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Vector, dot product

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the angles which the vector A = 3i -6j +2k makes with the coordinate axes




    3. The attempt at a solution
    Let a, b, c be the angles wich A makes with the positive x,y,z axes.
    A• i = (A)(i)cos(a) = 7*cos(a)


    The Solution says:
    Ai = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3

    And I do not understand how they get 3 as the answer.
     
  2. jcsd
  3. Jan 19, 2009 #2

    Doc Al

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    Staff: Mentor

    You need to take the product of the vector A with i, not the magnitude of A:

    [tex]\vec{A}\cdot \hat i = A \cos\theta[/tex]

    What do these equal?
    i • i = ?
    j • i = ?
    k • i = ?
     
    Last edited: Jan 19, 2009
  4. Jan 19, 2009 #3
    I was supposed to write A• i = (A)(1)cos(a) = sqrt(3**2 + (-6)**2 + 2**2) = 7*cos(a)

    second question:

    had to check the book and it says
    i• i = 1
    j• i = j• k = 0

    ah, thanks

    So
    A• i = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3 - 0 + 0 = 3 ?
     
  5. Jan 19, 2009 #4

    Doc Al

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    Staff: Mentor

    OK.

    Exactly. Now use this result to solve for the angle a in your first equation.
     
  6. Jan 19, 2009 #5
    Thanks
     
  7. Jan 19, 2009 #6

    Mark44

    Staff: Mentor

    There's nothing wrong with the above, as far as it goes. In addition to the coordinate definition of the dot product, there is the definition that involves the magnitudes of the vecctors and the angle between them.

    In this case cos(a) = (length of the projection of A onto the x-axis)/(magnitude of A) = 3/7.

    So A [itex]\cdot[/itex] i = 7 * cos(a) = 7 * 3/7 = 3
     
  8. Jan 19, 2009 #7

    Doc Al

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    Staff: Mentor

    That's true. (I could have explained things better in my first response.)
     
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