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When a plane passes through two points (2,2,1) and (- 1,1, - 1), it is perpendicular to another plane 2x-3y + Z = 3. Find the equation of the first plane
Let ax + by + CZ = D, then the normal direction of the plane can be expressed as n = (a, B, c)
Normal direction of plane 2x-3y + Z = 3: M = (2, - 3,1)
The two planes are perpendicular, so m * n = 0
The results show that 2a-3b + C = 0
Substituting two points (2,2,1) and (- 1,1, - 1) into ax + by + CZ = D, the equations are obtained
2A+2B+C=D
-A+B-C=D
And 2a-3b + C = 0
The solution is: a = 7b, B = 1, a = 7, C = - 11, d = 5
So the plane equation can be expressed as: 7x + y-11z = 5
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