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Multi-Variable Calculus: Cancellation of dot products

  1. Sep 7, 2011 #1

    Dembadon

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    1. The problem statement, all variables and given/known data

    In real-number multiplication, if uv1 = uv2 and u ≠ 0, then we can cancel the u and conclude that v1 = v2. Does the same rule hold for the dot product: If uv1 = uv2 and u ≠ 0, can you conclude that v1 = v2? Give reasons for your answer.

    2. Relevant equations



    3. The attempt at a solution

    If we let u = k<u1, u2> with u2 = 0 and scalar k, then the dot product of u with any other vector v = k<v1, v2> will simply be the component kv1 because u2 will make the ku2kv2 product always zero regardless of its value. Thus, v can be infinitely many different vectors and still have the same dot product with u.
     
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  3. Sep 7, 2011 #2

    Pyrrhus

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    The hint is can you compute an inverse of [itex]\vec{u}[/itex] ? and thus multiply both by that inverse in order that both vectors v1 and v2 are equal.
     
  4. Sep 7, 2011 #3

    lanedance

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    also a counter example might be good here, how about considering when
    u • v1 = u • v2 = 0

    what does this mean geometrically? using that it should be easy to find a counter example in 3D space
     
  5. Sep 7, 2011 #4

    vela

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    Just out of curiosity, why do you introduce the scalar k? Doesn't your argument work without the k's?
     
  6. Sep 7, 2011 #5

    Dembadon

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    Interesting! I hadn't thought of that. I'll have to play around with it.

    It would mean that both v1 and v2 are orthogonal to u, or that u is orthogonal to v1 and v2, right?

    You're absolutely right. Thank you for pointing that out. :smile:

    ----------------------------------------

    Thanks for the input, everyone. I appreciate your time.
     
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