U prostoru su date tačke A(1,1,1)A(1,1,1)A(1,1,1), B(−1,2,2)B(-1,2,2)B(−1,2,2) i C(0,0,3)C(0,0,3)C(0,0,3).
a) Naći ugao ∠BAC\angle BAC∠BAC.
b) Izračunati površinu trougla ABCABCABC.
v) Naći koordinate težišta (presečne tačke težišnih linija) trougla ABCABCABC.
Vektori: AB→=(−2,1,1)\overrightarrow{AB}=(-2,1,1)AB=(−2,1,1), AC→=(−1,−1,2)\overrightarrow{AC}=(-1,-1,2)AC=(−1,−1,2).
a) AB→⋅AC→=2−1+2=3\overrightarrow{AB}\cdot\overrightarrow{AC}=2-1+2=3AB⋅AC=2−1+2=3, ∣AB→∣=∣AC→∣=6|\overrightarrow{AB}|=|\overrightarrow{AC}|=\sqrt{6}∣AB∣=∣AC∣=6. Tada
cos∠BAC=36⋅6=12,∠BAC=60∘.\cos\angle BAC=\frac{3}{\sqrt{6}\cdot\sqrt{6}}=\frac12,\qquad\angle BAC=60^\circ.cos∠BAC=6⋅63=21,∠BAC=60∘.
b) AB→×AC→=(3,3,3)\overrightarrow{AB}\times\overrightarrow{AC}=(3,3,3)AB×AC=(3,3,3), njegova dužina je 27=33\sqrt{27}=3\sqrt{3}27=33. Znači S=12⋅33=332S=\dfrac12\cdot 3\sqrt{3}=\dfrac{3\sqrt{3}}{2}S=21⋅33=233.
v) T=A+B+C3=(1−1+03,1+2+03,1+2+33)=(0,1,2)T=\dfrac{A+B+C}{3}=\left(\dfrac{1-1+0}{3},\dfrac{1+2+0}{3},\dfrac{1+2+3}{3}\right)=(0,1,2)T=3A+B+C=(31−1+0,31+2+0,31+2+3)=(0,1,2).
Izvor: Do indeksa (autorski)