mtk minat 15-05-2020

2.a| = 2, |b| = 3
b.(a + b) = 12
=> b.a + b.b = 12
=> |b| . |a| . cos <(a,b) + |b|^2 = 12
=> 3.2.cos <(a,b) + (3)^2 = 12
=> 6 cos <(a,b) + 9 = 12
=> 6 cos <(a,b) = 3
=> cos <(a,b) = 1/2
=> <(a,b) = 60°
jd sudut antara vektor a dan b = 60°