Enunciado

Se

a = ( 3 , 1 , 2 )

b = ( - 1,1 , 0 )

c = ( 0 , 0 , - 4 )

Mostre que

a xx (b xx c) != (a xx b) xx c

Passo 1

Vamos começar calculando b xx c :

b xx c = |[[i,j,k],[-,1,1,0],[0,0,-,4]]| = |[[1,0],[0,-,4]]| i - |[[-,1,0],[0,-,4]]| j + |[[-,1,1],[0,0]]| k

b xx c = |[[i,j,k],[-,1,1,0],[0,0,-,4]]| = ( 1 . ( - 4 ) - 0 . 0 ) i - ( - 1 . ( - 4 ) - 0 . 0 ) j + ( - 1 . 0 - 1 . 0 ) k

b xx c = |[[i,j,k],[-,1,1,0],[0,0,-,4]]| - 4 i - 4 j + 0 k

b xx c = ( - 4 , - 4,0 )

Passo 2

Agora vamos para a xx (b xx c) :

a xx (b xx c) = |[[i,j,k],[3,1,2],[-,4,-,4,0]]| = |[[1,2],[-,4,0]]| i - |[[3,2],[-,4,0]]| j + |[[3,1],[-,4,-,4]]| k

a xx (b xx c) = |[[i,j,k],[3,1,2],[-,4,-,4,0]]| = ( 1 . 0 - 2 . ( - 4 ) ) i - ( 3 . 0 - 2 . ( - 4 ) ) j + ( 3 . ( - 4 ) - 1 . ( - 4 ) ) k

a xx (b xx c) = |[[i,j,k],[3,1,2],[-,4,-,4,0]]| = 8 i - 8 j - 8 k

a xx (b xx c) = ( 8 , - 8 , - 8 )

Passo 3

Agora vamos fazer a xx b :

a xx b = |[[i,j,k],[3,1,2],[-,1,1,0]]| = |[[1,2],[1,0]]| i - |[[3,2],[-,1,0]]| j + |[[3,1],[-,1,1]]| k

a xx b = |[[i,j,k],[3,1,2],[-,1,1,0]]| = ( 1.0 - 2.1 ) i - (3.0 - 2 . (- 1)) j + (3.1 - 1 . (- 1)) k

a xx b = |[[i,j,k],[3,1,2],[-,1,1,0]]| = - 2 i - 2 j + 4 k

a xx b = ( - 2 , - 2,4 )

Passo 4

E pra fechar, (a xx b) xx c :

(a xx b) xx c = |[[i,j,k],[-,2,-,2,4],[0,0,-,4]]| = |[[-,2,4],[0,-,4]]| i - |[[-,2,4],[0,-,4]]| j + |[[-,2,-,2],[0,0]]| k

(a xx b) xx c = |[[i,j,k],[-,2,-,2,4],[0,0,-,4]]| = (- 2 . (- 4) - 4.0) i - (- 2 . (- 4) - 4.0) j + (- 2.0 - (- 2) . 0) k

(a xx b) xx c = ( 8 , - 8,0 )

Assim, temos que

a xx (b xx c) != (a xx b) xx c

Resposta

Ei, a resposta está no passo a passo :)

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