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Vector question (scalar products)

  1. Feb 9, 2009 #1
    Find scalars s, t s.t. C - sA - tB is perpendicular to A and B

    A = i + j + 2k
    B = 2i - j + k
    C = 2i - j + 4k

    I took the cross product of A and B
    (1, 1, 2) x (2, -1, 1) =
    |i j k|
    |1 1 2|
    |2 -1 1|
    = 3i + 3j - 3k

    OK, that should be perpendicular to both vectors A and B I'm guessing...
    But how do I still determine the scalars s and t -- there is only one equation!
     
  2. jcsd
  3. Feb 9, 2009 #2

    danago

    User Avatar
    Gold Member

    Ill let Q = C - sA - tB.

    If Q is perpendicular to both A and B, and the vector (3,3,-3) is also perpendicular to both A and B (as you have calculated), then it should follow that Q is a scalar multiple of (3,3,-3) right?

    i.e.
    k(3,3,-3) = Q

    Now you have a system of 3 equations in 3 variables (s,t,k).

    ____

    Another way that you could solve it would be to realise that you need to find s,t such that Q.A = Q.B = 0 (a zero scalar product implies orthogonality), and form a system of two equations in two variables (s,t).
     
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