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

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To find the equation of a plane through the point P (3,3,1) that is perpendicular to the planes defined by x + y = 2z and 2x + z = 10, first identify the normal vectors of the given planes: <1,1,-2> for the first plane and <2,0,1> for the second. The desired plane's normal vector can be determined by calculating the cross product of these two vectors, which yields a vector perpendicular to both. Once the normal vector is found, the equation of the plane can be derived using the point-normal form of a plane equation. This method effectively leads to the solution for the problem.
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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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