How Can You Prove If Two Planes Are Perpendicular Analytically?

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SUMMARY

To prove if two planes are perpendicular analytically, one must utilize vectors and their normal equations. The normal vector of a plane, represented as (A, B, C) in the Cartesian equation Ax + By + Cz + D = 0, is crucial for this determination. The planes are perpendicular if the dot product of their normal vectors equals zero, which can be expressed mathematically as AU + BV + CW = 0 for planes Ax + By + Cz = P and Ux + Vy + Wz = Q. This method provides a straightforward analytical approach to understanding the relationship between planes in three-dimensional space.

PREREQUISITES
  • Understanding of Cartesian equations of planes (Ax + By + Cz + D = 0)
  • Knowledge of vector operations, specifically the dot product
  • Familiarity with normal vectors in three-dimensional geometry
  • Basic concepts of gradients and their relation to planes
NEXT STEPS
  • Study vector operations in depth, focusing on the dot product and its geometric interpretations
  • Explore the concept of normal vectors and their applications in 3D geometry
  • Learn about the equations of planes and their graphical representations
  • Investigate the relationship between gradients and planes in three-dimensional space
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who require a solid understanding of geometric relationships between planes, particularly in analytical geometry and vector calculus.

CalcDude
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i was wondering, how would you prove if two planes are perpendicular, analytically? don't you have to use vectors or something?
 
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You sort of have to use vectors, using the cartesian equations (Ax+By+Cz+D=0), take normals of planes as (A,B,C) and multiply them together with dot product to see if the the resultant scalar is zero. For all intents and purposes, the normal of a plane is like the gradient of a line in two dimensions, so this is probably as analytical as it can get if you wish to keep it simple. Of course, you can use partial differentiation and consider the gradients of the planes, I believe, but even that borders on using vectors, because "gradient" doesn't have much meaning in three space.
 
A plane is "given" by a point in it and its normal vector.

In particular, the plane through (x0, y0,z0) with normal vector <A, B, C> has equation A(x-x0)+ B(y-y0)+ C(z- z0)= 0 (which can also be written as Ax+ By+ Cz= Ax0+ By0+ Cz0).

Two planes are perpendicular if and only if their normal vectors are perpendicular which means their dot product must be 0.

The two planes Ax+ By+ Cz= P and Ux+ Vy+ Wz= Q are perpendicular if and only if AU+ BV+ CW= 0.
 

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