Intersection of two planes (without a given point)

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SUMMARY

The intersection of the planes defined by the equations 2x - 7y + 5z + 1 = 0 and x + 4y - 3z = 0 results in a line PQ. To find the general equation of a plane that passes through this line, one must utilize the cross product of the normals of the two planes to determine the direction vector. The general equation of the plane can be expressed as a*x + b*y + c*z + d = 0, where the coefficients a, b, c, and d must satisfy specific linear equations derived from the conditions of being perpendicular to the direction vector and containing a point on the intersection line.

PREREQUISITES
  • Understanding of vector cross products
  • Knowledge of plane equations in 3D space
  • Familiarity with parametric equations
  • Ability to solve linear equations
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  • Study the method of finding the cross product of vectors in 3D
  • Learn how to derive parametric equations from the intersection of two planes
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The Two Planes 2x-7y+5z+1=0 and x+4y-3z=0 intersect in a line PQ. What is the general equation of a plane through this line?

Have done cross product of the two normals to get vector form of the line, and parametric x,y and z. Would like advice on what to do next, Thanks
 
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Any other plane through that line would also have a normal that's perpendicular to the direction vector you found through the cross-product. It will also have to contain a point on the intersection line. Write the general plane as a*x+b*y+c*z+d=0 and translate those conditions into linear equations in a, b, c and d.
 
Another method would to be to eliminate one variable. Then just that variable be t, then find the other two variables in terms of t.

then r=(x,y,z)
 

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