Algebra linear Ajuda ae galera

Vamos supor

\(v1 = (1, 1, 0)\\ v2 = (1,0, 1)\\ v3 = (0, 2, 0)\)

Construindo uma base ortonormal:

\(w1 = v1 = (1, 1, 0) \\ ∥w1∥ =\sqrt{1^2+1^2+0^2}=\sqrt2\)

\(w2 =v2- \frac{<v2,w1>}{∥w1∥^2}w1\\ w2 = (1,0,1)-\frac{<(1,0,1),(1,1,0)>}{2}(1,1,0)\\ w2 = (1,0,1)-\frac{1}{2}(1,1,0)\\ w2 = (1,0,1)-(\frac{1}{2},\frac{1}{2},0)\\ w2 = (-\frac{1}{2},-\frac{1}{2},1)\\ ∥w2∥ =\sqrt{ (-\frac{1}{2})^2+(\frac{1}{2})^2+(1)^2}\\ ∥w2∥ =\sqrt{ \frac{3}{2}}\\\)

\(w3 =v3- \frac{<v3,w2>}{∥w2∥^2}w2\\ w3 = (0,2,0)-\frac{<(0,2,0),(\frac{-1}{2},\frac{-1}{2},1)>}{\frac{3}{2}}(\frac{-1}{2},\frac{-1}{2},1)\\ w3 = (1,0,1)-\frac{-1}{\frac{3}{2}}(\frac{-1}{2},\frac{-1}{2},1)\\ w3 = (1,0,1)+\frac{2}{3}(\frac{-1}{2},\frac{-1}{2},1)\\ w3 = (1,0,1)+(\frac{-1}{3},\frac{-1}{3},\frac{2}{3})\\ w3= (\frac{2}{3},\frac{-1}{3},\frac{5}{3})\\ ∥ w3 ∥=\sqrt{\frac{4}{9}+\frac{1}{9}+\frac{25}{9}}\\ ∥ w3 ∥=\sqrt{\frac{30}{9}}\\ ∥ w3 ∥=\sqrt{\frac{10}{3}} \)

Assim:

\(B=(1,1,0), (-\frac{1}{2},-\frac{1}{2},1),(\frac{2}{3},\frac{-1}{3},\frac{5}{3})\) é uma base ortogonal

normalizando dividindo pelas respectivas normas, encontramos a base ortonormal a seguir:

\(\boxed{B=(\frac{1}{\sqrt2},\frac{1}{\sqrt2},0), (-\frac{\sqrt2}{2\sqrt3},-\frac{\sqrt2}{2\sqrt3},-\frac{\sqrt2}{\sqrt3}),(\frac{2\sqrt3}{3\sqrt10},\frac{-\sqrt2}{3\sqrt10},\frac{5\sqrt3}{3\sqrt10})}\)