Find cartesian equation of hyperplane spanned by a set of vectors

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SUMMARY

The discussion focuses on finding the Cartesian equation of a hyperplane W in R4 spanned by the column vectors v1 = [3, 1, -2, -1], v2 = [0, -1, 0, 1], and v3 = [1, 2, 6, -2]. The final Cartesian equation derived is x2 + x4 = 0, which is confirmed as correct by participants. The conversation also touches on alternative methods, such as using parametric equations and the concept of cross products in lower dimensions, suggesting a more general approach for higher dimensions.

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  • Understanding of linear algebra concepts, specifically hyperplanes and vector spaces.
  • Familiarity with RREF (Reduced Row Echelon Form) and its application in solving linear equations.
  • Knowledge of parametric equations and their conversion to Cartesian forms.
  • Basic understanding of determinants and their role in vector operations.
NEXT STEPS
  • Study the process of finding the RREF of matrices in linear algebra.
  • Learn about parametric equations and their conversion to Cartesian equations in higher dimensions.
  • Explore the concept of determinants and cross products in vector spaces.
  • Investigate the generalization of hyperplane equations in Rn for various dimensions.
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Students and professionals in mathematics, particularly those specializing in linear algebra, as well as educators seeking to enhance their understanding of hyperplanes and vector spaces in higher dimensions.

Mr. Johnson
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Let W be a hyperplane in R4 spanned by the column vectors v1 , v2, and v3, where

Note that these are suppose to be COLUMN vectors:

v1 = [3,1, -2 , -1], v2 = [0, -1, 0 , 1], v3= [1,2 ,6, -2]

Find the Cartesian (i.e., linear) equation for W.

I'm not quite sure where to start or how to interpret this problem. I was thinking about first finding the span or column space?... But after that I would not know how to convert into a cartesian equation.

Any guidance or tips would be appreciated.

Thank you.
 
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Good morning, Mr. Johnson, how are you! :smile:

Hint: how would you find the Cartesian equation of the plane in R3 spanned by two column vectors v1 and v2 ? :wink:
 
Ok so I found an example but I feel like it isn't very intuitive and that there are better methods and approaches to this problem.

So here is what I did:

The hyperplane W is of the form Ax1 + Bx2 + Cx3 + Dx4 = 0 since it must pass through the origin.

rref ( 3 1 -2 -1) = [1 0 0 0]
(0 -1 0 1 ) [ 0 1 0 -1]
(1 2 6 -2 ) [0 0 1 0]

Thus A = C =0
B = D

=> 0x1 + Bx2 +0x3 + Bx4 =0

Dividing by B => x2 + x4 =0 which is the final answer for the cartesian equation for W.

Can anyone verify? I know there is a better approach to this problem.
 
not actually following what you've done there :redface:,

but your result x2 + x4 = 0 is obviously correct! :smile:

the more general method i was thinking of (for two vectors in ℝ3) was to find their cross product using a determinant …

can you see a 4D equivalent of that? :wink:
 
A little more direct way: Any vector in the span of those three vectors can be written
<x, y, z, t>= a<3, 1, -2, -1>+ b<0, -1, 0, 1>+ c<1, 2, 6, -2>= <3a+ c, a- b+ 2c, -2a+ 6c, -a+ b- 2c>

so that x= 3a+ c, y= a- b+ 2c, z= -2a+ 6c, t= -a+ b- 2c. Those are parametric equations for the plane, also there are three equations in the three "unknowns", a, b, and c. Can you solve the equations for a, b, and c, to get just a single equation in x, y, z.
 

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