Is there a way to extract the x, y,and z coefficients of a plane?

BUMPRW shared this question 4 years ago
Answered

Given two planes I can find the line of intersection, but see no way to manipulate the a: x+y+z= (whatever) to determine the normal vectors to the planes. I can easily do it manually by manually converting the equations to f(x,y) and g(x,y), but I am obviously missing something. I did try the coefficients command and many, many other things. I'm just trying to break out the x, y, and z components and their coefficients. I can do this in Python or Sage Math, but I'm lazy and Geogebra is generally much faster. I admit I'm probably missing something obvious.

Comments (7)

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hello

I see you have vectors, suposse w, then x(w) y(w) z(w) are the components of w

saludos

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The vectors came from the derivatives of f(x,y) and g(x,y) that I manually created from the plane equations. Nevertheless, I've thought it out and decided to use the ax+by+cz+d=0 format for the plane input format. I was just hoping there was something I missed.

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you can create automatically f(x,y,z) typing leftside[plane]

saludos

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I tried that and it gives me an "illegal argument: Plane a". In attached version I split the coefficients so I can manipulate the data. Another problem I ran into and worked around was when I took a Derivative and the resultant was a real number, that number could not be manipulated, specifically separate the numerator or denominator, or used as part of a point.

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hello

i am using the 5.0.265 and typing leftside[a] it works(I named h)


then h(1,0,0) h(0,1,0) h(0,0,1) are the coeffs


there is a little bug in leftside on CAS. try it in input line of GG

I'd like that Coefficients[a] works like Coefficients[conic]. I hope developers like too

saludos

Files: foro.ggb
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"...to determine the normal vectors to the planes..."

  1. PerpendicularVector[a]

;-)

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I'm on version 5.0.253 on a Mac. PerpendicularVector[a] did work without any errors. Thank you.

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