Jika a=5i+4j+2k,b=4i-5j+3k dan c=2i-j-2k.tentukan nilai a.b dan sudut antara vektor

diket
vektor a = 5i+4j+2k = (5,4,2)
vektor b = 4i-5j+3k = (4,-5,3)
vektor c = 2i-j-2k = (2,-1,-2)
jawab
a.b = a₁.b₁ + a₂.b₂ + a₃.b₃
a.b = 5x4 + 4x-5 + 2x3
a.b = 20 - 20 + 6
a.b = 6

cosθ = a.b/(IaI.IbI)
IaI = √a₁² + a₂² + a₃²
= √25 + 16 + 4
= √45
= 3√5
IbI = √b₁² + b₂² + b₃²
= √16 + 25 + 9
= √50
= 5√2
a.b = 6
cosθ = 6/(3√5x5√2)
cosθ = 6/15√10
cosθ = 0,4√10