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Math Help - Vector confusion.

  1. #1
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    Vector confusion.

    Ok,
    When you are finding an angle between two vectors, does it matter what direction their going in? I didn't think it did did...i mean an angle between two lines is an angle.
    So this was the question.
    ABCD is a parallelogram. Has position vectors A, B , C, D.
    A =(2, 1, 3) B = (3, -2, 0) c = (1, -1, -2)
    Determine to the nearest angle, ABC.
    So need to use formula a.b = /a//b/Cos C (/a/ means for modulus, not dividing)
    Where AB = a and BC = b
    So i worked out
    AB as (1, -3, -3)
    and
    BC as (-2, 1, -2)
    a.b = -2 -3 + 6 = 1
    /a/ = root 19
    /b/ = 3
    cos c = \frac{1}{3\sqrt{19}}
    and therefore c = 86 degrees.
    Well it was wrong and the answer was 94 degrees...and in the answer book it said i needed to do vector CB to get
    cos c = \frac{-1}{3\sqrt{19}}
    My question is why? Why does it matter if i use BC or CB and more importantly how will i know in an exam which one to use? No diagram was given...=(

    Thanks!
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Ok,
    When you are finding an angle between two vectors, does it matter what direction their going in? I didn't think it did did...i mean an angle between two lines is an angle.
    So this was the question.



    So need to use formula a.b = /a//b/Cos C (/a/ means for modulus, not dividing)
    Where AB = a and BC = b
    So i worked out
    AB as (1, -3, -3)
    and
    BC as (-2, 1, -2)
    a.b = -2 -3 + 6 = 1
    /a/ = root 19
    /b/ = 3
    cos c = \frac{1}{3\sqrt{19}}
    and therefore c = 86 degrees.
    Well it was wrong and the answer was 94 degrees...and in the answer book it said i needed to do vector CB to get
    cos c = \frac{-1}{3\sqrt{19}}
    My question is why? Why does it matter if i use BC or CB and more importantly how will i know in an exam which one to use? No diagram was given...=(

    Thanks!
    Draw a diagram (2-D) will do showing three points and the relevant vectors and their directions. Now the inner (dot) product of the unit vectors in those directions will give you the cosine of the angle between the vectors. Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have.

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  3. #3
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    Quote Originally Posted by Constatine11 View Post
    Draw a diagram (2-D) will do showing three points and the relevant vectors and their directions. Now the inner (dot) product of the unit vectors in those directions will give you the cosine of the angle between the vectors. Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have.

    .
    how will i know the directions of the vectors? There a clue in the question?
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    Quote Originally Posted by AshleyT View Post
    how will i know the directions of the vectors? There a clue in the question?
    The two rays of interest are BA and BC as B is the common point

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    Quote Originally Posted by Constatine11 View Post
    The two rays of interest are BA and BC as B is the common point

    .
    Yes, and that gives the angle AshleyT originally got. What reason do you have for saying "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have"? Why is that supplementary angle the angle ABC?
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    Quote Originally Posted by HallsofIvy View Post
    Yes, and that gives the angle AshleyT originally got. What reason do you have for saying "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have"? Why is that supplementary angle the angle ABC?
    1. The OP used the vectors AB and and BC, not BA and BC.

    2. "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have" that "which you did have" refers to the angle in the previous sentence. The angle we would have had from the proevous sentence was ABC, hence the supplementary angle is 180-ABC, which is what the OP had.

    XX XXX XXX XXXX XX

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    Quote Originally Posted by Constatine11 View Post
    1. The OP used the vectors AB and and BC, not BA and BC.

    2. "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have" that "which you did have" refers to the angle in the previous sentence. The angle we would have had from the proevous sentence was ABC, hence the supplementary angle is 180-ABC, which is what the OP had.

    XX XXX XXX XXXX XX

    .
    Yea i know thats what you do ,
    But what i mean is, (sorry to annoy) but why do change...why do you find the supplementary angle and not the acute angle? The question never said 'find supplementary angle' or 'acute angle'.
    How do you know its the supplementary angle that your mean to find, how do you know that?

    Thanks
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  8. #8
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    Quote Originally Posted by AshleyT View Post
    Yea i know thats what you do ,
    But what i mean is, (sorry to annoy) but why do change...why do you find the supplementary angle and not the acute angle? The question never said 'find supplementary angle' or 'acute angle'.
    How do you know its the supplementary angle that your mean to find, how do you know that?

    Thanks
    Did you draw a diagram?

    One angle is that between vectors AB and BC and the other is the angle between vectors BA and BC.

    (The angle in question is the angle between the unit vectors when then are drawn pointing away from a common point)

    .
    Attached Thumbnails Attached Thumbnails Vector confusion.-gash.png  
    Last edited by Constatine11; January 11th 2009 at 01:51 AM.
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  9. #9
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    Quote Originally Posted by Constatine11 View Post
    1. The OP used the vectors AB and and BC, not BA and BC.
    Oh, my goodness, I missed that completely! Thanks.

    2. "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have" that "which you did have" refers to the angle in the previous sentence. The angle we would have had from the proevous sentence was ABC, hence the supplementary angle is 180-ABC, which is what the OP had.

    XX XXX XXX XXXX XX

    .
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  10. #10
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    Quote Originally Posted by Constatine11 View Post
    Did you draw a diagram?

    One angle is that between vectors AB and BC and the other is the angle between vectors BA and BC.

    (The angle in question is the angle between the unit vectors when then are drawn pointing away from a common point)

    .
    Thank-you very, very much!! For all patience and replies
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