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anonymous

  • one year ago

if v1=(2,-5) and v2=(4,-3) then the angle between the two vectors is_____ degrees. (Round your answer to two decimal places)

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  1. anonymous
    • one year ago
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    @Hero can you help me?

  2. IrishBoy123
    • one year ago
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    dot product them

  3. anonymous
    • one year ago
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    would it be -23

  4. IrishBoy123
    • one year ago
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    that's part of the process but where did you get minus sign?

  5. anonymous
    • one year ago
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    well when you add-8 and -15, you get -23, or did I do it wrong?

  6. IrishBoy123
    • one year ago
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    -8 and - 15?? same again, where are the signs coming from ?!

  7. IrishBoy123
    • one year ago
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    2*4 + (-5)*(-3)

  8. anonymous
    • one year ago
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    what I did was (-2)(4)+(5)(-3) and that is how I -23

  9. IrishBoy123
    • one year ago
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    \(\vec a \bullet \vec b = <a_{x}, a_{y}>\bullet <b_{x},b_{y}> = a_{x}*b_{x}+a_{y}*b_{y}\)

  10. IrishBoy123
    • one year ago
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    you have \(\vec v_1=(2,-5)\) and \( \vec v_2=(4,-3)\)

  11. anonymous
    • one year ago
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    actually they are (-2,5) and v2=(4,-3),

  12. IrishBoy123
    • one year ago
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    can you just check that again because you posted differently above...

  13. anonymous
    • one year ago
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    oh never mind, I am going stupid, sorry about that, I got mixed up!

  14. IrishBoy123
    • one year ago
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    so is the opening post right, then?

  15. IrishBoy123
    • one year ago
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    or is it (-2,5) and v2=(4,-3) ?

  16. anonymous
    • one year ago
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    yeah the opening post is that right one, soory again!

  17. IrishBoy123
    • one year ago
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    so the dot prod is 23, right?!?! next, the dot product is also this: \(\vec a \bullet \vec b = |\vec a | \ | \vec b| \ cos \theta\), right? so you need \( |\vec a |\) and \(| \vec b|\)....

  18. IrishBoy123
    • one year ago
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    \(|\vec a| = \sqrt {a_x^2 + a_y ^2}\)

  19. IrishBoy123
    • one year ago
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    do that for v1, and v2

  20. IrishBoy123
    • one year ago
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    and then \(\large \theta = cos^{-1}( \frac {23}{|\vec v_1| \ \vec v_2|})\) gotta dash...

  21. anonymous
    • one year ago
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    This is hard i am getting really confused!

  22. anonymous
    • one year ago
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    Wait for v1 and v2 in the bottom of the equation, what do i sub them in for?

  23. IrishBoy123
    • one year ago
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    i showed you above how to calc \(|\vec a|\), which is the magnitude of \(\vec a\) you do that for v1 and v2 then put them into the equation i'll do v1 \( |\vec v_1| = \sqrt { 2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt {29}\) can you do the same for v2?

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