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.

- anonymous

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- ganeshie8

mag = \(\sqrt{x^2+y^2}\)
angle = \(\tan^{-1} \frac{y}{x}\)

- anonymous

I know, but I'm confused with the "counter clockwise from the x-axis" part of the problem.

- ganeshie8

angle is always measured in counter clockwise, so dont wry about it

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## More answers

- ganeshie8

|dw:1381000651274:dw|

- ganeshie8

btw, ur magnitudes are correct,
for angle just plug it in the formula

- anonymous

Is the angle for the second problem correct?

- ganeshie8

nope

- ganeshie8

clearly, that 2nd vector is in 4th quadrant, so we shud get angle >180

- anonymous

Hmmm, that's where I'm getting confused

- ganeshie8

yeah :) just add 180 to that angle, as tan period is \(\pi\)

- ganeshie8

|dw:1381001032077:dw|

- 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

yeah, add 180 to that negative :) whats the big deal !

- ganeshie8

sorry wait, negative angle is Clockwise,
so from positive x direction, it wud be 360-\(\theta\)

- ganeshie8

|dw:1381001359771:dw|

- anonymous

But how can it be in the 4th quadrant when -.5,1 is in 2nd quadrant?

- anonymous

wouldn't 360--thetha give me an angle over 360?

- anonymous

theta

- ganeshie8

look at the figure,
make out whats happening

- ganeshie8

|dw:1381001694372:dw|

- ganeshie8

the whole thing is 360
the vector is \(\theta\) fro positive x-axis clockwise direction

- ganeshie8

so the vector wud be \(360-\theta\) from positive x-axis counter clockwise direction

- anonymous

Hmmm okay I guess that makes sense. Thank you!

- ganeshie8

in \(-\theta\),
minus just tells us the direction is clockwise

- anonymous

oh okay!!!!

- 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

?

- ganeshie8

|dw:1381001958479:dw|

- anonymous

Yea, its in the 2nd quadrant

- ganeshie8

angle = \(\large \tan^{-1}\frac{y}{x}\)
\(\large \tan^{-1}\frac{-1}{.5}\)
\(-63.43\)
\(296.57\)

- ganeshie8

add/subtract 180 to get to 2nd quadrant as tan period is \(\pi\)

- anonymous

So when do I use 180 v 360? Does the counter clock wise from +x affect that?

- ganeshie8

when the angle is NEGATIVE,
that means its measured in CLOCKWISE
so subtract it from 360 to get COUNTER-CLOCKWISE

- ganeshie8

|dw:1381002350796:dw|

- anonymous

And 180?

- ganeshie8

180 comes from the fact that period of tan is 180
tan(x+180) = tan(x)

- anonymous

But when do you need to add 180?

- 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

yes

- ganeshie8

since adding/subtracting 180 to tan wont change its value,
add/subtract 180 to get to 2nd quadrant

- ganeshie8

|dw:1381002627185:dw|

- ganeshie8

|dw:1381002685399:dw|

- anonymous

That makes sense!

- anonymous

Thank you so much for your help! I understand this now!

- ganeshie8

np :)

- anonymous

:)

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