Find cartesian equation of hyperplane spanned by a set of vectors

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Homework Help Overview

The discussion revolves around finding the Cartesian equation of a hyperplane in R4 spanned by three given column vectors. Participants express uncertainty about the initial steps and interpretation of the problem, particularly regarding the transition from span or column space to a Cartesian equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various methods to derive the Cartesian equation, including the use of reduced row echelon form (rref) and parametric equations. Questions arise about the validity of these approaches and the potential for more intuitive methods.

Discussion Status

Some participants have provided hints and alternative methods, while others have expressed confusion about the steps taken. There is recognition of the correctness of certain results, but no consensus on the best approach has been reached.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the types of solutions or methods discussed. The problem's complexity in transitioning from vector representation to a Cartesian equation is acknowledged.

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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