Question

Luas segitiga yang titik-titik sudutnya a(0,1,-2) b(-1,0,2) c(-1,-1,0) adalah..
a.1/2(√9)
b.1/2(√11)
c.1/2(√41)
d.1/2(√137)
e.1/2(√145)

Answer 1

Tentukan panjang masing-masing
(Akar pangkat dua dari jumlah kuadrat komponen antar vektor):
AB = √[(0-(-1))²+(1-0)²+(-2-2)²] = √[1+1+16] = √18 = 3√2
AC = √[(0-(-1))²+(1-(-1))²+(-2-0)²] = √[1+4+4] = 3
BC = √[(-1-(-1))²+(0-(-1))²+(2-0)²] = √[0+1+4] = √5

s = 1/2 (AB+AC+BC)
s = 1/2 (3+3√2+5)

Maka, luasnya:
L
= √[s(s-AB)(s-AC)(s-BC)]
= √[(1/2)⁴ (3+3√2+√5)(3-3√2+√5)(-3+3√2+√5)(3+3√2-√5)]
Jabarkan secara mendalam
= 1/4 √[(9-18)+√5(3+3√2+3-3√2)+5][(-9+18)+√5(3-3√2+3+3√2)-5]
= 1/4 √[-9+6√5+5][9+6√5-5]
= 1/4 √[6√5-4][6√5+4]
= 1/4 √[180 - 16]
= 1/4 √164
= 1/4 x 2√41
= 1/2 √41 [C]