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Math Help - 3D vector addition / algebra

  1. #1
    Junior Member froodles01's Avatar
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    3D vector addition / algebra

    I have 3 point charges, each with a 3D vector notation.
    A is at (0, 2a, 0)
    B is at (3a/2, 0, 0)
    C is at (-3a/2, 0,0)

    I'm usng Coulombs Law (not gone into here) & the working I have been given is;

    F = (1/ 4*pi*E0) * (q2 / [rA - rB])3 (rA - rB)

    F = (1/ 4*pi*E0) * (q2 / [(-3a/2, 2a, 0)]3) * (-3a/2, 2a, 0)

    Which is OK for me, standard, really. It's the next bit which I don't get how to get to, with the a moving from everywhere to one place.

    F = (q2 / 4*pi*E0*a2) * (-3/2, 2, 0)/(9/4 + 4)3/2


    Could someone please just go through it. I'm sure I'm missing something elementary.
    I included a pic of the question if that makes it easier to see.

    Attached Thumbnails Attached Thumbnails 3D vector addition / algebra-3d-vector-notation-problem.jpg  
    Last edited by froodles01; September 25th 2012 at 09:34 AM. Reason: Adding detail & sceenshot of question
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  2. #2
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    earboth's Avatar
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    Re: 3D vector addition / algebra

    Quote Originally Posted by froodles01 View Post
    I have 3 point charges, each with a 3D vector notation.
    A is at (0, 2a, 0)
    B is at (3a/2, 0, 0)
    C is at (-3a/2, 0,0)

    I'm usng Coulombs Law (not gone into here) & the working I have been given is;

    F = (1/ 4*pi*E0) * (q2 / [rA - rB])3 (rA - rB)

    F = (1/ 4*pi*E0) * (q2 / [(-3a/2, 2a, 0)]3) * (-3a/2, 2a, 0)

    Which is OK for me, standard, really. It's the next bit which I don't get how to get to, with the a moving from everywhere to one place.

    F = (q2 / 4*pi*E0*a2) * (-3/2, 2, 0)/(9/4 + 4)3/2


    Could someone please just go through it. I'm sure I'm missing something elementary.
    I included a pic of the question if that makes it easier to see.

    1.  \langle \frac32 a, 2a, 0 \rangle = a \cdot \langle \frac32, 2, 0 \rangle

    2. I assume that

    \left[ \langle \frac32 a, 2a, 0 \rangle \right]

    is used to caculate the absolute value of \langle \frac32 a, 2a, 0 \rangle

    If so:

    \left[ \langle \frac32 a, 2a, 0 \rangle \right]^3 = a^3 \cdot \left[ \langle \frac32 , 2, 0 \rangle \right]^3

    Now cancel the factor a in the numerator and the denominator which will leave you with a^2 in the denominator.

    3. The absolute value of || \langle \frac32, 2, 0 \rangle || = \sqrt{\langle \frac32, 2, 0 \rangle \cdot \langle \frac32, 2, 0 \rangle} = \sqrt{\frac94 +4} = \left( \frac94 +4 \right)^\frac12

    which can be simplified to \frac52
    Thanks from froodles01
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