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Homework Help: Two Vectors of Equal Magnitudes

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Two vectors A and B have precisely equal magnitudes. For the magnitude of A + B to be 25 times greater than the magnitude of A - B, what must be the angle between A and B?

    Hint Given:
    Without loss of generality, you could assume that A and B are unit vectors. Also, the orientation of the vectors is irrelevant, so you're free to assume for example that one vector lies along the x axis, or that they're symmetrically located about an axis.
    How would you draw triangles to represent the two vector combinations, and how are the angles in these two triangles related?

    2. Relevant equations

    3. The attempt at a solution

    I really don't where to start with this one. I set up an isosceles triangle with the two legs as the vectors, but I can't find any angles. I also don't know how to apply the fact that A+B is 25 times A-B.
  2. jcsd
  3. Jan 18, 2010 #2
    Why don't you just use the law of cosines?
  4. Jan 18, 2010 #3
    I don't have three sides.
  5. Jan 18, 2010 #4
    The third side is supposed to be the resultant(A-B & A+B).
  6. Jan 18, 2010 #5
    I don't know what you call it but
    Resultant= sqrt(A^2 +B^2 + 2ABcosine(theta) )
    Use first the + sign for A+B and - for A-B.
  7. Jan 18, 2010 #6


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    Hi danielatha4! :smile:

    Draw two triangles together, one with sides A and B, the other with A and -B.

    In other words, a triangle PQR where the midpoint of QR is O, OP = A, OQ = B, OR = -B. :wink:
  8. Jan 18, 2010 #7
    Okay tinytim, I took your suggestion and drew two triangles with the sharing side of A. The base of the right triangle is B and the base of the left triangle is -B. The right side of the big triangle is A+B and the left side of the big triangle is A-B.

    I need the angle across from A+B, as that is the angle between vectors A and B. And since it is an isosceles triangle I would think

    [tex]\alpha[/tex]= arccos( 1 - [tex]\frac{(A+B)^2}{2A^2}[/tex])

    However this reduces to arccos(-1) = 180

    I do know that the other angle (between A and -B) is 180-[tex]\alpha[/tex]

    And I'm stuck again with applying that A+B and A-B have a relation.
  9. Jan 18, 2010 #8


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    Hi danielatha4! :smile:

    (Sorry, I haven't quite followed what you've done. :redface:)

    Hint: Since A = B, the triangle is in a circle, and the double-B side is a diameter.

    So … ? :smile:
  10. Jan 18, 2010 #9
    I'm sorry, I still can't figure anything out. I've applied the law of cosines to just about every angle and I can't come up with any answers that are of use to me.

    Here's a picture of what I drew

    Edit: Where x=(A+B)


    Attached Files:

  11. Jan 18, 2010 #10


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    It's a triangle in a circle, and the base is a diameter of the circle, so the angle opposite the base is … ? :smile:
  12. Jan 18, 2010 #11
    Use the hint.
    If you assume that
    A=(1,0) and

    Then |A|=|B| implies that

    Then, by using both facts as well as the fact that

    You can easily solve for x and y and finding the angle is easy. No diagrams are needed. You guys are making it way to hard.
  13. Jan 18, 2010 #12
    I don't understand your approach Pinu7

    So the angle across the base should be 90 degrees right?

    I then solved for beta in my diagram and found it to be arctan(25)=87.8 therefore alpha must be 2.2. Therefore on the right triangle, theta = 180 - 2(2.2) = 175.6

    This isn't right.
  14. Jan 18, 2010 #13


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    Yup! :smile:
    (i make arctan(25)=87.7)

    How about the supplementary angle, theta = 2(2.2) or 2(2.3) ?
  15. Jan 18, 2010 #14
    If Beta is 87.7, then that means alpha is 2.3, right?

    Theta is 180 - 2(2.3) = 175. This isn't right either.

    How do you assume theta is 2(2.3)?
  16. Jan 18, 2010 #15
    I agree with Pinu7, using the law of cosines twice (once for [itex]|\mathbf{A}+\mathbf{B}|[/itex] and the other for [itex]|\mathbf{A}-\mathbf{B}|[/itex] ) makes this a lot easier since we know that [itex]|\mathbf{A}|=\mathbf{B}|[/itex]:


    where [itex]c[/itex] is the normal hypotenuse and [itex]\gamma[/itex] the angle between [itex]a[/itex] and [itex]b[/itex].
  17. Jan 18, 2010 #16
    We can ASSUME that
    A=(1,0) in a particular coordinate system.
    So if

    we must have x^2+y^2=1 (since the magnitude of B is 1)

    |A+B|=25|A-B| or

    x and y can be solved by simple algebra(remember that x^2+y^2=1 so we have 2 equations and 2 variables)
    And then, arctan(y/x) would be the angle.
  18. Jan 18, 2010 #17
    shouldn't it be arctan(y/25x)?
  19. Jan 18, 2010 #18
    No, since A=(1,0) lies on the horizontal axis, then the angle between A and B is the angle B makes with the horizontal so

    tan(angle)=vertical side/horizontal side=y/x
    (the horizontal component of B is x, not 25x, I do not know where that came from)
  20. Jan 18, 2010 #19
    Touche...not sure why, but I was erroneously thinking that you'd need to take into account the 25 from the ratio of a-b and a+b. Plus the fact that the denominator should contain the 25 made me think it'd need to be there also.

    I am usually wrong before I get it right. Though, I think in this case using the two laws of cosines for isosceles triangles would be easier than your method.
  21. Jan 19, 2010 #20


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    (just got up :zzz: …)
    The angle between A and B can be either the angle on the left or the angle on the right.

    (They add up to 180º, of course.)

    A question that asks "what must be the angle between …" usually means the "principal" value, ie the value < 180º. :wink:
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