MO = √(800^2 + (-600)^2) = 1000 N·m
Assuming Fx = 200 N, Fy = 0, Fz = 0 (since the force is acting in the x-direction) MO = √(800^2 + (-600)^2) = 1000 N·m
Calculate the moment of the force about point O. Fy = 0
MO = (3(0) - 4(0))i - (2(0) - 4(200))j + (2(0) - 3(200))k = 0i + 800j - 600k MO = √(800^2 + (-600)^2) = 1000 N·m
rOA = 2i + 3j + 4k
Calculate the position vector from O to A.