anonymous
  • anonymous
Calc 3..Find an xyz equation for the plane that is perpendicular to {2,-1,-1} and goes through the point {1,-1,0}
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
(2,-1,-1) is the normal vector? or just a point?
anonymous
  • anonymous
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
anonymous
  • anonymous
seems right except that that would be containing the given vector, as opposed to being orthogonal.

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anonymous
  • anonymous
and yes sorry that is the normal vector

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