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Math Help - Qustion about vector identity

  1. #1
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    Qustion about vector identity

    Hi! I have a problem understanding a vector identity:

    (AxB) dot (AxB) = A dot (Ax(AxB)),

    where A and B are orthogonal vectors. Help me understand this please.
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  2. #2
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    What do you not understand exactly?

    Well, if you need a picture, draw A and B orthogonal to each other and then draw \vec{A}\times \vec{B} and then \vec{A}\times(\vec{A}\times \vec{B}) and see where that will take you.

    If you need a proof, we need to know what tools you are allowed to use, since there are many ways to prove it.

    Then, you have the identity itself. You should know that if \vec{V} is a vector then \vec{V}\cdot\vec{V} =|\vec{V}|^2
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  3. #3
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    Well I would like some algebrically understandings, not the geometrical. I cant see how i can move the cross product that way. If you have some way to prove it, id be glad. Doesnt really matter which way you prove it, but choose the one you'd think be easiest.
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  4. #4
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    Well, you can go here Vector Identities to see the various identities. I think your identity is wrong. Here is the vector triple product identity:

    \displaystyle \mathbf{A\times}\left(\mathbf{B}\times\mathbf{C}\r  ight)=\left(\mathbf{A}\cdot\mathbf{C}\right)\mathb  f{B}-\left(\mathbf{A}\cdot\mathbf{B}\right)\mathbf{C}

    However, instead of ABC, on your RHS we have AAB and then take the dot product with A to get:

    \displaystyle \mathbf{A}\cdot\left(\mathbf{A\times}\left(\mathbf  {A}\times\mathbf{B}\right)\right)=\mathbf{A}\cdot\  left(\left(\mathbf{A}\cdot\mathbf{B}\right)\mathbf  {A}-\left(\mathbf{A}\cdot\mathbf{A}\right)\mathbf{B} \right)

    \displaystyle =\left(\mathbf{A}\cdot\mathbf{A}\right)\left(\math  bf{A}\cdot\mathbf{B}\right)-\left(\mathbf{A}\cdot\mathbf{B}\right)\left(\mathb  f{A}\cdot\mathbf{A}\right) \right) = 0

    Then on the LHS use the following identity

    \displaystyle \mathbf{\left(A\times B\right)\cdot}\left(\mathbf{C}\times\mathbf{D}\rig  ht)=\left(\mathbf{A}\cdot\mathbf{C}\right)\left(\m  athbf{B}\cdot\mathbf{D}\right)-\left(\mathbf{B}\cdot\mathbf{C}\right)\left(\mathb  f{A}\cdot\mathbf{D}\right)

    with ABCD replaced by ABAB, giving

    \displaystyle \mathbf{\left(A\times B\right)\cdot}\left(\mathbf{A}\times\mathbf{B}\rig  ht)=\left(\mathbf{A}\cdot\mathbf{A}\right)\left(\m  athbf{B}\cdot\mathbf{B}\right)-\left(\mathbf{B}\cdot\mathbf{A}\right)\left(\mathb  f{A}\cdot\mathbf{B}\right)

    \displaystyle =\left|\mathbf{A}\right|^2 \left|\mathbf{B}\right|^2-\left(\mathbf{A}\cdot\mathbf{B}\right)^2

    which can clearly be non-zero.
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  5. #5
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    Thanks for answer! Helped
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