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Math Help - vector dot product question

  1. #1
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    vector dot product question

    If I define vectors \vec{CA}=\vec{V}+\vec{U} and     \vec{CB}=\vec{V}-\vec{U} is the following true?

    \vec{CA}\cdot\vec{CB}=\vec{V}^{2}-\vec{U}^{2}

    thanks for the help!
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  2. #2
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    The dot product obeys the usual foiling rule for multiplying out. In addition, the dot product is commutative. Those two facts should enable you to tell me the answer to your question.
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  3. #3
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    I thought the dot product gave a number, not a vector. So I am not sure.
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  4. #4
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    Quote Originally Posted by duaneg37 View Post
    I thought the dot product gave a number, not a vector. So I am not sure.
    It does. But the square of a vector (the square of its length) is also a number. So there at least a fighting chance that you're right. Show me what you get when you foil out the dot product in the usual way.
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  5. #5
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    You are correct. The dot product is a scalar.
    I think your mistake is in notation.
    \vec{CA}\cdot\vec{CB}=\vec{V}\cdot\vec{V}-\vec{U}\cdot\vec{U}. Do see why that is the case?
    Now \vec{V}\cdot\vec{V}=\|\vec{V}\|^2, the length squared.
    So we have \|\vec{V}\|^2-\|\vec{U}\|^2, a number.
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  6. #6
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    Reply to Plato:

    Perhaps. But I've seen notation abused in this manner all the time, especially by physicists.
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Reply to Plato:
    Perhaps. But I've seen notation abused in this manner all the time, especially by physicists.
    I did not see your reply before posting. Sorry.
    But I think it is still an important point.
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  8. #8
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    Suppose that AB is the diameter of a circle with center O, and that C is a point on one of the two arcs joining A and B. Show that \vec{CA} and \vec{CB} are orthogonal. \vec{V} is from O to C. \vec{U} is from O to B. \vec{-U} is from O to A.

    I know if their dot product is zero, they are orthogonal. If the dot product is \vec{U}^{2}-\vec{V}^{2} it would have to be zero because their magnitudes are the same.

    Is this enough proof? I think it is.
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  9. #9
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    Had to think about it for a while, but yes, I think your proof works fine.
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