Вектора
AB(-1;5;-5)
AC(1;1;-3)
S(ABC) = 1/2 | ABxAC | =
1/2 √( ( (5*(-3) - 1*(-5) )^2 + ((-1)*(-3)-1*(-5))^2 + ((-1)*1 -1*5)^2 ) =
1/2 √( (-15+5)^2 + (3+5)^2 + (-1-5)^2 ) =
1/2 √(100+64+36) = 5√2
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Verified answer
Вектора
AB(-1;5;-5)
AC(1;1;-3)
S(ABC) = 1/2 | ABxAC | =
1/2 √( ( (5*(-3) - 1*(-5) )^2 + ((-1)*(-3)-1*(-5))^2 + ((-1)*1 -1*5)^2 ) =
1/2 √( (-15+5)^2 + (3+5)^2 + (-1-5)^2 ) =
1/2 √(100+64+36) = 5√2