Diketahui vektor a = 4 i + 2 j + 2 k dan vektor b = 3 i + 3 j. Besari sudut antara vektor a dan b adalah

Cos theta = A.B/(|A||B|)
A = 4i + 2j + 2k; |A| = 2*akar(6)
B = 3i + 3j + 0k; |B| = 3
A.B = 12 + 6 = 18
Cos theta = 18/(6*akar(6))
Cos theta = 2/akar(6) = 1/3*akar(6)
Arccos (1/3*akar(6)) = 0.61547970867...