diketahui titik A(-3,1,2) B(2,0,1) dan C( 1,2,3) . kosinus sudut antara AB dan AC adalah.... tolong dibantu

diketahui:
A (-3, 1, 2)
B (2, 0, 1)
C (1, 2, 3)

ditanya:
cos θ antara AB dan AC = ....?

jawab:
AB = B - A
= (2, 0, 1) - (-3, 1, 2)
= (2-(-3), 0-1, 1-2)
= (5, -1, -1)

AC = C - A
= (1, 2, 3) - (-3, 1, 2)
= (1-(-3), 2-1, 3-2)
= (4, 1, 1)

AB.AC = (5, -1, -1).(4, 1, 1)
= 20+(-1)+(-1)
= 18

|AB| = √5²+(-1)²+(-1)²
= √25+1+1
= √27 = √9×3 = 3√3

|AC| = √4²+1²+1²
= √16+1+1
= √18 = √9×2 = 3√2

|AB|.|AC| = 3√3 × 3√2 = 9√6

cos θ = AB.AC / |AB|.|AC|
cos θ = 18/9√6
= 2/√6

jadi kosinus sudut antara AB dan AC adalah 2√6