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Math Help - Vectors.

  1. #1
    Newbie Altami's Avatar
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    Vectors.

    Hey everybody I know I have been posting a lot and I am sorry, I just really want to be sure that I am doing my H.W. right and getting the right answers.

    Vector A = <1,-4> and vector B = <5,7>

    a) Find the vector C=2A-B

    For C I got <-3,1>

    b) Find the magnitude of C

    For this I got /square root 10/

    c) Find 'Theta' the direction of C such that 0degrees is 'less than or equal to' "theta" < 360 degrees.

    d) Find the angle between the vectors A and B...

    The first two I just need verification but the last two I couldn't get

    If anybody can help me...thanks.
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  2. #2
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    Your \mathbf{C} is wrong.

    \mathbf{C} = 2\mathbf{A} - \mathbf{B}

     = 2(1, -4) - (5, 7)

     = (2, -8) - (5, 7)

     = (2-5, -8-7)

     = (-3, -15).


    Now fix up everything that follows.
    Last edited by Prove It; July 24th 2010 at 09:59 PM.
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  3. #3
    Newbie Altami's Avatar
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    Ok, so I ended up getting

    Like you said <-3,-11> for C and the magnitude I got /square root 130/

    I think the magnitude is right?

    But again I don't really know how to do the other two?
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Your \mathbf{C} is wrong.

    \mathbf{C} = 2\mathbf{A} - \mathbf{B}

     = 2(1, -4) - (5, 7)

     = (2, -4) - (5, 7) <=== unfortunately you made a small mistake here

     = (2-5, -4-7)

     = (-3, -11).


    Now fix up everything that follows.
     = 2(1, -4) - (5, 7)

     = (2, -8) - (5, 7) = (-3, -15)

    and |(-3, -15)| = \sqrt{234}
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  5. #5
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    Quote Originally Posted by earboth View Post
     = 2(1, -4) - (5, 7)

     = (2, -8) - (5, 7) = (-3, -15)

    and |(-3, -15)| = \sqrt{234}
    Yes I did, Gah! Thanks. Editing now.
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  6. #6
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    Jesus this is a great forum, you guys really help each other out. I thank you guys for helping me as well. I was actually wondering about why only the "1" had gotten multiplied by two and not the "-4". Anyways thanks guys for helping me!

    But again, the last two I don't really know how even start...if you guys could give me any tips?

    By last two I mean "C and D"
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  7. #7
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    Quote Originally Posted by Altami View Post
    Jesus this is a great forum, you guys really help each other out. I thank you guys for helping me as well. I was actually wondering about why only the "1" had gotten multiplied by two and not the "-4". Anyways thanks guys for helping me!

    But again, the last two I don't really know how even start...if you guys could give me any tips?

    By last two I mean "C and D"
    To be honest: I don't understand the question C)

    To D: Use the definition of the dot product:

    \vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\angle(\vec a, \vec b)~\implies~\cos(\angle(\vec a, \vec b) = \dfrac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}
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  8. #8
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    I don't now how you guys use that perfect numerical abbreviations but anyways.

    For D). I got 77.2degrees....

    Would that be right?
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  9. #9
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    Ok I spent like 10 minutes trying to figure out how to use Latex...anyways question C supposed to look like this



    Find  \theta the direction of C such that 0\deg \leq \theta < 360\deg
    Last edited by Altami; July 24th 2010 at 10:39 PM.
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  10. #10
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    Quote Originally Posted by Altami View Post
    I don't now how you guys use that perfect numerical abbreviations but anyways.
    Have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html

    For D). I got 77.2degrees....

    Would that be right?
    I don't know how you got this result but here is what I've done:

    \cos(\angle(\vec a, \vec b)) = \dfrac{(1, -4) \cdot (5, 7)}{\sqrt{17} \cdot \sqrt{74}} = \dfrac{-23}{\sqrt{1258}}

    which yields \angle(\vec a, \vec b) \approx 130.426^\circ

    If you use absolute values you'll get an angle of 49.574^\circ
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  11. #11
    Newbie Altami's Avatar
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    Everybody thanks so much, this is one question that I really didn't get at all.

    But just for last verification

    I got
    A)= (-3,-15)=C
    B) Magnitude= \sqrt234
    C) Find \theta, the direction of C such that 0\deg \leq \theta < 360\deg I couldn't find this one, if someone has any last tips.
    D) Angle between vectors A and B = 130.43\deg
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  12. #12
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    Try drawing the vector on a set of axes [remember that it goes from the origin to (-3, -15)]. You should be able to find the angle made with the x axis using trigonometry.
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  13. #13
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    For some reason I just can't get this...could someone just tell me the answer, I hate asking but as you can see I do try to do the problems but this one I think I just need to see what to do in order to actually fully understand.

    11.8 degrees can anybody verify this?
    Last edited by Altami; July 25th 2010 at 08:34 PM.
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  14. #14
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    Quote Originally Posted by Prove It View Post
    Try drawing the vector on a set of axes [remember that it goes from the origin to (-3, -15)]. You should be able to find the angle made with the x axis using trigonometry.
    Quote Originally Posted by Altami View Post
    For some reason I just can't get this...could someone just tell me the answer, I hate asking but as you can see I do try to do the problems but this one I think I just need to see what to do in order to actually fully understand.

    11.8 degrees can anybody verify this?
    That is nearly the angle between the negative y-axis and the vector. But again: As long as you don't post your working we can't show you what you've done wrong and what you should do instead.

    As ProveIt suggested draw the vector. The direction is determined by the angle between the positive x-axis and the vector, measured anti-clockwise.
    The head of the vector \vec c (the point C(-3, -15)) is situated in the 3rd quadrant, thus the angle in question must be greater than 180.

    Use the indicated right triangle to calculate the angle \angle(x, v):

    \tan^{-1}\left(\frac{-15}{-3}\right) = \tan^{-1}(5)= 78.69^\circ = 258.69^\circ = 438.69^\circ = ...

    Choose the correct result.
    Attached Thumbnails Attached Thumbnails Vectors.-richtg_vektor.png  
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  15. #15
    Newbie Altami's Avatar
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    Oh, yea I ended up putting 11.3\deg. What I had done was \tan^-^1[-5/15] But I forgot that tan is opp/adj...but thanks everybody for your help.
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