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|dw:1440642477745:dw|

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

yep!

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

that last arrow I drew is the vector A+B

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

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

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

The diagonal would have to be 3 m, then.

no it's going to be a bit longer

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

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

since 60+30 = 90

That makes sense..

|dw:1440643005737:dw|

|dw:1440643058820:dw|

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

we have this triangle
|dw:1440643113916:dw|

use the law of cosines to find x

Whoops, sorry, that's what I meant!

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

3 times the square root of two :)

three, sorry

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

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

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

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

then form the diagonal
|dw:1440643523787:dw|

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

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

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

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

you're welcome