Finding the Line of Intersection for Two Planes

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To find the line of intersection for the planes defined by the equations 7x - 2y + 3z = -2 and -3x + y + 2z + 5 = 0, one method involves calculating the cross product of the normal vectors of the planes. Alternatively, the equations can be manipulated by solving for two variables in terms of a parameter. By substituting and simplifying, the line of intersection can be expressed as x = -7t - 12, y = -23t - 41, and z = t. This approach avoids the need for row reduction while providing a clear parameterization of the intersection line. Understanding these methods is crucial for solving similar problems in three-dimensional geometry.
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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.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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