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Find vector C, using A and B

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Given vectors [tex]\bold{\vec{A}}=-4.8\bold{\hat{i}}+6.8\bold{\hat{j}}[/tex] and [tex]\bold{\vec{B}}=9.6\bold{\hat{i}}+6.7\bold{\hat{j}}[/tex], determine the vector [tex]\bold{\vec{C}}[/tex] that lies in the xy plane perpendicular to [tex]\bold{\vec{B}}[/tex] whose dot product with [tex]\bold{\vec{A}}[/tex] is 20.0

    2. Relevant equations

    [tex]\theta_{\vec{B}}=\arctan{\frac{6.7}{9.6}}=35 degrees[/tex]

    Dot prod. of perpendicular vectors = 0, therefore
    [tex]\vec{B}{\cdot}\vec{C}=B_{i}C_{i}+B_{j}C_{j}=0[/tex]
    [tex]\vec{B}{\cdot}\vec{C}=9.6C_{i}+6.7C_{j}=0[/tex]

    [tex]\vec{A}{\cdot}\vec{C}=A_{i}C_{i}+A_{j}C_{j}=20[/tex]
    [tex]\vec{A}{\cdot}\vec{C}=-4.8C_{i}+6.8C_{j}=20[/tex]

    3. The attempt at a solution

    I'm not sure where to start, that's why I'm here :)
     
  2. jcsd
  3. Oct 11, 2009 #2
    It looks like you already have your 2 independent equations set up to solve for your 2 unknowns Ci and Cj. Just solve by a quick substitution and you will have vector C's i and j components. Its k component is 0 since it lies in the xy plane.
     
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