anonymous
  • anonymous
Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I got this much so far, is it right? u=sqrt. 2^2+-4^2 v=sqrt.3^2+-8^2 u=4.5 v=8.5 dot=38.25
IrishBoy123
  • IrishBoy123
look at your dot again
anonymous
  • anonymous
isnt dot multiplying u and v?

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More answers

IrishBoy123
  • IrishBoy123
2x3 + (-4)x(-8)
IrishBoy123
  • IrishBoy123
=?
anonymous
  • anonymous
so 38?
IrishBoy123
  • IrishBoy123
yes
anonymous
  • anonymous
okay and now what do i do?
Michele_Laino
  • Michele_Laino
better is if we write the lengths of our vectors, like this: \[\Large \begin{gathered} \left\| u \right\| = \sqrt {4 + 16} = \sqrt {20} \hfill \\ \left\| v \right\| = \sqrt {9 + 64} = \sqrt {73} \hfill \\ \end{gathered} \]
IrishBoy123
  • IrishBoy123
\(\large cos \theta = \frac{\vec u \bullet \vec v}{|\vec u||\vec v|} \)
IrishBoy123
  • IrishBoy123
find arccos use calculator!
anonymous
  • anonymous
Im so confused lol what do I put for cos theta= like what goes in the u and v?
IrishBoy123
  • IrishBoy123
calculate the RHS first then, ie using your numbers: \(\large \frac{38}{4.5 \times 8.5}\) what do you get we NEED TO optimise later but do this for now
anonymous
  • anonymous
.9934640523 arccos=6.55
IrishBoy123
  • IrishBoy123
yeah! that's a correct answer for "your" numbers but as @Michele_Laino pointed out, your rounding is going to let you down use these instead \(\huge \frac{38}{\sqrt{20}\times \sqrt{73}}\) your calculator will now compute these square roots for you to a very high dgeree of accuracy, and you will get a much more accurate answer
anonymous
  • anonymous
I got .9945054529
IrishBoy123
  • IrishBoy123
yep that should give a very accurate answer if you do not round yourself into oblivion! what do you get for \(\theta\) now?
anonymous
  • anonymous
What how do I find theta again?
IrishBoy123
  • IrishBoy123
you say \(cos \theta = .9945054529\) so ask your calculator for the inverse cosine of \( .9945054529\) like you did last time....
anonymous
  • anonymous
.1023736009
IrishBoy123
  • IrishBoy123
the issue here might be calculator this is what mine looks like and it is a fairly basic one
IrishBoy123
  • IrishBoy123
see how all information is in the calculator, so i am not rounding anything. that guarantees accuracy
IrishBoy123
  • IrishBoy123
i've also inadvertently given you the answer (!!) .... but anyways compare that to the 6.55 you originally calculated. the question wants you to be accurate to within 1/10 th of a degree, so 6.55 would have failed...
anonymous
  • anonymous
Yeah whoa accuracy haha yeah my calculator doesn't read what yours reads I wonder why
anonymous
  • anonymous
thank you so much can you help me in a last question though?
IrishBoy123
  • IrishBoy123
sure but make a new thread for each quetsion
anonymous
  • anonymous
okie dokie!

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