anonymous
  • anonymous
Find the angle between the given vectors to the nearest tenth of a degree. u = <6, -1>, v = <7, -4>?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i think it is 0+2kpi
anonymous
  • anonymous
thats no possiable as one of the given solutions..
amistre64
  • amistre64
theres a relationship between cos and dot products, do you recall it?

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anonymous
  • anonymous
yes.
amistre64
  • amistre64
then the angle is simple to be found knowing:\[|u||v|cos~\alpha=u \cdot v\]
amistre64
  • amistre64
solve for alpha :)
anonymous
  • anonymous
i remember the formula how do i plug the numbers into the formula?
amistre64
  • amistre64
you know how to do a dot product right? and to find lengths of vectors?
anonymous
  • anonymous
yeah
amistre64
  • amistre64
those are the numbers you need then, what do you get for: u.v and the lengths of u and v ?
anonymous
  • anonymous
38
amistre64
  • amistre64
u = <6, -1>; sqrt(37) v = <7, -4>; sqrt(65) --------------- u.v 42+5 = 49 |u||v| = sqrt(37*65) .... what does 38 mean?
amistre64
  • amistre64
\[\cos~\alpha=\frac{u.v}{|u||v|}\] \[\cos~\alpha=\frac{49}{\sqrt{37*65}}\] \[\alpha=cos^{-1}\frac{49}{\sqrt{37*65}}\]
anonymous
  • anonymous
the 38 came from 42+(-4)
amistre64
  • amistre64
lol ... i do tend to add when i need to multiply when doing that
amistre64
  • amistre64
-1*-4 is +4, and i did a +5 :)
anonymous
  • anonymous
okay thank you i got 20.28?
amistre64
  • amistre64
i got something closer to 10
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=arccos%2848%2Fsqrt%2837*65%29%29

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