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Another question vector angle proofs.

  1. Sep 10, 2007 #1
    First, thanks for all the help so far everyone!

    vectors a and b exist in the x,y plane and make angles (alpha) and (beta) with x.

    (Ill use A as alpha and B as beta)

    prove: cos (A-B) = cos(A)cos(B)+sin(A)sin(B)

    prove: sin (A-B) = sin(A)cos(B) - cos(A)sin(B)


    I think there is some relationship (for the first one) to this:
    r.s= rscos(theta), but I really don't know where to start with this one..
     
  2. jcsd
  3. Sep 10, 2007 #2

    learningphysics

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    Hint: what is the angle between vectors a and b ?

    Use dot product to get cos(A-B)... Use cross product to get sin(A-B).
     
  4. Sep 10, 2007 #3
    The angle is some angle (alpha) I guess. I suppose it doesn't matter where you apply each symbol.
     
  5. Sep 10, 2007 #4

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    Yes, but what is the angle between the two vectors? If one vector makes an angle A with the x-axis. The other vector makes an angle B... what is the angle between the two vectors.

    draw a picture.
     
  6. Sep 10, 2007 #5
    use the dot product on cos(A)cos(B)+sin(A)sin(B) you mean?
     
  7. Sep 10, 2007 #6

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    No, on the two vectors a and b. Use a.b = |a||b|costheta

    What is the angle theta between the two vectors?
     
  8. Sep 10, 2007 #7
    the angle between them would be 180 - (the absolute value of alpha and beta)
     
  9. Sep 10, 2007 #8

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    Did they give a picture... if you measure the angles counterclockwise from the positive x axis... then the angle would be A-B (assuming A>B).
     
  10. Sep 10, 2007 #9
    no they didn't provide a picture...
    when I drew it I drew (just randomly) alpha in the positive x section and beta in the negative x section. I drew an angle down to the x on both ( so the angle between crosses through the y-axis).
     
  11. Sep 10, 2007 #10

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    Ah... I see... draw them both from the positive x direction make A the bigger angle... then A-B is the angle between them.
     
  12. Sep 10, 2007 #11
    ok, lol that seems like a vague question. so A-B is my angle giving:

    lal . lbl = lal.lbl cos (A-B)
     
  13. Sep 10, 2007 #12

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    it should be:

    a . b = lal.lbl cos (A-B)

    try to write a in the form (x,y)... use cosA sinA etc... along with |a|...

    then try the same with b... but using cosB sinB and |b|...
     
  14. Sep 10, 2007 #13
    am i considering each vector independently or together?

    i.e. vector a and its angle A to x, or vector a and the angle (A-B) ?
     
  15. Sep 10, 2007 #14

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    to get the x-component and y-component of vector a... so you'd use the angle A to x.
     
  16. Sep 10, 2007 #15
    so, like before I have extra info...
    using the equation: (vectors) a.b = axbx + ayby
    I substituted:

    cosA(a)cosB(b)+ sinA(a)sinB(b)

    I'm close to the answer...
     
  17. Sep 10, 2007 #16

    learningphysics

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    yes, very you're close... to be precise you should use :
    ax = |a|cosA, ay = |a|sinA, bx = |b|cosB, by = |b|sinB.

    and get a.b using those...

    then substitute your expression into the left side of:
    a.b = |a||b|cos(A-B)

    then a little simplification and you get the result.
     
  18. Sep 10, 2007 #17
    doesn't a.b = cosA(a)cosB(b)+ sinA(a)sinB(b) ?
     
  19. Sep 10, 2007 #18

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    remember a and b are vectors... we need magnitudes on the right hand side.

    The x-component of a vector is the magnitude of the vector times costheta from the positive x-axis.

    The y-component of a vector is the magnitude of the vector times sintheta from the positive x-axis.

    I'll give an example... take the vector (-3,4)... It has a magnitude of 5, and an angle with the positive x-axis of 143.1

    so (-3,4) = (5cos143.1, 5sin143.1)

    take another vector (1,1) = (1.41cos45, 1.41sin45)

    (-3,4).(1,1) =
    (5cos143.1,5sin143.1).(1.41cos45,1.41sin45) =

    5cos143.1*1.41cos45 + 5sin143.1*1.41sin45

    The reason I did this example, is to show that it is the magnitudes 5 and 1.41 that appear in the formula...

    So a = (|a|cosA,|a|sinA)
    b = (|b|cosB,|b|sinB)

    so:

    a.b = |a|cosA|b|cosB + |a|sinA|b|sinB
     
  20. Sep 10, 2007 #19
    comprendez.

    Thanks alot man
     
  21. Sep 10, 2007 #20

    learningphysics

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    no prob.
     
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