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Math Help - properties of cross and dot products

  1. #1
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    Cool properties of cross and dot products

    let u be a nonzero vector in space and let v and w be any two vectors in space. if u.v = u.w and u x v = u x w, can you conclude that v=w? give reason for your answer.

    my answer ::

    v is not necessary equal to w
    since u is nonzero vector thus the cross product of the vector u with v and w will produce another vector and it is consider the same if theirs magnitude and directions are same while their dot product will produce a scalar which has no indication of direction. thus v and w is not necessary same vector.

    is my argument is true... or can anyone improve it and prove it with example... your help is highly appreciated.... thanx
    Last edited by bobey; September 20th 2010 at 07:23 AM.
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  2. #2
    A Plied Mathematician
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    since u is nonzero vector thus the cross product of the vector u with v and w will produce another vector and it is consider the same theirs magnitude and directions are same and their dot product will produce a scalar which has no indication of direction. thus v and w is not necessary same vector.
    This is not good English. Please re-write using complete sentences with no run-on sentences and correct punctuation.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    This is not good English. Please re-write using complete sentences with no run-on sentences and correct punctuation.
    its not about the language... its about math!!! no matter how terrible your language is but as long as it has a mathematical logic then it is accepted... i guess... huhuhu
    yes v = w.
    u.v = v.u and u x v = u x w implies that u.(v-w)=0 and u x (v-w) = 0 implies that u perpendicular with (v-w) and u parallel to (v-w) implies that u = 0 or (v-w) = 0

    thus v=w since u is not a zero vector


    IS MY ARGUMENT CORRECT NOW????
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  4. #4
    A Plied Mathematician
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    its not about the language... its about math!!! no matter how terrible your language is but as long as it has a mathematical logic then it is accepted... i guess... huhuhu
    I'm afraid you're in for a rude awakening when you interview for a job. Communication skills (including correct grammar, spelling, punctuation, capitalization, etc.) are more important in the workplace than problem-solving skills. And the reason for that is not hard to find. It used to be that if you wanted to design a steering wheel for a car, you'd get 10 engineers together, and they'd just go at it. Nowadays, they have these "functional groups", or some such title. They consist of an engineer, someone in finance, someone in manufacturing, someone in marketing, etc. So the engineer has to be able to communicate effectively with people who are not technically oriented. If you can't do that, you're not nearly as useful. I'm not saying the problem-solving skills are useless: far from it. But they have to be accompanied by communication skills, or they become much less useful.

    In the context of this forum, you might check out Rules 11 and 12 in the rules sticky.

    So, for example, if I were to write up this problem with your solution, it would look something like the following.

    Problem: Let \mathbf{u} be a nonzero vector and let \mathbf{v} and \mathbf{w} be any two vectors. If \mathbf{u}\cdot\mathbf{v}=\mathbf{u}\cdot\mathbf{w  } and \mathbf{u}\times\mathbf{v}=\mathbf{u}\times\mathbf  {w}, can you conclude that \mathbf{v}=\mathbf{w}? Give the reason for your answer.

    Answer: Since \mathbf{u}\cdot\mathbf{v}=\mathbf{u}\cdot\mathbf{w  }, it follows that \mathbf{u}\cdot\mathbf{v}-\mathbf{u}\cdot\mathbf{w}=\mathbf{u}\cdot(\mathbf{  v}-\mathbf{w})=0. Since \mathbf{u}\times\mathbf{v}=\mathbf{u}\times\mathbf  {w}, it follows that \mathbf{u}\times\mathbf{v}-\mathbf{u}\times\mathbf{w}=\mathbf{u}\times(\mathb  f{v}-\mathbf{w})=0. Therefore, \mathbf{u} is both perpendicular, and either parallel or anti-parallel, to the vector \mathbf{v}-\mathbf{w}. The only way for that to happen is if either \mathbf{u}=0 or \mathbf{v}-\mathbf{w}=0. Since \mathbf{u}\not=0 by assumption, it must be that \mathbf{v}-\mathbf{w}=0, or \mathbf{v}=\mathbf{w}.

    Now, in critiquing this proof, I would say that the implication

    Therefore, \mathbf{u} is both perpendicular, and either parallel or anti-parallel, to the vector \mathbf{v}-\mathbf{w}. The only way for that to happen is if either \mathbf{u}=0 or \mathbf{v}-\mathbf{w}=0.
    might need a bit more work. It's certainly plausible, and I'm convinced, but unless you have a theorem to quote somewhere, it might be considered a weak link in the chain of reasoning. I suppose it also depends, to some extent, on the level of rigor required in your course.

    Make sense?
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