Find Perpendicular Planes Intersecting at a Point

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SUMMARY

This discussion focuses on finding the equations of two perpendicular planes that intersect at a specific point, using the example of the plane defined by 3x+y+z=10 and the intersection point (1,2,5). The key takeaway is that for two planes to be perpendicular, their normal vectors must satisfy the condition of having a scalar product of zero. Additionally, the vector product of two normal vectors yields the third normal vector, ensuring that all three planes intersect at the specified point.

PREREQUISITES
  • Understanding of plane equations in the form Ax+By+Cz+D=0
  • Knowledge of vector operations, including scalar and vector products
  • Familiarity with the concept of normal vectors
  • Basic skills in solving systems of equations
NEXT STEPS
  • Study the properties of normal vectors in three-dimensional geometry
  • Learn how to derive equations of planes from given points and normal vectors
  • Explore the application of vector products in determining perpendicularity
  • Investigate the geometric interpretation of intersections of multiple planes
USEFUL FOR

This discussion is beneficial for students and professionals in mathematics, physics, and engineering who are working with three-dimensional geometry, particularly in fields involving spatial analysis and vector calculus.

mathman99
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How to find (equations of) two perpendicular planes intersecting to a plane (say 3x+y+z=10) in a point (say 1,2,5 ).
All the three planes are perpendicular to each other and intersecting at a single point (say (1,2,5) in this example)
If possible explain it in vector form and non-vector forms.

Thanks in advance.
 
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So you got three perpendicular planes:

Ax+By+Cz+D=0
Ex+Fy+Gz+H=0
Ix+Jy+Kz+W=0

If two planes are perpendicular to each other then, then their normal vectors are also perpendicular to each other.

The normal vectors are (A,B,C), (E,F,G) and (I,J,K). This means that the scalar product (A,B,C) o (E,F,G) = 0 , (A,B,C) o (I,J,K) =0 , (E,F,G) o (I,J,K)=0

Also the vector product of two of them gives us the third vector.
For ex. (A,B,C) x (E,F,G) = (I,J,K)

The intersection point is standing on all of the planes. So that if it is [itex](x_1,y_1,z_1)[/itex] then:
[tex]Ax_1+By_1+Cz_1+D=0[/tex]
[tex]Ex_1+Fy_1+Gz_1+H=0[/tex]
[tex]Ix_1+Jy_1+Kz_1+W=0[/tex]Regards.
 

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