anonymous
  • anonymous
if v1=(2,-5) and v2=(4,-3) then the angle between the two vectors is_____ degrees. (Round your answer to two decimal places)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Hero can you help me?
IrishBoy123
  • IrishBoy123
dot product them
anonymous
  • anonymous
would it be -23

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IrishBoy123
  • IrishBoy123
that's part of the process but where did you get minus sign?
anonymous
  • anonymous
well when you add-8 and -15, you get -23, or did I do it wrong?
IrishBoy123
  • IrishBoy123
-8 and - 15?? same again, where are the signs coming from ?!
IrishBoy123
  • IrishBoy123
2*4 + (-5)*(-3)
anonymous
  • anonymous
what I did was (-2)(4)+(5)(-3) and that is how I -23
IrishBoy123
  • IrishBoy123
\(\vec a \bullet \vec b = \bullet = a_{x}*b_{x}+a_{y}*b_{y}\)
IrishBoy123
  • IrishBoy123
you have \(\vec v_1=(2,-5)\) and \( \vec v_2=(4,-3)\)
anonymous
  • anonymous
actually they are (-2,5) and v2=(4,-3),
IrishBoy123
  • IrishBoy123
can you just check that again because you posted differently above...
anonymous
  • anonymous
oh never mind, I am going stupid, sorry about that, I got mixed up!
IrishBoy123
  • IrishBoy123
so is the opening post right, then?
IrishBoy123
  • IrishBoy123
or is it (-2,5) and v2=(4,-3) ?
anonymous
  • anonymous
yeah the opening post is that right one, soory again!
IrishBoy123
  • IrishBoy123
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 \(| \vec b|\)....
IrishBoy123
  • IrishBoy123
\(|\vec a| = \sqrt {a_x^2 + a_y ^2}\)
IrishBoy123
  • IrishBoy123
do that for v1, and v2
IrishBoy123
  • IrishBoy123
and then \(\large \theta = cos^{-1}( \frac {23}{|\vec v_1| \ \vec v_2|})\) gotta dash...
anonymous
  • anonymous
This is hard i am getting really confused!
anonymous
  • anonymous
Wait for v1 and v2 in the bottom of the equation, what do i sub them in for?
IrishBoy123
  • IrishBoy123
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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