Question

Даны три точки A (-2.3.1) B(1.-2.3) C(4.-1.0). Найти |2AB+3BC|

Answer 1

AB = (1-(-2), -2-3, 3-1) = (3,-5,2)
BC = (4-1, -1-(-2), 0-3) = (3,1,-3)
2AB+3BC = 2*(3,-5,2) + 3*(3,1,-3) = (6,-10,4) + (9,3,-9) = (15,-7,-5)
|2AB+3AB| = sqrt (225 + 49 + 25) = sqrt (299)