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Homework Help: Properties of cross and dot products

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations



    3. The attempt at a solution
    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 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.

    is my argument is true... or can anyone improve it and prove it with example... your help is highly appreciated.... thanx
     
  2. jcsd
  3. Sep 20, 2010 #2

    HallsofIvy

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    Science Advisor

    I have no clue what that means. You start by saying v is not necessarily the same as w, say a couple of general things about cross product and dot product and conclude that v and w are not necessarily the same. What, in what you said about cross product and dot product, led to that conclusion? When you say " it is consider the same their magnitude and directions are same" what vectors are you talking about having the same magnitude and direction? What does "their" refer to?
    I think you would be better off using specific formulas. In particular I would recommend using the fact that [itex]u\cdot v= |u||v|cos(\theta)[/itex] and [itex]|u\times v|= |u||v| sin(\theta)[/itex].

    It will help to know that for [itex]\theta[/itex] and [itex]\phi[/itex] both between 0 and 180 degrees, [itex]tan(\theta)= tan(\phi)[/itex] implies [itex]\theta= \phi[/itex].
     
  4. Sep 21, 2010 #3
    why you think to use this fact : [itex]u\cdot v= |u||v|cos(\theta)[/itex] and [itex]|u\times v|= |u||v| sin(\theta)[/itex] in your arguments??? I CAN'T SEE IT!!!

    this is my new argument ::

    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???? or any other ideas...


    THANX 4 D RESPONSE HallsofIvy... :tongue:
     
  5. Sep 21, 2010 #4
    Yes, that is right.
     
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