
when the angle is NEGATIVE, that means its measured in CLOCKWISE so subtract it from 360 to get COUNTERCLOCKWISE
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381002350796:dw
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
180 comes from the fact that period of tan is 180 tan(x+180) = tan(x)
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
But when do you need to add 180?
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
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 ?
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
since adding/subtracting 180 to tan wont change its value, add/subtract 180 to get to 2nd quadrant
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381002627185:dw
 6 months ago

ganeshie8Best ResponseYou've already chosen the best response.1
dw:1381002685399:dw
 6 months ago

Study23Best ResponseYou've already chosen the best response.1
Thank you so much for your help! I understand this now!
 6 months ago
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