Finding the Line of Intersection for Two Planes

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SUMMARY

The line of intersection for the planes defined by the equations 7x - 2y + 3z = -2 and -3x + y + 2z + 5 = 0 can be determined using the parameterization method. By solving the equations for two variables and expressing the third as a parameter, the intersection line is represented as x = -7t - 12, y = -23t - 41, and z = t. This method avoids row reduction and directly provides the parametric equations for the line of intersection.

PREREQUISITES
  • Understanding of linear equations in three dimensions
  • Familiarity with the concept of normal vectors
  • Knowledge of parameterization techniques
  • Basic skills in vector operations, including cross products
NEXT STEPS
  • Study the method of finding intersections of planes using normal vectors
  • Learn about parameterization of lines in three-dimensional space
  • Explore vector operations, specifically the cross product and its applications
  • Review systems of linear equations and alternative solving methods
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Students and professionals in mathematics, particularly those studying geometry, linear algebra, or engineering, will benefit from this discussion on finding the line of intersection between two planes.

theorist
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Hello,

How do I find the line of intersections of the two planes 7x - 2y + 3z = -2 and -3x + y + 2z + 5 =0, without having to resort to solving it by row reduction?
 
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Calculate the cross product of the two normal vectors of the planes. Then plug in one point of intersection to yield the line of intersection.

cookiemonster
 
Or, if that is too much trouble, solve the two equations for two of the variables, leaving the third as parameter: from 7x - 2y + 3z = -2 and -3x + y + 2z + 5 =0, multiply the second equation by 2 and add to get
(7+2(-3))x+ (-2+2)y+ (3+2(2))z+ 2(5)= -2 or x+ 7z+ 10= -2. From that,
x= -7z- 12 and then, using the second equation, -3(-7z-12)+ y+ 2z+ 5= 0 so
y= -23z- 41 or, writing the parameter as "t"
x= -7t- 12, y= -23t- 41, z= t is the line of intersection.
 

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