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Find the components of the vector by rotating the origin

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data

    Hello, I just started my summer class "Applied Linear Algebra I"
    Last math class of my under-grad and its been 1 year since I've last taken Calc3, so I'm trying my best to get back into the groove...anyways.

    Our first topic we are discussing in class is vectors.

    [1]
    This is all the information give to me.

    Consider the Vector v = [x y] or <x, y > (which ever notation you prefer)

    Find the components of the vector obtained by rotating v about the origin counter clockwise through an angle of 30 degrees.


    2. Relevant equations


    3. The attempt at a solution


    Vector v lives in Quadrant 1, and therefore all I know is that x, and y are positive.

    So I rotate vector v by 30 degrees counter clock wise.
    (Attached rough sketch)

    I'm not sure what to do next but this is what I believe.

    cos 30 = (square root of (3) )/ 2
    sin 30 = 1/2

    cos 30 = x / 1 (CAH)
    sin 30 = y /1 (SOH)

    Answer 1]
    Components of vector v = <x,y>

    Answer 2]
    Components of vector v = < cos30, sin30 >

    Can someone please tell me if im on the right track.
    thanks any guidance towards the right direction is greatly appreciated. I am here to learn.
     

    Attached Files:

  2. jcsd
  3. Jun 16, 2015 #2

    Dick

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    You haven't rotated the vector <x,y> by 30 degrees, you've rotated the vector <1,0> by 30 degrees. Don't you have a formula for rotating a vector by an angle somewhere in your book or notes?
     
  4. Jun 16, 2015 #3

    Ray Vickson

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    Figure out the new vectors ##\vec{e}_x', \, \vec{e}_y'\,##, obtained by rotating the vectors ##\vec{e}_x = \langle 1,0 \rangle## and ##\vec{e}_y = \langle 0,1 \rangle## through 30 degrees. Now use that facts that ##\vec{v} = \langle x,y \rangle = x \vec{e}_x + y \vec{e}_y## and the rotation operation is linear.

    Basically, this method replaces geometry by algebra!
     
  5. Jun 16, 2015 #4

    SammyS

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    This is the figure you posted. It doesn't correspond to the exercise that you're trying to complete. See Dick's response.
     
  6. Jun 16, 2015 #5

    RUber

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    Rotation can be thought of as an operator, and thus represented by a matrix.
    If your trig is better than your linear algebra, you could also do the following:

    If you start with polar coordinates, ## x = rcos \theta, y = r sin \theta ## then the image after rotation by ##\alpha ## is ##x'= r cos (\theta + \alpha), y' = r sin(\theta + \alpha) .##
    Using sum angle formulae ## \sin(a+b) = \sin a \cos b + \cos a \sin b, \cos(a+b) = \cos a \cos b - \sin a \sin b,## you get
    ## x' = r \cos \theta \cos \alpha - r \sin \theta \sin \alpha, y' =r \sin \theta \cos \alpha + r\cos \theta \sin \alpha##
    Substitute back in your original x and y to get x' and y' in terms of x, y, and alpha.
    Now, build a matrix ##M(\alpha)## to multiply ##\pmatrix{x\\y}## by, such that ##M(\alpha)\pmatrix{x\\y} =\pmatrix{x'\\y'} ##, and you will have the general form for any rotation (counterclockwise).
     
  7. Jun 16, 2015 #6
    I see what you're saying.
    I've only had one class session so far, and for the first week he is teaching out of the book...
    So Im using google and here for learning this for now.
     
  8. Jun 16, 2015 #7

    Where did <1,0> and <0,1> come from?
    is it because since we are in R2. X, and Y are standard basis vectors
    therefore x = <1,0> and y = <0,1>

    So that means when I graph it i graph the x and y vector correct?

    and the next to step it to rotate it by 30 degrees counter clock wise?
     
  9. Jun 16, 2015 #8

    SammyS

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    You had them in that figure you included. They don't actually belong in this problem.
     
  10. Jun 16, 2015 #9
    Okay thanks for pointing that out. I just picked random position in Quadrant I since I wasn't given the x or y coordinates of vector v.
     
  11. Jun 16, 2015 #10

    SammyS

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    You can't just pick the magnitude of the vector to be 1 .
     
  12. Jun 16, 2015 #11
    I don't think I did, in my mind this is the steps i did.

    I picked a random x, y position for vector v.(in quad 1)
    then rotated it by 30 degrees counter clock wise.
    Formed a right triangle(30,60,90), and made the hypotenuse 1, so i can attempt to solve for x and y.

    I have a very similar example in my notes,
    however in my example, it actually has a numerical value for the two components. such as vector m = <1,0>
     
  13. Jun 16, 2015 #12

    SammyS

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    Making the hypotenuse equal to 1 is the same as making the magnitude of the vector equal to 1 .
     
  14. Jun 16, 2015 #13
    True, I just thought about it You are right.
    What would be the first step you suggest to take to begin solving this problem?
     
  15. Jun 16, 2015 #14

    RUber

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    Ray was implying that if you know how to transform the basis vectors, you can make the general form for the operation.
    For example, a rotation by 90 degrees takes (1,0) to (0,1) and (0,1) to (-1,0).
    Stacking the resulting vectors into a matrix ##\pmatrix{ 0, -1\\ 1, 0 } ## gives a matrix operation that does the job.
    (edited matrix format)
    You can verify this by doing the operation:
    ##\pmatrix{ 0, -1\\ 1, 0 }\pmatrix{ 1\\0 } =\pmatrix{ 0\\1 }, \pmatrix{ 0, -1\\ 1, 0 }\pmatrix{ 0\\1 } = \pmatrix{ -1\\0 }.##
    In general, this is true for any (x,y) so:
    ##\pmatrix{ 0, -1\\ 1, 0 }\pmatrix{ x\\y } =\pmatrix{ -y\\x } . ##
    Is the form for rotation by 90 degrees.
     
  16. Jun 16, 2015 #15
    thanks for explaining that to me, it makes better sense.


    How do i approach this problem since I dont know what x or y is in vector v?
     
  17. Jun 16, 2015 #16

    RUber

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    *edited to show correct order for basis vectors*
    So what does the rotation of (1,0) look like? Call that (a,b)
    How about the rotation of (0,1)? Call that (c,d)
    Make a matrix ##\pmatrix{ a, c \\ b, d } ##.
    Multiply ##\pmatrix{ a, c \\ b, d }\pmatrix{ x \\ y } ##.
    Boom! You're done.
     
    Last edited: Jun 16, 2015
  18. Jun 16, 2015 #17

    SammyS

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    This is for a rotation of 90° .
     
  19. Jun 16, 2015 #18
    Looks easy enough, I understand the intuition behind your answer, and why you chose arbitrary variables to represent the position it rotated to.
    But my main concern is where did (0,1) and (1,0) come from?
    x = (0,1)
    and y =(1,0) ?
     
  20. Jun 16, 2015 #19

    RUber

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    As you suggested earlier, this is due to the standard basis used for (x, y) .
     
  21. Jun 16, 2015 #20

    RUber

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    Oops, your basis should be in the right order, so (1,0) should go first, then (0,1) to make the right operation for (x,y).
     
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