anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1440642477745:dw|
jim_thompson5910
  • jim_thompson5910
so B is pointing directly north? and has an angle of 90 degrees?
anonymous
  • anonymous
yep!

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jim_thompson5910
  • 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
  • jim_thompson5910
I basically slid the vectors over to form the other missing sides
jim_thompson5910
  • 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
  • jim_thompson5910
that last arrow I drew is the vector A+B
anonymous
  • anonymous
I see. So I would solve for the diagonal to get the sum of the two vectors?
jim_thompson5910
  • jim_thompson5910
yeah you form a parallelogram and draw the diagonal to get the sum of the two vectors
jim_thompson5910
  • jim_thompson5910
http://hom.wdfiles.com/local--files/roberval/ParallelogramLaw2.PNG
anonymous
  • anonymous
The diagonal would have to be 3 m, then.
jim_thompson5910
  • jim_thompson5910
no it's going to be a bit longer
anonymous
  • anonymous
Oh...How would I calculate that then? I understand the law, but..
jim_thompson5910
  • jim_thompson5910
this angle is 60 degrees |dw:1440642977013:dw|
jim_thompson5910
  • jim_thompson5910
since 60+30 = 90
anonymous
  • anonymous
That makes sense..
jim_thompson5910
  • jim_thompson5910
|dw:1440643005737:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1440643058820:dw|
anonymous
  • anonymous
Ah, okay! Now I see. So then I would use the law of sines?
anonymous
  • 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
  • jim_thompson5910
we have this triangle |dw:1440643113916:dw|
jim_thompson5910
  • jim_thompson5910
use the law of cosines to find x
anonymous
  • anonymous
Whoops, sorry, that's what I meant!
jim_thompson5910
  • 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
  • anonymous
3 times the square root of two :)
anonymous
  • anonymous
three, sorry
anonymous
  • anonymous
whoops. three times the square root of 3, i meant
jim_thompson5910
  • jim_thompson5910
yes 3*sqrt(3) is the exact length of A+B
anonymous
  • anonymous
That makes sense. How would I then start on figuring out A-B?
jim_thompson5910
  • 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
  • jim_thompson5910
and you add A and -B just like before first form a parallelogram |dw:1440643496154:dw|
jim_thompson5910
  • jim_thompson5910
then form the diagonal |dw:1440643523787:dw|
anonymous
  • anonymous
I'm assuming the obtuse angles in here again are 120 degrees, like in the last part?
jim_thompson5910
  • jim_thompson5910
my drawing is way off though geogebra is showing me this (see attached)
anonymous
  • anonymous
Okay...so it's still the same as A + B?
jim_thompson5910
  • 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
  • anonymous
Okay!! I think I got it now. Thank you!!
jim_thompson5910
  • jim_thompson5910
you're welcome

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