I need some help with displacement and vectors!!
The displacement vectors A and B shown both have magnitudes of 3 m. The direction of vector A is theta=30 degrees. Find graphically:
(a) A + B
(b) A - B
(c) B - A
(d) A-2B
Report all angles counterclockwise from the positive x-axis.

- anonymous

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- anonymous

|dw:1440642477745:dw|

- jim_thompson5910

so B is pointing directly north? and has an angle of 90 degrees?

- anonymous

yep!

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## More answers

- jim_thompson5910

one way is through the parallelogram method
use the two vectors A and B to construct a parallelogram like so
|dw:1440642698082:dw|

- jim_thompson5910

I basically slid the vectors over to form the other missing sides

- jim_thompson5910

then you draw a vector from the original tail to where the other vectors meet up
along the diagonal
|dw:1440642758813:dw|

- jim_thompson5910

that last arrow I drew is the vector A+B

- anonymous

I see. So I would solve for the diagonal to get the sum of the two vectors?

- jim_thompson5910

yeah you form a parallelogram and draw the diagonal to get the sum of the two vectors

- jim_thompson5910

http://hom.wdfiles.com/local--files/roberval/ParallelogramLaw2.PNG

- anonymous

The diagonal would have to be 3 m, then.

- jim_thompson5910

no it's going to be a bit longer

- anonymous

Oh...How would I calculate that then? I understand the law, but..

- jim_thompson5910

this angle is 60 degrees
|dw:1440642977013:dw|

- jim_thompson5910

since 60+30 = 90

- anonymous

That makes sense..

- jim_thompson5910

|dw:1440643005737:dw|

- jim_thompson5910

|dw:1440643058820:dw|

- anonymous

Ah, okay! Now I see. So then I would use the law of sines?

- anonymous

Since two sides are 3 m and from focusing on the one triangle, I've found that the other two angles are 30 degrees each.

- jim_thompson5910

we have this triangle
|dw:1440643113916:dw|

- jim_thompson5910

use the law of cosines to find x

- anonymous

Whoops, sorry, that's what I meant!

- jim_thompson5910

law of cosines
c^2 = a^2 + b^2 - 2ab*cos(C)
x^2 = 3^2 + 3^2 - 2*3*3*cos(120)
x = ???

- anonymous

3 times the square root of two :)

- anonymous

three, sorry

- anonymous

whoops. three times the square root of 3, i meant

- jim_thompson5910

yes 3*sqrt(3) is the exact length of A+B

- anonymous

That makes sense. How would I then start on figuring out A-B?

- jim_thompson5910

A - B = A + (-B) = A + (-1*B)
so you start with vector A and add on the opposite of vector B
vector -B = -1*B points in the complete opposite direction of B
|dw:1440643443046:dw|
so in this case directly south. The magnitude or length of -B is the same as +B

- jim_thompson5910

and you add A and -B just like before
first form a parallelogram
|dw:1440643496154:dw|

- jim_thompson5910

then form the diagonal
|dw:1440643523787:dw|

- anonymous

I'm assuming the obtuse angles in here again are 120 degrees, like in the last part?

- jim_thompson5910

my drawing is way off though
geogebra is showing me this (see attached)

##### 1 Attachment

- anonymous

Okay...so it's still the same as A + B?

- jim_thompson5910

yeah more or less, you should find that angle ADF = 60 degrees from the attachment I posted above
you'll use the law of cosines again
c^2 = a^2 + b^2 - 2ab*cos(C)
x^2 = 3^2 + 3^2 - 2*3*3*cos(60)
x = ??

- anonymous

Okay!! I think I got it now. Thank you!!

- jim_thompson5910

you're welcome

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