Study23
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.
Delete
Share
This Question is Closed
ganeshie8
Best Response
You've already chosen the best response.
1
mag = \(\sqrt{x^2+y^2}\)
angle = \(\tan^{-1} \frac{y}{x}\)
Study23
Best Response
You've already chosen the best response.
1
I know, but I'm confused with the "counter clockwise from the x-axis" part of the problem.
ganeshie8
Best Response
You've already chosen the best response.
1
angle is always measured in counter clockwise, so dont wry about it
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381000651274:dw|
ganeshie8
Best Response
You've already chosen the best response.
1
btw, ur magnitudes are correct,
for angle just plug it in the formula
Study23
Best Response
You've already chosen the best response.
1
Is the angle for the second problem correct?
ganeshie8
Best Response
You've already chosen the best response.
1
nope
ganeshie8
Best Response
You've already chosen the best response.
1
clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180
Study23
Best Response
You've already chosen the best response.
1
Hmmm, that's where I'm getting confused
ganeshie8
Best Response
You've already chosen the best response.
1
yeah :) just add 180 to that angle, as tan period is \(\pi\)
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381001032077:dw|
Study23
Best Response
You'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????
ganeshie8
Best Response
You've already chosen the best response.
1
yeah, add 180 to that negative :) whats the big deal !
ganeshie8
Best Response
You've already chosen the best response.
1
sorry wait, negative angle is Clockwise,
so from positive x direction, it wud be 360-\(\theta\)
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381001359771:dw|
Study23
Best Response
You've already chosen the best response.
1
But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant?
Study23
Best Response
You've already chosen the best response.
1
wouldn't 360--thetha give me an angle over 360?
Study23
Best Response
You've already chosen the best response.
1
theta
ganeshie8
Best Response
You've already chosen the best response.
1
look at the figure,
make out whats happening
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381001694372:dw|
ganeshie8
Best Response
You've already chosen the best response.
1
the whole thing is 360
the vector is \(\theta\) fro positive x-axis clockwise direction
ganeshie8
Best Response
You've already chosen the best response.
1
so the vector wud be \(360-\theta\) from positive x-axis counter clockwise direction
Study23
Best Response
You've already chosen the best response.
1
Hmmm okay I guess that makes sense. Thank you!
ganeshie8
Best Response
You've already chosen the best response.
1
in \(-\theta\),
minus just tells us the direction is clockwise
Study23
Best Response
You've already chosen the best response.
1
oh okay!!!!
ganeshie8
Best Response
You'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?
```
Study23
Best Response
You've already chosen the best response.
1
?
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381001958479:dw|
Study23
Best Response
You've already chosen the best response.
1
Yea, its in the 2nd quadrant
ganeshie8
Best Response
You've already chosen the best response.
1
angle = \(\large \tan^{-1}\frac{y}{x}\)
\(\large \tan^{-1}\frac{-1}{.5}\)
\(-63.43\)
\(296.57\)
ganeshie8
Best Response
You've already chosen the best response.
1
add/subtract 180 to get to 2nd quadrant as tan period is \(\pi\)
Study23
Best Response
You've already chosen the best response.
1
So when do I use 180 v 360? Does the counter clock wise from +x affect that?
ganeshie8
Best Response
You'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 COUNTER-CLOCKWISE
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381002350796:dw|
Study23
Best Response
You've already chosen the best response.
1
And 180?
ganeshie8
Best Response
You've already chosen the best response.
1
180 comes from the fact that period of tan is 180
tan(x+180) = tan(x)
Study23
Best Response
You've already chosen the best response.
1
But when do you need to add 180?
ganeshie8
Best Response
You'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 ?
Study23
Best Response
You've already chosen the best response.
1
yes
ganeshie8
Best Response
You'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
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381002627185:dw|
ganeshie8
Best Response
You've already chosen the best response.
1
|dw:1381002685399:dw|
Study23
Best Response
You've already chosen the best response.
1
That makes sense!
Study23
Best Response
You've already chosen the best response.
1
Thank you so much for your help! I understand this now!
ganeshie8
Best Response
You've already chosen the best response.
1
np :)
Study23
Best Response
You've already chosen the best response.
1
:)