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Math Help - Orthogonal Vectors in a plane

  1. #1
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    Orthogonal Vectors in a plane

    Find two orthogonal vectors in the plane containing the points A(-5,3,-4) B(-4,-3,-1) and C(-2,5,3)

    Ok so i have that
    AB = <1,-6,3>
    AC = <3,2,7>

    AB X AC = n = determinant of the 3x3 matrix = -48i + 2j +20k

    so i need u and v such that u . v = 0 and u . n = 0 and v . n =0
    i have a feeling it should be easy but i can't figure it out?
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  2. #2
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    Re: Orthogonal Vectors in a plane

    Hey linalg123.

    Hint: If you need two orthogonal vectors that lie in the plane I suggest you use one of the vectors (like AB), then given the plane normal take AB X N and get the second vector that is both orthogonal to AB and the Normal vector.
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  3. #3
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    Re: Orthogonal Vectors in a plane

    thanks i knew it was gonna be obvious!
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  4. #4
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    Re: Orthogonal Vectors in a plane

    As an alternative method, use the fact that any vector in the plane of the triangle can be expressed as some linear combination of (say) \underline{AB} \text{ and } \underline{AC}.

    Suppose then that a suitable vector is \alpha\underline{AB}+\beta\underline{AC}.

    For this to be perpendicular to \underline{AB}, its scalar product with \underline{AB} has to equal zero.

    The result from that allows you to write either \alpha \text{ or } \beta in terms of the other.

    Remember that your answer is not going to be unique, if you have a vector that is perpendicular to \underline{AB}, then any (non-zero) scalar multiple will also be perpendicular to \underline{AB}, so you don't need an actual value for \alpha \text{ or } \beta.
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