Besar sudut antara antar vektor a = -i+2j-3k b=3i+j+2k adalah

Materi : Vektor

Pembahasan :
Cos β = (a . b)/|a| |b|
Cos β = [(-1 2 -3) . (3 1 2)]/[√14 √14]
Cos β = [(-1).3 + 2.1 + (-3).2]/14
Cos β = (-3 + 2 - 6)/14
Cos β = -7/14
Cos β = -1/2
Cos β = Cos 120°
β = 120°

DIRGAHAYU RI KE 72 !!!
Jayalah Indonesiaku !!!

a*b = |a||b| cos a
|a|= √(-1)²+2²+(-3)² = √14
|b| = √9+1+4 = √14

cos a = ab/|a||b|
= (-1)*3 + 2(1) + (-3)*2/√14 *√14
= -3 + 2 - 6 / 14
= -1/2

A = arc cos -1/2
A = 120°