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R2 Rotation about (0,0)

  1. Jun 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A rotation ρ about the origin in ℝ2 drives the point P = (4,3) to the point ρ(P) = (3,4).
    Find the angle of rotation as well as its matrix notation.

    2. Relevant equations

    Ok so I made a sketch and I realised I needed to find θ = θ1 - θ2 where θ1 and θ2 equal arctan(4/3) and arctan(3/4) respectively.

    I guess it's correct but then how would I work out the sin and cos of that angle θ without a calculator?

    SO. I worked out the cosine angle between the vectors 0P and 0ρ(P) and got cosθ = 24/2√25

    But now I don't know how to work out the sine!

    My goal is to replace everything into the ℝ2 rotation matrix (row 1; row 2):
    (cosα -sinα; sinα cosα)

    and express as g(x) = Ax + v

    Thanks a lot!
     
  2. jcsd
  3. Jun 29, 2013 #2

    Simon Bridge

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    A point at ##P=(a,o)## is at the end of the hypotenuse of a right-angled triangle with adjacent side ##a## and opposite side ##o## ... so the hypotenuse has length ##h=\sqrt{a^2+o^2}## and you can find sine cosine and tangent by SOH CAH TOA.
     
  4. Jun 29, 2013 #3

    tiny-tim

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    hi phyzz! :smile:
    sec2 = 1 + tan2

    cos2 = 1/sec2

    sin = cos*tan :wink:
     
  5. Jun 29, 2013 #4
    How do you know that this is a right angle though?
     

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  6. Jun 29, 2013 #5
    Ohhhhh my bad I didn't see it at first. I just used sin^2 + cos^2 = 1
    and the cos of the angle I found between the two vectors.

    Thanks!
     
  7. Jun 29, 2013 #6

    Simon Bridge

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    ... in fact, P=(3,4) and Q=(4,3) each make a special kind of triangle.
    The rotation angle is the difference between complimentary angles in this triangle - if you recall your special triangles, you don't need the trig.

    Of course, since you know the start and finish, you could just expand the matrix equation into a pair of simultanious equations and solve them?
     
    Last edited: Jun 29, 2013
  8. Jun 29, 2013 #7
    Thank you Simon Bridge!

    In other news, when I work out (in this case a is a constant) [tex] \left \| T_{\gamma}(t) \right \| = \left \| \frac{1}{\sqrt{1 + a^{2}}} (-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t)) \right \|[/tex]

    I know I can just do [tex]\left \|(-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t)) \right \|[/tex] and then multiply by the constant [tex]\frac{1}{\sqrt{1 + a^{2}}} [/tex]at the very end.

    My question:

    Given that [tex] N_{\gamma}(t) = \frac{1}{\sqrt{1 + a^{2}}} (-acos(t),-\sqrt{1+a^{2}}sin(t),-cos(t)) [/tex]

    and I want to work out [tex] T_{\gamma}(t) \times N_{\gamma}(t) [/tex]

    can I just do [tex]\frac{1}{\sqrt{1 + a^{2}}} \frac{1}{\sqrt{1 + a^{2}}}[/tex] multiplied by the result of [tex](-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t))\times (-acos(t),-\sqrt{1+a^{2}}sin(t),-cos(t))[/tex]? Thanks so much again!
     
  9. Jun 29, 2013 #8

    Simon Bridge

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    Don't see the connection of all that with your question... in ##T_\gamma(t)\times N_\gamma(t)## - is that ##\times## supposed to be a cross product?

    FYI: putting a backslash in front of the function name turns ##sin\theta## into ##\sin\theta##
     
  10. Jun 29, 2013 #9
    Yes, cross product

    Thanks for the tip!
     
  11. Jun 29, 2013 #10

    Simon Bridge

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    If ##\vec{u}=a(x,y,z)## and ##\vec{v}=b(p,q,r)## then is ##\vec{u}\times\vec{v}=ab## ... is that your question?
    Does a cross product usually produce a scalar?
     
  12. Jun 29, 2013 #11
    What I meant was if we have constants in front of two vectors we want to cross product, are we allowed to multiply the constants together and perform the cross product to the vector bits?
    Sorry for the vagueness!
     
  13. Jun 29, 2013 #12

    HallsofIvy

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    Yes, the "associative" law holds for the cross product: [itex](a\vec{u}\times (b\vec{v})= (ab)(\vec{u}\times\vec{v})[/tex]
     
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