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Help with dot product for vectors

  1. Oct 4, 2009 #1
    Question
    Let vectors [tex]\vec{A} =(2,1,-4), \vec{B}=(-3,0,1), \vec{C}=(-1,-1,2).[/tex]

    What is the angle (in radians) [tex]\theta_{AB} [/tex] between [tex]\vec{A}[/tex] and [tex]\vec{B}[/tex]?


    Important equations
    [tex]\vec{A} \cdot \vec{B} = \vert \vec{A} \vert \,\vert \vec{B}\vert \cos(\theta)[/tex], where [tex]\theta[/tex] is the angle between [tex]\vec{A}[/tex] and [tex]\vec{B}.[/tex]

    My attempt

    [tex]\vec{A}\cdot\vec{B} = (2*-3) + (1*0) + (-4*1) = -6 - 4 = -10[/tex]

    [tex]|\vec{A}||\vec{B}| = (2*3) + (1*0) + (4*1) = 10[/tex]

    therefore, [tex]-10 = 10*cos(\theta)[/tex]

    [tex]-1 = cos(\theta)[/tex]
    [tex]arccos(-1) = \theta[/tex]
    [tex]180 = \theta[/tex]
    [tex]180 = \pi[/tex] [tex]radians[/tex]

    but it says [tex]\pi[/tex] isnt the right answer

    But I don't think that sounds right, because [tex]\vec{A}[/tex] is at [tex]\arctan{\frac{1}{2}}[/tex] (26.7) degrees, and [tex]\vec{B}[/tex] runs along the negative X axis at 180 degrees, So thats a difference of 153.3 degrees. And if you convert that to radians, you get approximately [tex]\frac{17\pi}{20}[/tex] radians, which isnt right either.
     
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2
    You did your dot product right but the magnitude of the vectors is wrong. Recall that the magnitude of a vector is square root of the sum of its individual components squared...the length of the vector.

    For example, the magnitude of vector A would be |A| = sqrt(2^2 + 1^2 + -4^2) = sqrt(21)

    Hope that helps.
     
  4. Oct 5, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This is wrong. Find |A| and |B| separately and then multiply!

    [itex]|A|= \sqrt{2^2+ 1^2+ (-4)^2}[/itex]

     
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