Question

Jika diketahui kordinat titik A (3,2,1) B (4,3,2) dan C (1,2,5) maka luas segitiga ABC sama dengan

Answer 1

Vektor AB = b - a = (4,3,2) - (3,2,1) = (1,1,1)
panjang vektor AB = |AB| = √(1²+1²+1²) = √3

Vektor AC = c - a = (1,2,5) - (3,2,1) = (-2,0,4)
panjang vektor AC = |AC| = √((-2)²+0²+4²) = √20 = 2√5

cos(AB, AC) = cos(α) =  (AB . AC)/(|AB|*|AC|) = (1,1,1) . (-2,0,4)/(√3 * 2√5) = (-2 + 0 + 4)/(2√15) = 1/√15

sin² α = 1 - cos² α = 1 - (1/√15) = 1 -  1/15 = 14/15
sin α = √(14/15)  = √14/√15

dengan demikian,
Luas segi3 ABC = 1/2 |AB| |AC| sin α
= 1/2  * √3 * 2√5 * √14/√15
= √14