Vectors Question (Orthogonality)

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SUMMARY

The discussion focuses on applying the Gram-Schmidt process to the vectors u1 = [2, -2, 2], u2 = [-2, 2, 1], and u3 = [0, 1, 2] to create an orthogonal basis ß = {v1, v2, v3} for R3. The first vector is correctly identified as v1 = [2, -2, 2]. The second vector, v2, is calculated as [-1, 1, 2], which is confirmed to be correct. However, the third vector v3 was incorrectly computed as [1, 2/3, 4/3], leading to confusion regarding its orthogonality with v1 and v2, as the dot products do not equal zero.

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  • Familiarity with vector operations, including dot products
  • Knowledge of orthogonal vectors and bases
  • Basic linear algebra concepts
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Homework Statement



Let u1 = [2 -2 2 ] u2 = [-2 2 1], u3 = [0 1 2]Use the Gram-Schmidt process to u1, u2, u3, in this order. The resulting vectors are:

v1 = [___ ___ ___], v2 = [___ ___ ___], v3 = [___ ___ ___]

And ß = {v1, v2, v3} is an orthongal basis for R3.

Homework Equations



v1 = u1

v2 = u2 - u2 . v1 v1
------------ v1 . v1
v3= u3 - u3 . v1 v1 - u3 . v2 v2
---------------v1 . v1 v2 . v2

The Attempt at a Solution



Well, the first one is easy.

v1 = [2 -2 2]

I solved v2 and got [-1, 1, 2] which was also right.

I tried v3 based on the equation and got [1, 2/3, 4/3] which was apparently incorrect.
 
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Why do you say "apparently" incorrect? You should have checked yourself that <2, -2, 2].[1, 2/3, 4/3]= 2- 4/3+ 8/3= 2+ 4/3 and [-1, 1, 2].[1, 2/3, 4/3]= -1+ 2/3+ 8/3= -1+ 10/3, neither of which is 0.

If you show what you did, that is show use exactly how you used that formula, we might be able to point out an error.
 

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