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anonymous

  • one year ago

Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

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  1. anonymous
    • one year ago
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    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

  2. IrishBoy123
    • one year ago
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    look at your dot again

  3. anonymous
    • one year ago
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    isnt dot multiplying u and v?

  4. IrishBoy123
    • one year ago
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    2x3 + (-4)x(-8)

  5. IrishBoy123
    • one year ago
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    =?

  6. anonymous
    • one year ago
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    so 38?

  7. IrishBoy123
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    okay and now what do i do?

  9. Michele_Laino
    • one year ago
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    better is if we write the lengths of our vectors, like this: \[\Large \begin{gathered} \left\| u \right\| = \sqrt {4 + 16} = \sqrt {20} \hfill \\ \left\| v \right\| = \sqrt {9 + 64} = \sqrt {73} \hfill \\ \end{gathered} \]

  10. IrishBoy123
    • one year ago
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    \(\large cos \theta = \frac{\vec u \bullet \vec v}{|\vec u||\vec v|} \)

  11. IrishBoy123
    • one year ago
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    find arccos use calculator!

  12. anonymous
    • one year ago
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    Im so confused lol what do I put for cos theta= like what goes in the u and v?

  13. IrishBoy123
    • one year ago
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    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

  14. anonymous
    • one year ago
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    .9934640523 arccos=6.55

  15. IrishBoy123
    • one year ago
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    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

  16. anonymous
    • one year ago
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    I got .9945054529

  17. IrishBoy123
    • one year ago
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    yep that should give a very accurate answer if you do not round yourself into oblivion! what do you get for \(\theta\) now?

  18. anonymous
    • one year ago
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    What how do I find theta again?

  19. IrishBoy123
    • one year ago
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    you say \(cos \theta = .9945054529\) so ask your calculator for the inverse cosine of \( .9945054529\) like you did last time....

  20. anonymous
    • one year ago
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    .1023736009

  21. IrishBoy123
    • one year ago
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    the issue here might be calculator this is what mine looks like and it is a fairly basic one

  22. IrishBoy123
    • one year ago
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    see how all information is in the calculator, so i am not rounding anything. that guarantees accuracy

  23. IrishBoy123
    • one year ago
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    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...

  24. anonymous
    • one year ago
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    Yeah whoa accuracy haha yeah my calculator doesn't read what yours reads I wonder why

  25. anonymous
    • one year ago
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    thank you so much can you help me in a last question though?

  26. IrishBoy123
    • one year ago
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    sure but make a new thread for each quetsion

  27. anonymous
    • one year ago
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    okie dokie!

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