Find an Equation of Plane Through P (3,3,1) Perpendicular to Planes

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SUMMARY

The discussion focuses on finding the equation of a plane that passes through the point P (3,3,1) and is perpendicular to the planes defined by the equations x + y = 2z and 2x + z = 10. The normal vector for the first plane is <1,1,-2>, while the normal vector for the second plane is <2,0,1>. To determine the required plane's normal vector, the cross product of these two vectors is utilized, leading to the derivation of the plane's equation using the point-normal form.

PREREQUISITES
  • Understanding of 3D coordinate systems
  • Knowledge of vector operations, specifically cross product
  • Familiarity with the equation of a plane in point-normal form
  • Basic skills in linear algebra
NEXT STEPS
  • Study the properties of normal vectors in 3D geometry
  • Learn how to compute the cross product of vectors
  • Explore the derivation of plane equations from normal vectors
  • Practice solving problems involving multiple planes in 3D space
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Giuseppe
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Hello, can anyone help me with this problem?

Find an equation of the plane through P (3,3,1) that is perpendicular to the planes x +y = 2z and 2x +z =10

Thanks
 
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First take the two planes you have. The normal to the first is the vector <1,1,-2>. The normal to the second is <2,0,1>. Since both of these will essentially be parallel to the plane you are looking for, you want to find the vector perpendicular to that plane. In other words, the binormal vector to <1,1,-2> and <2,0,1>. How do you do that in a 3 dimensional coordinate system? Try using the cross product, and assuming you know how to get the equation of a plane from the normal vector and a point, you shouldn't have any trouble arriving at an answer.
 

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