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Math Help - Proving Vector Dot Product Properties

  1. #1
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    Unhappy Proving Vector Dot Product Properties

    I have a big hole in my math skills in that I don't really no where to begin in proving properties. I can use the dot product to work out things just not sure how to prove the properties.

    Let U = (Ux, Uy, Uz), Let V = (Vx, Vy, Vz), Let W = (Wx, Wy, Wz), Let c and k be scalars. Prove the following dot properties.

    A. U . V = V . U
    B. U . (V + W) = U . V + U . W
    C. k (U . V) = (kU) . V = U . (kV)
    D. V . V = || V ||^2
    E. 0 . V = 0

    I have worked out D as:-

    V . V = Vx Vx + Vy Vy + Vz Vz
    = Vx^2 + Vy^2 + Vz^2
    = ( SQRT( Vx^2 + Vy^2 + Vz^2) )^2
    = ( ||V|| )^2

    With A I know that,

    U . V = Ux Vx + Ux Vy + Uz Vz
    V . U = Vx Ux + Vy Uy + Vz Uz

    I just don't know how to prove U . V = V . U without using values. My problem is the same for the remaining questions.

    What topic do I need to study further to improve this skill I'm lacking?
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  2. #2
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    Quote Originally Posted by RedKMan View Post
    I just don't know how to prove U . V = V . U without using values.
    Just notice that (U_x)(V_x)=(V_x)(U_x).
    That is true for the other two components as well.
    Multiplication is commutive.
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  3. #3
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    So I could simply put:-

    (Ux Vx + Uy Vy + Uz Vz) = (Vx Ux + Vy Uy + Vz Uz) = U . V = V . U

    This would be enough to prove that U . V = V . U ?
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  4. #4
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    Quote Originally Posted by RedKMan View Post
    So I could simply put:-

    (Ux Vx + Uy Vy + Uz Vz) = (Vx Ux + Vy Uy + Vz Uz) = U . V = V . U

    This would be enough to prove that U . V = V . U ?
    Yes it is exactly that.
    U . V =(Ux Vx + Uy Vy + Uz Vz) = (Vx Ux + Vy Uy + Vz Uz) = V . U
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  5. #5
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    For this property, U . (V + W) = U . V + U . W

    I have,

    U . ( V + W ) = ( Ux(Vx + Wx) + Uy (Vy + Wy) + Uz (Vz + Wz)

    = (Ux Vx + Uy Vy + Uz + Vz) + (Ux Wx + Uy Wy + Uz Wz) = U . V + U.W

    Do I have the right idea here?
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  6. #6
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    Quote Originally Posted by RedKMan View Post
    For this property, U . (V + W) = U . V + U . W

    I have,

    U . ( V + W ) = ( Ux(Vx + Wx) + Uy (Vy + Wy) + Uz (Vz + Wz)

    = (Ux Vx + Uy Vy + Uz + Vz) + (Ux Wx + Uy Wy + Uz Wz) = U . V + U.W

    Do I have the right idea here?
    Yes, that works nicely.
    If would be nice if you could learn to use LaTex.

    [tex](U_x V_x + U_y V_y + U_z V_z) + (U_x W_x + U_y W_y + U_z W_z) = U\cdot V +U\cdot W[/tex]
    gives
    (U_x V_x + U_y V_y + U_z  V_z) + (U_x W_x + U_y W_y + U_z W_z) = U\cdot V +U\cdot W
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  7. #7
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    I just wanted to confirm this is right:-

    k(U \cdot V) = (kU) \cdot V = U \cdot (kV)

    (kU_x kV_x + kU_y kV_y + kU_z kV_z) = k(U \cdot V)  = ((kU_x) V_x + (kU_y) V_y + (kU_z) V_z) = (kU) \cdot V  = (U_x (kV_x) + U_y (kV_y) + U_z (kV_z) = U \cdot (kV)
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  8. #8
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    Quote Originally Posted by RedKMan View Post
    I just wanted to confirm this is right:-

    k(U \cdot V) = (kU) \cdot V = U \cdot (kV)

    (kU_x kV_x + kU_y kV_y + kU_z kV_z) = k(U \cdot V) = ((kU_x) V_x + (kU_y) V_y + (kU_z) V_z) = (kU)  \cdot V = (U_x (kV_x) + U_y (kV_y) + U_z (kU_z) = U \cdot (kV)
    This is correct.
     k(U \cdot V) = ((kU_x) V_x + (kU_y) V_y + (kU_z) V_z) = (kU)  \cdot V = (U_x (kV_x) + U_y (kV_y) + U_z (kU_z) = U \cdot (kV)
    You have too many k's in (kU_x kV_x + kU_y kV_y + kU_z kV_z)
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  9. #9
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    That has confused me a bit as I thought the scalar k would apply to both the U and the V vectors. I thought both vectors would be multiplied by the scalar k.

    I was working from the idea that 2(a + b) = (2 * a) + (2 * b).
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  10. #10
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    Quote Originally Posted by RedKMan View Post
    That has confused me a bit as I thought the scalar k would apply to both the U and the V vectors. I thought both vectors would be multiplied by the scalar k.
    I was working from the idea that 2(a + b) = (2 * a) + (2 * b).
    \begin{gathered}<br />
  k\left( {U + V} \right) = kU + kV \hfill \\<br />
  k\left( {U \cdot V} \right) = \left( {kU} \right) \cdot V = U \cdot \left( {kV} \right) \hfill \\ <br />
\end{gathered}
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