Jika vektor a=(-1,2) dan vektor b=(1,3) besar sudut antara vektor a dan vektor b adalah
a.b = (-1 2).(1 3) = -1+6 = 5
|a| = √x² + y² = √(-1)² + 2²
= √1 + 4
= √5
|b| = √x² + y² = √1² + 3²
= √1 + 9
= √10
Cos∅ = a.b / |a|.|b|
= 5 / 50
= 5 / 5√2
= 1 / √2 . √2 / √2
= 1/2 √2
Cos∅ = 1/2 √2
∅ = arc cos (1/2 √2)
= (180⁰ - 45⁰)
= 135⁰
Ket: ∅= teta
/= per (satu per dua)
√= akar
[answer.2.content]
|a| = √x² + y² = √(-1)² + 2²
= √1 + 4
= √5
|b| = √x² + y² = √1² + 3²
= √1 + 9
= √10
Cos∅ = a.b / |a|.|b|
= 5 / 50
= 5 / 5√2
= 1 / √2 . √2 / √2
= 1/2 √2
Cos∅ = 1/2 √2
∅ = arc cos (1/2 √2)
= (180⁰ - 45⁰)
= 135⁰
Ket: ∅= teta
/= per (satu per dua)
√= akar
[answer.2.content]