41 Use a definição de produto escalar, a-b=a b cos theta e o fato de que a cdot b=ax bx+ay by+az bz para calcular o ângulo entre os vetores a=3,0 i+3,0 j+3,0 k e b=2,0 i+1,0 j+3,0 k.

Passo 1

O enunciado pede o ângulo theta entre
a=3,0i+3,0j+3,0k e
b=2,0i+1,0j+3,0k
sabendo que
acdotb=|a|,|b|costheta
e
acdotb=axbx+ayby+azbz.

Passo 2

Calculamos o produto escalar
acdotb =3,0cdot2,0 ;+;3,0cdot1,0;+;3,0cdot3,0 =6,0;+;3,0;+;9,0 =18,0

Passo 3

Calculamos as normas
|a| =raiz3,0^2+3,0^2+3,0^2 =raiz27,0 =3raiz3,
|b| =raiz2,0^2+1,0^2+3,0^2 =raiz14,0

Passo 4

Aplicamos a definição de cosseno
costheta =acdotb|a|,|b| =18,03raiz3,raiz14,0 =6raiz42

Passo 5

Finalmente,
theta=arccos!l(tfrac6raiz42r)approx22,3^circ

Resposta

thetaapprox22,3^circ

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