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stranger7

  • 3 years ago

Calc 3..Find an xyz equation for the plane that is perpendicular to {2,-1,-1} and goes through the point {1,-1,0}

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  1. edr1c
    • 3 years ago
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    (2,-1,-1) is the normal vector? or just a point?

  2. eliassaab
    • 3 years ago
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    The equation of the plane is 2 x -y -z = d\\ to find d. The plane has to go through the point (1,-1,0} 2(1) -(-1) -0 =d 2 + 1 =d d=3 The equation of the plane is 2 x -y -z =3

  3. stranger7
    • 3 years ago
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    seems right except that that would be containing the given vector, as opposed to being orthogonal.

  4. stranger7
    • 3 years ago
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    and yes sorry that is the normal vector

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