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Finding unit vectors

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all unit vectors in the plane determined by u= (3, 0, 1) and v=(1, -1, 1) that are perpendicular to the vector w= (1, 2, 0).


    2. Relevant equations



    3. The attempt at a solution
    I'm basically having trouble with the concept and visualizing the problem. Well first I found the normal of the points u and v, which i found n= (1, -2, -3). After this, I'm supposed to take the cross product of n and of w..why would this be? At the end, i found (6, -3, 4) and (-6,3,-4) but the answers say the same but all the x,y,z values are divided by the square root of 61...?? Could someone please clarify this? Do orthogonal solutions have infinitely many solutions?? Thanks a bunch.
     
  2. jcsd
  3. Nov 7, 2009 #2

    Mark44

    Staff: Mentor

    You're not looking at this the right way. u and v are vectors, not points. The plane is the one that is made of of all linear combinations of these two vectors. IOW, the set of vectors au + bv, where a and b are arbitrary real numbers.

    There is no need to find the normal to the plane. What you want are the vectors in this plane that are perpendicular to w = (1, 2, 0).

    To find these vectors, take the dot product of the vectors in the plane and w and set the dot product to zero.
    (au + bv).w = 0

    Solve for a and b. Since you have one equation in two unknowns, there won't be a unique solution.

    The vectors you found aren't unit vectors. To make unit vectors of them, multiply each by the reciprocal of its magnitude.
     
  4. Nov 7, 2009 #3
    The only thing i don't get is why we would add the two vectors in the beginning, like, why don't we calculate it by subtracting the two vectors? Also, why do we place scalars in front of u and v? Thanks again.
     
  5. Nov 7, 2009 #4
    The subspace spanned by two vectors u and v is defined to be all vectors of the form au + bv for all scalars a and b. Geometrically, this is the set of vectors coplanar with both u and v, as was required in the problem statement. Negations are covered by the scalar -1. so using - instead of + generates the same set.
     
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