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.

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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.

Mathematics
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|dw:1440642477745:dw|
so B is pointing directly north? and has an angle of 90 degrees?
yep!

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one way is through the parallelogram method use the two vectors A and B to construct a parallelogram like so |dw:1440642698082:dw|
I basically slid the vectors over to form the other missing sides
then you draw a vector from the original tail to where the other vectors meet up along the diagonal |dw:1440642758813:dw|
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?
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.
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?
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
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?
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 = ??
Okay!! I think I got it now. Thank you!!
you're welcome

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