Question

Найдите периметр ΔABC, если точка А(7;1;-5) точка В(4;-3;-4) точка С(1;3;-2;)

Answer 1

А(7;1;-5)  В(4;-3;-4) С(1;3;-2)
$AB^2 = (X_B-X_A)^2+(Y_B-Y_A)^2+(Z_B-Z_A)^2=$
= (4 - 7)² + (-3 - 1)² +(-4 + 5)² = 9 + 16 + 1 = 26
AB = √26

$AC^2 = (X_C-X_A)^2+(Y_C-Y_A)^2+(Z_C-Z_A)^2=$
= (1 - 7)² + (3 - 1)² + (-2 + 5)² = 36 + 4 + 9 = 49 
AC = √49 = 7

$CB^2 = (X_B-X_C)^2+(Y_B-Y_C)^2+(Z_B-Z_C)^2=$
= (4 - 1)² + (-3 - 3)² + (-4 + 2)² = 9 + 36 + 4 = 49
CB = √49 = 7

$P_{ABC}$ = √26 + 7 + 7 = 14 + √26