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Math Help - cartesian equation

  1. #1
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    cartesian equation

    Find a cartesian equation for the plane containing the point (3, 1, 5) and perpendicular to the line with
    parametric form x = 5t + 1, y = 6t − 2, z = 8t, t 2 R.
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  2. #2
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    Re: cartesian equation

    Quote Originally Posted by rcarron View Post
    Find a cartesian equation for the plane containing the point (3, 1, 5) and perpendicular to the line with
    parametric form x = 5t + 1, y = 6t − 2, z = 8t, t 2 R.
    The direction vector of the line is the normal vector of the plane: \overrightarrow{n_P}= (5,6,8)

    Let A(3,1,5) denote a point located in the plane with it's stationary vector \vec a and \vec p = (x,y,z) the stationary vector of any point in the plane then the equation of the plane P is:

    P: \overrightarrow{n_P} \cdot ( \vec p - \vec a) = 0

    Using the given values you have:

    P: (5,6,8)((x,y,z)-(3,1,5))=0~\implies~\boxed{5x+6y+8z-61=0}
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    Re: cartesian equation

    Again, find the direction vector first => <5,6,8>
    C = <5,6,8> dot product <x,y,z>
    C = <5,6,8> dot product <3,1,5> = 61

    Therefore our plane is 5x + 6y + 8z = 61
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  4. #4
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    Re: cartesian equation

    thanks guys
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